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removed curly braces: Simpſon’s Algebra, 8vo. {3d Edit. \\ 1ſt Edit.} # _Lond_. 1767 \\ _ibid_. 1745 removed "/tb" and "tb" in the index

THE TWO BOOKS OF APOLLONIUS PERGÆUS, CONCERNING TANGENCIES, As they have been Reſtored by FRANCISCUSVIET A and MARINUSGHETALDUS. WITH A SUPPLEMENT.
~~
~~

~~By JOHN LAWSON, B. ~~
~~D. ~~
~~Rector of Swanſcombe in Kent.~~
~~
~~

THE SECOND EDITION.
TO WHICH IS NOW ADDED, A SECOND SUPPLEMENT, BEING Monſ. FERMAT’S Treatiſe on Spherical Tangencies.
LONDON: Printed by G. BIGG, Succeſſor to D. LEACH.
And ſold by B. White, in Fleet-Street; L. Davis, in Holborne, J. Nourse, in the Strand; and T. Payne, near the Mews-Gate.
MDCCLXXI.

PREFACE.
~~
~~~~
~~~~
~~

~~OF the twelve Analytical Treatiſes recited by Pappus in ~~
his Preface to the 7th Book of his Mathematical Col-

lections, we have very little of the Originals remaining, viz.

~~only Euclid’s Data, and part of Apollonius’s Conics. ~~
~~The ~~
loſs of the reſt is very much to be lamented by all Lovers of

the Mathematics.

~~“ Valde quidem dolendum eſt quod re-~~
liqui tractatus Veterum Analytici, a Pappo memorati,

aut perierint, aut nondum lucem conſpexerint.

~~Nam ~~
minime dubito quin eorum nonnulli, Arabicè ſaltem

verſi, alicubi terrarum lateant, pulvere magis quam tene-

bris ſuis involuti.

~~” Dr. ~~
Halley’s Preface to his Apol-

LONIUS DE Sectione Rationis et Spatii.~~
~~

Some ingenious men have attempted, from the account of them given by Pappus, to reſtore ſome of theſe loſt Trea-

tiſes.

~~Snellius has endeavoured to give us the Books ~~
De

Sectione Rationis, De Sectione Spatii, and De Sec-

TIONE Determinata. ~~Fermat and Schooten have laboured ~~
in the Treatiſe

De Locis Planis; ~~and Marinus Ghetaldus ~~
in that

De Inclinationibus. ~~But thoſe who have ſuc-~~
ceeded beſt, and done the moſt this way, are two incom-

~~
parable Mathematicians of our own Country, Dr. ~~
~~Halley ~~
and Dr.

~~Simſon, to whom the World is very much obliged ~~
for their Geometrical Labours.

~~The firſt of theſe, from an ~~
Arabic MS in the Bodleian Library, has reſtored the Books

De Sectione Rationis; ~~and from his own Sagacity ſup-~~
plied thoſe

De Sectione Spatii: ~~and the other has with ~~
equal pains and ingenuity completed thoſe

De Loc is

~~As to the Treatiſe ~~
De Tactionibus, which I now give the Engliſh Reader, it has been reſtored by Vieta under the

Title of Apollonius Gallus, and his Deficiencies ſupplied by

Marinus Ghetaldus.

~~I have endeavoured to do Juſtice to my ~~
Authors by all poſſible Care both in the Text and in the

Figures;

~~and have added a few Propoſitions of my own, by ~~
way of Supplement, in which I have propoſed Ghetaldus’s

Problems over again without a Determination, and have

found the

Locus of the center of the circle required, which I have not ſeen done before in any Author.

~~
~~

EXTRACT from PAPPUS’s Preſace to his Seventh Book in Dr. HALLEY’s Tranſlation.
DE TACTIONIBUS II.
~~
~~

~~HIS ordine ſubnexi ſunt libri duo ~~
DE Tactionibus, in quibus plures ineſſe propoſitiones videntur;

~~ſed & ~~
~~ex ~~
his unam etiam faciemus, ad hunc modum ſe habentem.

~~“ E ~~
punctis rectis &

~~circulis, quibuſcunque tribus poſitione ~~
datis, circulum ducere per ſingula data puncta, qui, ſi fieri

poſſit, contingat etiam datas lineas.

~~” Ex hac autem ob mul-~~
titudinem in Hypotheſibus datorum, tam ſimilium quam diſſi-

milium

GENERUM, fiunt neceſſario decem propoſitiones di-verſæ;

~~quia ex tribus diſſimilibus generibus fiunt diverſæ ~~
triades inordinatæ numero decem.

~~Data etenim eſſe poſſunt ~~
vel tria puncta;

~~vel tres rectæ; ~~
~~vel duo puncta & ~~
~~recta; ~~
~~vel ~~
duæ rectæ &

~~punctum; ~~
~~vel duo puncta & ~~
~~circulus; ~~
~~vel duo ~~
circuli &

~~punctum; ~~
~~vel duo circuli & ~~
~~recta; ~~
~~vel punctum, ~~
recta &

~~circulus; ~~
~~vel duæ rectæ & ~~
~~circulus; ~~
~~vel tres circuli. ~~
~~Horum duo quidem prima problemata oſtenduntur in libro ~~
quarto primorum Elementorum.

~~Nam per tria data puncta, ~~
quæ non ſint in linea recta, circulum ducere, idem eſt ac

circa datum triangulum circumſcribere.

~~Problema autem in ~~
tribus datis rectis non parallelis, ſed inter ſe occurrentibus,

idem eſt ac dato triangulo circulum inſcribere.

~~Caſus vero ~~
duarum rectarum parallelarum cum tertiâ occurrente, quaſi

~~
pars eſſet ſecundæ ſubdiviſionis, cæteris permittitur. ~~
~~Deinde ~~
proxima ſex problemata continentur in primo libro.

~~Reliqua ~~
duo, nempe de duabus rectis datis &

~~circulo, & ~~
~~de tribus ~~
datis circulis, ſola habentur in ſecundo libro;

~~ob multas di-~~
verſaſque poſitiones circulorum &

~~rectarum inter ſe, quibus ~~
fit ut etiam plurium determinationum opus ſit.

~~Prædictis his ~~
Tactionibus congener eſt ordo problematum, quæ ab edito-

ribus omiſſa fuerant.

~~Nonnulli autem priori horum librorum ~~
illa prefixerunt:

~~Compendioſus enim & ~~
~~introductorius erat ~~
tractatus ille, &

~~ad plenam de Tactionibus doctrinam abſol-~~
vendam maxime idoneus.

~~Hæc omnia rurſus una propoſitio ~~
complectitur, quæ quidem quoad Hypotheſim magis quam

præcedentia contracta eſt, ſuperaddita autem eſt conditio ad

conſtructionem:

~~eſtque hujuſmodi. ~~
~~“ E punctis, rectis, vel ~~
circulis, datis duobus quibuſcunque, deſcribere circulum

magnitudine datum, qui tranſeat per punctum vel puncta

data, ac, ſi fieri poſſit, contingat etiam lineas datas.

~~” Con-~~
tinet autem hæc propoſitio ſex problemata:

~~ex tribus enim ~~
quibuſcunque diverſis generibus fiunt Duades inordinatæ di-

verſæ numero ſex.

~~Vel enim datis duobus punctis, vel duabus ~~
rectis, vel duobus circulis, vel puncto &

~~rectâ, vel puncto & ~~
circuìo, vel rectâ &

~~circulo, opportet circulum magnitudine ~~
datum deſcribere,

QUI DATA CONTINGAT; ~~hæc autem re-~~
ſolvenda ſunt &

~~componenda ut & ~~
~~determinanda juxta Caſus. ~~
~~Liber primus ~~
Tactionum problemata habet ſeptem; ~~ſe-~~
cundus vero quatuor.

~~Lemmata autem ad utrumque librum ~~
ſunt XXI;

~~Theoremata LX.~~
~~
~~

PROBLEMS CONCERNING TANGENCIES.
PROBLEM I.
~~
~~~~
~~~~
~~

~~THROUGH two given points A and B to deſcribe a circle whoſe radius ~~
ſhall be equal to a given line Z.

~~
~~

Limitation. ~~2 Z muſt not be leſs than the diſtance of the points A and B.~~
~~
~~

Construction. ~~With the centers A and B, and diſtance Z, deſcribe two ~~
arcs cutting or touching one another in the point E, ( which they will neceſſarily

do by the Limitation ) and E will be the center of the circle required.

~~
~~

PROBLEM II.
~~
~~~~
~~~~
~~~~
~~

Having two right lines AB CD given in poſition, it is required to draw 2 circle, whoſe Radius ſhall be equal to the given line Z, which ſhall alſo touch

both the given lines.

~~
~~

Case 1ſt. ~~Suppoſe AB and CD to be parallel.~~
~~
~~

Limitation. ~~2Z muſt be equal to the diſtance of the parallels, and the ~~
conſtruction is evident.

~~
~~

Case 2d. ~~Suppoſe AB and CD to be inclined to each other, let them be ~~
produced till they meet in E, and let the angle BED be biſected by EH, and

through E draw EF perpendicular to ED, and equal to the given line Z;

~~through ~~
F draw FG parallel to EH, meeting ED in G, and through G draw GH paral-

lel to EF.

~~I ſay that the circle deſcribed with H center, and HG radius, ~~
touches the two given lines:

~~it touches CD, becauſe EFGH is a Parallelogram,
~~
~~
and hence the angle FEG is equal to EGH, but FEG is a right one by Con-~~
ſtruction.

~~Let now HI be drawn from H perpendicular to AB: ~~
~~then the two ~~
triangles EHI and EHG having two angles in one HEI and EIH reſpectively

equal to two angles in the other HEG and EGH, and alſo the ſide EH com-

mon, by Euc.

~~I. ~~
~~26. ~~
~~HI will be equal to HG, and therefore the circle will ~~
touch alſo the other line AB:

~~and HG or HI equals the given line Z, becauſe ~~
EF was made equal to Z, and HG and EF are oppoſite ſides of a paral-

lelogram.

~~
~~

PROBLEM III.
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~
~~
~~~~
~~~~
~~~~
~~~~
~~

Having two circles given whoſe centers are A and B, it is required to draw another, whoſe Radius ſhall be equal to a given line Z, which ſhall alſo touch

the two given ones.

~~
~~

This Problem has various Caſes, according to the various poſition of the given circles, and the various manner of deſcribing the circle required:

~~but there ~~
are ſix principal ones, and to the conditions of theſe all the reſt are ſubject.

~~
~~

Case 1ſt. ~~Let the circle to be deſcribed be required to be touched outwardly ~~
by the given circles.

~~
~~

Limitation. ~~Then it is neceſſary that 2Z, or the given Diameter, ſhould ~~
not be leſs than the ſegment of the line joining the centers of the given circles

which is intercepted between their convex circumferences, viz.

~~not leſs than CD ~~
in the Figure belonging to Caſe 1ſt.

~~
~~

Case 2d. ~~Let the circle to be deſcribed be required to be touched inwardly by ~~
the given circles.

~~
~~

Limitation. ~~Then it’s Diameter muſt not be given leſs than the right line, ~~
which drawn through the centers of the given circles, is contained between their

concave circumferences;

~~viz. ~~
~~not leſs than CD.~~
~~
~~

Case 3d. ~~Let the circle to be deſcribed be required to be touched outwardly ~~
by one of the given circles, and inwardly by the other.

~~
~~

Limitation. ~~Then it’s Diameter muſt not be given leſs than the ſegment ~~
of the right line, joining the centers of the given circles, which is intercepted

between the convex circumference of one and the concave circumference of the

other;

~~viz. ~~
~~not leſs than CD.~~
~~
~~

Case 4th. ~~Let one of the given circles include the other, and let it be re-~~
quired that the circle to be deſcribed be touched outwardly by them both.

~~
~~

Limitation. ~~Then it’s Diameter muſt not be given greater than the greater ~~
ſegment of the right line, joining the centers of the given circles, which is in-

tercepted between the concave circumference of one and the convex circumference

of the other;

~~nor leſs than the leſſer ſegment; ~~
~~viz. ~~
~~not greater than CD, nor ~~
leſs than MN.

~~
~~

Case 5th. ~~Let one of the given circles include the other, and let be required ~~
that the circle to be deſcribed be touched outwardly by one of the given circles,

and inwardly by the other.

~~
~~

Limitation. ~~Then it’s Diameter muſt not be given greater than the greater ~~
ſegment of the right line, joining the centers of the given circles, which is in-

tercepted between the two concave circumferences of the ſaid circles, nor leſs

than the leſſer ſegment;

~~viz. ~~
~~not greater than CD, nor leſs than MN.~~
~~
~~

Case 6th. ~~Let the two given circles cut each other, and let it be required ~~
that the circle to be deſcribed, and to be touched by them both, ſhall alſo be

included in each of them.

~~
~~

Limitation. ~~Then it’s Diameter muſt not be given greater than the ſeg-~~
ment of the right line, joining the centers of the given circles, intercepted by

their concave circumferences, which lies in the ſpace common to both the given

circles;

~~viz. ~~
~~not greater than CD.~~
~~
~~

There may be alſo three other Caſes of this Problem, when the given circles cut each other;

~~but becauſe they are ſimilar to the 1ſt, 2d, and 4th Caſes already ~~
propoſed, and ſubject to juſt the ſame Limitations;

~~except that which is ſimilar ~~
to the 1ſt, which is ſubject to no Limitation at all, they are here omitted;

~~as are ~~
likewiſe thoſe Caſes where the given circles touch each other;

~~becauſe they are ~~
the ſame as the preceding, and ſolved in the ſame manner.

~~
~~

~~Join the given centers A and B, and where the Caſe requires, let AB be pro-~~
duced to meet the given circumferences in C and D:

~~and let CI and DH be ~~
taken equal to the given line Z:

~~and let two circles be deſcribed; ~~
~~one with cen-~~
ter A and diſtance AI, and the other with center B and diſtance BH:

~~and theſe ~~
two circles will neceſſarily cut or touch each other by the Limitations given.

~~Let ~~
the point of concourſe be E:

~~from E draw the right line EAF cutting the circle ~~
whoſe center is A in F;

~~as alſo EBG cutting the circle whoſe center is B in G: ~~
~~then with center E and diſtance EF deſcribe a circle FK, this will be the circle ~~
required:

~~becauſe AF and AC are equal as alſo AI and AE; ~~
~~therefore FE ~~
and CI are alſo equal:

~~but CI was made equal to Z, therefore FE is equal to ~~
Z.

~~Again, becauſe BD and BG are equal, as alſo BH and BE, therefore DH ~~
and EG are alſo equal:

~~but DH was made equal to Z, therefore EG is equal to ~~
Z.

~~Hence it appears that the circle FK, paſſing through F will alſo paſs thro’ ~~
G, and likewiſe that it will alſo touch the given circles in F and G, becauſe EAF

and EBG are right lines paſſing through the centers.

~~
~~

PROBLEM IV.
~~
~~
~~
~~~~
~~

Having a given point A, and a given right line BC, it is required to draw a circle, whoſe Radius ſhall be equal to a given line Z, which ſhall paſs through

the given point, and alſo touch the given line.

~~
~~

Limitation. ~~2Z muſt not be given leſs than the perpendicular let fall from ~~
the given point A upon the given line BC.

~~
~~

From the point A let AD be drawn perpendicular to BC, and in this perpen-dicular take DE equal to the given line Z:

~~and through E draw EF parallel to ~~
BC, and from A upon this line EF ſet off AF equal to Z, which may be done,

for by the Limitation Z is not leſs than AE:

~~then with center F and diſtance ~~
FA deſcribe a circle, and I ſay it will touch the line BC:

~~for through F drawing ~~
FG parallel to AD, FGDE will be a Parallelogram, and FG will be equal to

DE, that is to Z, and at right angles to BC.

~~
~~

PROBLEM V.
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~

Having a given point A, and alſo a given circle whoſe center is B, it is re-quired to draw a circle whoſe Radius ſhall be equal to a given line Z, which ſhall

paſs through the given point, and alſo touch the given circle.

~~
~~

This Problem has three Caſes, each of which is ſubject to a Limitation.~~
~~

Case Iſt. ~~Let the circle to be doſcribed be required to be touched outwardly ~~
by the given circle.

~~
~~

Limitation. ~~Then the Diameter muſt not be given leſs than the ſegment ~~
of the right line, joining the given point and the center of the given circle,

which is intercepted between the given point and the convex circumference;

~~viz. ~~
~~not leſs than AC.~~
~~
~~

Case 2d. ~~Let the circle to be deſcribed be required to be touched inwardly ~~
by the given circle.

~~
~~

Limitation. ~~Then the Diameter muſt not be given leſs than the right line ~~
which, drawn from the given point through the center of the given circle, is con-

tained between the given point and the concave circumſerence;

~~viz. ~~
~~not leſs ~~
than AC.

~~
~~

Case 3d. ~~Let the given point lie in the given circle.~~
~~
~~

Limitation. ~~Then a diameter of the given circle being drawn through the ~~
given point, it is divided into two ſegments by the ſaid point, and the Diameter

of the circle required muſt not be given greater than the greater of them, nor

leſs than the leſſer;

~~viz. ~~
~~not greater than AC, nor leſs than AG.~~
~~
~~

Let A and B be joined, and in the line AB take CF equal to Z, and then with center A and diſtance Z, let an arc be drawn, and with center B, and

diſtance BF let another be drawn, which by the Limitations will neceſſarily

either touch or cut the former;

~~let the point of their concourſe be D; ~~
~~then with ~~
D center and DA diſtance let a circle be drawn, and I ſay it will touch the

given circle whoſe center is B:

~~for DB being drawn meeting the circumference ~~
of the circle whoſe center is B in E, BC is equal to BE, and hence CF equals

ED, and they are both equal to the given line Z.

~~
~~

PROBLEM VI.
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~

Having a right line given BC, and alſo a circle whoſe center is A, it is re-quired to draw another circle, whoſe Radius ſhall be equal to a given right line

Z, and which ſhall touch both the given line and alſo the given circle.

~~
~~

This Problem has alſo three caſes, each of which is ſubject to a Limitation.~~
~~

Case Iſt, Let the circle to be deſcribed be required to be touched outwardly by the given circle.

~~
~~

Limitation. ~~Then the Diameter of the circle required muſt not be given ~~
leſs than the ſegment of a line, drawn from the center of the given circle, per-

pendicular to the given line, which is intercepted between the ſaid line and the

convex circumference;

~~viz. ~~
~~not leſs than BD.~~
~~
~~

Case 2d. ~~Let the circle to be deſcribed be required to be touched inwardly by ~~
the given circle.

~~
~~

Limitation. ~~Then the given line muſt not be in the given circle, neither ~~
muſt the Diameter of the circle required be given leſs than that portion of the

perpendicular, drawn from the center of the given circle to the given line, which

is intercepted between the ſaid line and the concave circumference;

~~viz. ~~
~~not leſs ~~
than BD.

~~
~~

Case 3d. ~~Let the circle to be deſcribed be required to be both touched and ~~
included in the given circle.

~~
~~

Limitation. ~~Then the given right line muſt be in the given circle, and ~~
when a Diameter of this given circle is drawn cutting the given line at right an-

gles, the Diameter of the circle required muſt not be given greater than the

greater ſegment ;

~~viz. ~~
~~not greater than BD.~~
~~
~~

From A draw AB perpendicular to BC, cutting the given circumference in D; ~~and in this perpendicular let BG and DF be taken each equal to the given line ~~
Z;

~~and through G draw GE parallel to BC; ~~
~~and with center A and diſtance ~~
AF let an arc be ſtruck, which by the Limitations will neceſſarily either touch

or cut GE;

~~let the point of concourſe be E, let AE be joined, and, if neceſſary, ~~
be produced to meet the given circumference in H;

~~then with E center and ~~
EH diſtance deſcribe a circle, and I ſay it will be the required circle;

~~it is evi-~~
dent it will touch the given circle:

~~and becauſe AD and AH are equal, as alſo ~~
AF and AE, therefore DF (which was made equal to Z) will be equal to HE:

let now EC be drawn perpendicular to BC, then GBCE will be a Parallelogram,

and EC will be equal to GB, which was alſo made equal to Z:

~~hence the ~~
circle will alſo touch the given line BC, becauſe the angle ECB is a right

one.

~~
~~

PROBLEM VII.
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~

Having two points A and B given in poſition, and likewiſe a right line EF given in poſition, it is required to find the center of a circle, which ſhall paſs

through the given points and touch the given line.

~~
~~

Case Iſt. ~~When the points A and B are joined, ſuppoſe AB to be parallel to ~~
EF:

~~then biſecting AB in D, and through D drawing DC perpendicular to it, ~~
DC will alſo be perpendicular to EF:

~~draw a circle therefore which will paſs ~~
through the three points A, B, andC, (by Euc.

~~IV. ~~
~~5.) ~~
~~and it will be the circle ~~
required:

~~(by a Corollary from Euc. ~~
~~III. ~~
~~1. ~~
~~and another from Euc. ~~
~~III. ~~
~~16.)~~
~~
~~

Case 2d. ~~Suppoſe AB not parallel to EF, but being produced meets it in ~~
E:

~~then from EF take the line EC ſuch, that its ſquare may be equal to the ~~
rectangle BEA, and through the points A, B, C, deſcribe a circle, and it will be

the circle required by Euc.

~~III. ~~
~~37.~~
~~
~~

This is Vieta’s Solution. ~~But Mr. ~~
~~Thomas Simpſon having conſtructed this, ~~
and ſome of the following, both in the Collection of Problems at the end of his

Algebra, and alſo among thoſe at the end of his Elements of Geometry, I ſhall

add one of his Conſtructions.

~~
~~

Let A and B be the points given, and CD the given line: ~~drawing AB and ~~
biſecting it in F, through E let EF be drawn perpendicular to AB and meeting

CD in F:

~~and from any point H in EF draw HG perpendicular to CD, and ~~
having drawn BF, to the ſame apply HI = HG, and parallel thereto draw BK

meeting EF in K:

~~then with center K and radius BK let a circle be deſcribed, ~~
and the thing is done:

~~join KA, and draw KL perpendicular to CD, then be-~~
cauſe of the parallel lines, HG:

~~HI:~~
~~: KL: ~~
~~KB; ~~
~~whence as HG and HI are ~~
equal, KL and KB are likewiſe equal.

~~But it is evident from the Conſtruction ~~
that KA = KB, therefore KB = KL = KA.

~~
~~

Because two equal lines HI and Hi may be applied from H to BF each equal to HG, the Problem will therefore admit of two Solutions, as the Figure

ſhews:

~~except in the caſe when one of the given points, A for inſtance, is given ~~
in the line CD, for then the Problem becomes more ſimple, and admits but of

one conſtruction, as the center of the circle required muſt be in the line EF con-

tinued, as alſo in the perpendicular raiſed from A to CD, and therefore in their

common interſection:

~~and this is the limit of poſſibility; ~~
~~for ſhould the line CD ~~
paſs between the given points, the Problem is impoſſible.

~~
~~

~~N. ~~
~~B. ~~
Tho’ Vieta does not take notice that this Problem is capable of two anſwers, yet this is as evident from his conſtruction, as from Mr.

~~Simpſon’s, for ~~
EC (the mean proportional between E B and EA) may be ſet off upon the given

line EF either way from the given point E.

~~
~~

LEMMA I.
~~
~~~~
~~

Apoint A being given between the two right lines BC and DE, it is requir-ed through the point A to draw a line cutting the two given ones at equal

angles.

~~
~~

If the given lines be parallel, then a perpendicular to them through the point A is the line required.

~~But if not, then let them be produced to meet in the ~~
point F:

~~and let FG be drawn biſecting the Angle BFD, and through A draw ~~
a perpendicular to FG, and it will be the line required by Euc.

~~I. ~~
~~26.~~
~~
~~

PROBLEM VIII.
~~
~~~~
~~

Having a point A given, and alſo two right lines BC and DE, to draw a circle which ſhall paſs through the given point, and touch both the given right lines.

~~
~~

By the preceding Lemma draw a line IAH’ through the point A, which ſhall make equal angles with the two given lines BC and DE:

~~biſect IH in K; ~~
~~and ~~
taking KL = KA, by means of the preceeding Problem draw a circle which ſhall

paſs through the points A and L, and likewiſe touch one of the given lines, BC

for inſtance, in the point M.

~~I ſay this circle will alſo touch the other given ~~
line DE:

~~for from the center N letting fall the perpendicular NO, and joining ~~
NI, NH, NM;

~~in the triangles NKH, NKI, NK being common, and HK = ~~
KI, and the Angles at K right ones, by Euc.

~~I. ~~
~~4. ~~
~~NH = NI likewiſe the ~~
angle NHK = angle NIK, from hence it follows:

~~that the angle NHM = angle ~~
NIO;

~~and the angles at M and O, being both right, and NH being proved ~~
equal to NI, NM will be equal to NO by Euc.

~~I. ~~
~~26.~~
~~
~~

Mr. Simpſon conſtructs the Problem thus.
~~
~~

Let BD and BC be the given lines meeting in B, and A the given point, join AB, and draw BN biſecting the given angle DBC:

~~and from any point E ~~
in BN upon BC let fall the perpendicular EF, and to BA apply EG = EF, pa-

rallel to which draw AH meeting BN in H:

~~then from center H with Interval ~~
AH let a circle be deſcribed, and the thing is done.

~~Upon BC and BD let fall ~~
the perpendiculars HI, HK, which are manifeſtly equal, becauſe by Conſtruction

the angle HBI = HBK;

~~moreover as EF: ~~
~~EG:~~
~~: HI: ~~
~~HA: ~~
~~but EF and EG ~~
are equal, therefore alſo HI and HA.

~~
~~

PROBLEM IX.
~~
~~
~~
~~

Having a circle whoſe center is A given in magnitude and poſition, and alſo two right lines BD and ZC given in poſition, to draw a circle which ſhall touch

all three.

~~
~~

From A draw two perpendiculars to the right lines DB, ZC; ~~viz. ~~
~~ADF and ~~
AZX;

~~and in theſe perpendiculars take DF, ZX, on “either ſide of D and Z, ~~
equal to the Radius of the given circle:

~~and through F~~
rallel to DB, ZC;

~~viz. ~~
~~FG, XH; ~~
~~and then by the preceding Problem draw ~~
a circle which ſhall paſs through the given center A and touch the two lines

FG, XH;

~~and E the center of this circle will alſo be the center of the circle ~~
required, as appears by ſubtracting equals from equals in Figure 1:

~~and by adding ~~
equals to equals in Figure 2.

~~
~~

LEMMA II.
~~
~~~~
~~~~
~~

If the two circles CEB and CED cut one another C, then I ſay a line drawn from the point of ſection CBD cutting both circles, will cut off diſſimilar ſeg-

ments from thoſe circles.

~~
~~

~~1ſt. ~~
Suppose CB to be the Diameter of one of them: ~~then draw to the ~~
other point of ſection E the line CE, and joining EB, ED, the angle CEB will

be a right one, and the angle CED either greater or leſs than a right one, and

conſequently CD cannot be a Diameter of the other.

~~
~~

~~2dly. ~~
Suppose CBD not to paſs through the center of either: ~~then through ~~
C draw a Diameter CAG, and continue it to meet the other circle in F, and

join BG, DF:

~~then the angle CBG is a right one, and the angle CDF is either ~~
greater or leſs than a right one:

~~and therefore the lines BG and DF are not ~~
parallel:

~~let H be the center of the other circle, and let a Diameter CHI be ~~
drawn:

~~draw DI and continue it meet to meet CG in K: ~~
~~then DIK will be pa-~~
rallel to BG:

~~hence CB: ~~
~~CD:~~
~~: CG: ~~
~~CK. ~~
~~But CI and CK are unequal, ~~
(being both applied from the ſame point in a right angle) and therefore it cannot

be 2s CB:

~~CD:~~
~~: CG: ~~
~~CI: ~~
~~and hence it appears that the Segments CB and ~~
CD are diſſimilar.

~~
~~

LEMMA III.
~~
~~~~
~~

If through the legs of any triangle EDF (ſee Figure to Problem 10.) ~~a line ~~
BI be

triangles about the ſame vertex;

~~and a circle be circumſcribed about each of ~~
theſe triangles;

~~theſe circles will touch one another in the common vertex E.~~
~~
~~

It is plain that they will either touch or cut each other in the point E: ~~if ~~
they cut each other, then by the preceding Lemma the Segments BE and DE

would be diſſimilar;

~~but they are ſimilar, and they muſt therefore touch each ~~
other.

~~
~~

PROBLEM X.
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~

Having a point A, and alſo a right line BC, given in poſition; ~~together ~~
with a circle whoſe center is G given both in m

~~to deſcribe
~~
~~
a circle which ſhall paſs through the given point, and likewiſe touch both the ~~
given line and the given circle.

~~
~~

The given right line may either 1ſt cut, touch, or be entirely without the given circle, and the given point be without the ſame or in the circum-

ference;

~~or 2dly, it may cut the given circle, and the given point be within ~~
the ſame, or in the circumference.

~~
~~

Case Iſt. ~~Suppoſe the given line BC to cut, touch, or fall entirely with-~~
out the given circle;

~~and the given point A to be without the ſame, or in the ~~
circumference:

~~through G the center of the given circle draw DGFC per-~~
pendicular to BC, and joining DA, take DH a 4th proportional to DA, DC,

DF, ſo that DA X DH = DC X DF:

~~then through the points A and H draw ~~
a circle touching the line CB by Problem VII, and I ſay it will alſo touch the

given circle.

~~
~~

~~Draw DB cutting the given circle in E, and join FE. ~~
~~Now becauſe the ~~
triangles DEF, DCB are ſimilar, DF:

~~DE:~~
~~: DB: ~~
~~DC, and therefore DC ~~
X DF = DB X DE.

~~But DC X DF = DA X DH by Conſtruction. ~~
~~Hence ~~
DB X DE = DA X DH, and therefore the points B, E, H, A, will be alſo in

a circle:

~~but the point E is alſo in the given circle; ~~
~~therefore theſe circles either ~~
touch or cut one another in that point.

~~Let now BI be drawn from the point ~~
of contact B perpendicular to the touching line BC to meet the circumference

again in I, and it will be a Diameter:

~~and let EI be joined: ~~
~~then becauſe the ~~
angles FED and BEI are vertical and each of them right ones, FEI will be a

continued ſtraight line:

~~and it appears that the two circles will touch each ~~
other by the preceding Lemmas.

~~
~~

Case 2d. ~~Suppoſe the given line BC to cut the given circle, and the given ~~
point to be within the ſame, or in the circumference;

~~the Conſtruction and De-~~
monſtration are exactly the ſame as before, except that the angles FED and

BEI are not vertical but coincident, and ſo EI is coincident with EF.

~~
~~

~~N.~~
~~B. ~~
~~In either of theſe caſes if the point A coincide with E or be given in ~~
the circumference, draw DEB, and erect BI perpendicular to CB to meet FE in

I, then upon BI as diameter deſcribe a circle, and the thing will manifeſtly be done.

~~
~~

PROBLEM XI.
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~

Having two circles given in magnitude and poſition, whoſe centers are A and B, as likewiſe a right line CZ;

~~to draw a circle which ſhall touch all three.~~
~~
~~

From the center of the leſſer circle B let BZ be drawn perpendicular to CZ, and in BZ (or in BZ continued as the caſe requires) let be taken ZX = AL the

Radius of the other circle;

~~and through X let XH be drawn parallel to CZ, ~~
and with center B and Radius BG, equal to the difference ( or ſum as the Caſe

requires) of the Radii of the two given circles, let a circle be deſcribed;

~~and laſtly
~~
~~
let another circle be deſcribed, touching this laſt, and alſo the line XH and paſſ-~~
ing through the point A by Problem X, and I ſay that E the center of this circle

will alſo be the center of the circle required;

~~as will appear by taking equals from ~~
equals, or adding equals to equals, as the aſſigned Caſe and Data ſhall require.

~~
~~

The Caſes are four, though Vieta makes but three.~~
~~

Case Iſt. ~~If it be required that the circle ſhould touch both the others ex-~~
ternally then BG muſt be taken equal to the difference of the Semidiameters of

the two given circles, and ZX muſt be taken in BZ produced.

~~
~~

Case 2d. ~~If it be required that the circle ſhall touch and include both the ~~
given ones;

~~then BG muſt be taken equal to the difference, as in Caſe 1ſt, but ~~
ZX muſt be taken in BZ itſelf.

~~
~~

Case 3d. ~~If it be required that the circle ſhould touch and include the greater ~~
of the given circles, and touch externally the other whoſe center is B;

~~then BG ~~
muſt be taken equal to the ſum of the Radii of the given circles, and ZX muſt

be taken in BZ itſelf.

~~
~~

Case 4th. ~~If it be required that the circle ſhould touch the greater of the ~~
given circles externally, and touch and include the leſſer;

~~then BG muſt be taken ~~
equal to the ſum of the Radii, and ZX muſt be taken in BZ produced.

~~
~~

PROBLEM XII .
~~
~~~~
~~~~
~~There are other Conſtructions of this Problem in Hugo de Omerique, Simpſon, and the Mathematician. See alſo Monthly Review for Oct. 1764, where the Reviewer is pleaſed to ſpeak favourably of the 1ſt Edi-tion of this work, but wiſhes that ſome modern ſolutions of theſe Problems had been inſerted, which, he ſays, are more conciſe and elegant than any which are to be met with in the works of the Antients. The Editor acknowledges that the conſtruction there given is more ſimple than Vieta’s; but Vieta is not an Antient, and he knows of no others that exceed his.

Having two points given B and D, and like wiſe a circle whoſe center is A; ~~to de-~~
ſcribe another circle which ſhall paſs through the given points, and touch the

given circle.

~~
~~

Let DB be joined, as alſo AB, and let AB be produced to cut the given circle in the points I and K, then let BH be taken a 4th proportional to DB,

BK, BI;

~~ſo that BD X BH = BI X BK: ~~
~~from H let a Tangent HF be drawn ~~
to the given circle;

~~and BF be joined and cut the circle again in G: ~~
~~and let DG ~~
be drawn cutting the given circle again in E, and laſtly through the points D,

B, G, let a circle be drawn, I ſay it will touch the given circle in G.

~~
~~

For joining EF; ~~becauſe the rectangle DBH = the Rectangle KBI, i.~~
~~e. ~~
~~the ~~
Rectangle GBF, thereforethefourpoints D, H, F, G, are in a circle;

~~and hence the ~~
angle HFB = the angle GDB:

~~(for in the two firſt ſigures one is theexternal angle ~~
of a quadrilateral figure, and the other is the internal and oppoſite, and in the two

laſt figures theſe angles are in the ſame ſegment.)

~~But the angle HFB = the angle ~~
GEF by Eu.

~~III. ~~
~~32. ~~
~~hence GEF = GDB: ~~
~~therefore the triangles GEF and ~~
GDB are ſimilar and under the ſame vertex, and therefore by Lemma 3.

~~the cir-~~
cles deſcribed about them will touch each other in the common vertex G.

~~
~~

LEMMA IV.
~~
~~~~
~~

Having two circles ABCI and EFGH given, it is required to find a point M, in the line joining their centers, or in that line continued, ſuch, that any

line drawn through the ſaid point M, cutting both the circles, ſhall always cut

off ſimilar ſegments.

~~
~~

Let the line KL joining the centers be ſo cut or produced, that KM may be to LM in the given ratio of the Radii AK to EL [in the caſe of producing

KL we muſt make it as AK-EL:

~~EL:~~
~~: KL: ~~
~~LM, for then by compoſition it ~~
will be AK:

~~EL:~~
~~: KM: ~~
~~LM] and then I ſay that the point M will be the ~~
point required.

~~For from it drawing any line MGFCB cutting the circle ~~
ABCI in B and C, and the circle EFGH in F and G;

~~and let B and F be the ~~
correſpondent points moſt diſtant from M, and C and G the correſpondent

points that are nearer;

~~and let be joined BK, CK, FL, GL, and thereby two ~~
triangles will be formed BKC, FLG.

~~Now becauſe by Conſtri~~
~~LM ~~
:

~~: KB: ~~
~~LF, KB and LF will be parallel by Euc. ~~
~~VI. ~~
~~7. ~~
~~and for the ſame rea-~~
ſon KC and LG will be parallel;

~~and therefore the triangles BKC and FLG ~~
will be equiangled, and hence the ſegment BC will be ſimilar to the ſegment FG.

~~
~~

LEMMA V.
~~
~~~~
~~

The point M being found as in the preceding Lemma, I ſay that it is a pro-perty of the ſaid point, that MG x MB = MH x MA:

~~as alſo that MF x ~~
MC = ME x MI.

~~
~~

For joining CI and GH, it is evident that theſe lines will alſo be parallel. ~~Hence MI: ~~
~~MC:~~
~~: MH: ~~
~~MG, but MI: ~~
~~MC:~~
~~: MB: ~~
~~MA, therefore ~~
MH:

~~MG:~~
~~: MB: ~~
~~MA, and MG x MB = MH x MA. ~~
~~Again MH: ~~
~~MG ~~
:

~~: MF: ~~
~~ME, but MH: ~~
~~MG:~~
~~: MI: ~~
~~MC, therefore MF: ~~
~~ME:~~
~~: MI: ~~
~~MC ~~
and MF x MC = ME x MI.

~~
~~

PROBLEM XIII.
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~

Having two circles given in magnitude and poſition, whoſe centers are K and L, and alſo a point D;

~~to draw a circle which ſhall touch the two given ~~
ones, and alſo paſs through the point D.

~~
~~

Join the given centers by drawing KL, and in KL or KL produced ſind the point M (by Lemma 4.)

~~ſuch, that all the lines drawn from it cutting the given ~~
circles ſhall cut off ſimilar ſegments;

~~and let KL cut the circumferences, one ~~
of them in the points A and I, and the other in the points E and H;

~~and join-~~
ing MD, make it as MD:

~~MA:~~
~~: MH: ~~
~~MN. ~~
~~Then through the points D and ~~
N draw a circle which ſhall alſo touch the given circle whoſe center is K, by

Prob.

~~XII. ~~
~~and I ſay that this circle will alſo touch the other given circle whoſe ~~
center is L.

~~For let B be the point of contact and BM be drawn cutting the ~~
circle K in C, and the circle L in F and G;

~~then by Lemma 5. ~~
~~MB x MG =
~~
~~
MA x MH: ~~
~~but MA x MH = MD x MN by Conſtruction; ~~
~~therefore MB x ~~
MG = MD x MN, and the points B, G, N, D, are in a circle.

~~But the point ~~
G is alſo in the circle L;

~~therefore theſe circles either touch or cut each other in ~~
the point G.

~~Now the circles BND and BCI touch one another in B by con-~~
ſtruction;

~~therefore the ſegment BC is ſimilar to the ſegment BG; ~~
~~and alſo by ~~
conſtruction the ſegment BC is ſimilar to the ſegment FG;

~~and therefore the ~~
ſegment FG is ſimilar to the ſegment BG;

~~and hence the circles FGE and BGD ~~
touch one another in the point G.

~~
~~

The Caſes are three.~~
~~

Case Iſt. ~~Iſ the circle be required to touch and include both the given ones; ~~
~~then M muſt be taken in KL produced; ~~
~~and MN muſt be taken a fourth pro-~~
portional to MD, MA, MH, A being the moſt diſtant point of interſection in

the circle K, and H the neareſt point of interſection in the circle L.

~~
~~

Case 2d. ~~If the circle be required to touch both the given ones externally; ~~
~~then alſo M muſt be taken in KL produced; ~~
~~and MN taken a fourth proportional ~~
to MD, MA, MH, A being the neareſt point of interſection in the circle K,

and H the moſt diſtant in the circle L.

~~
~~

Case 3d. ~~If the circle be required to touch and include the circle K, and to ~~
touch L externally;

~~then M muſt be taken in KL itſelf; ~~
~~and MN a fourth pro-~~
portional to MD, MA, MH, A being the moſt diſtant point in K, and H the

neareſt in L.

~~
~~

PROBLEM XIV.
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~

Having three circles given whoſe centers are A, B, and D; ~~to draw a fourth ~~
which ſhall touch all three.

~~
~~

Let that whoſe center is A be called the 1ſt, that whoſe center is B the 2d, and that whoſe center is D the 3d.

~~Then with center B, and radius equal to ~~
the differcnce, or ſum, as the caſe requires, of the ſemidiameters of the 1ſt and

2d circles, let an auxiliary circle be deſcribed;

~~and likewiſe with D center, and ~~
radius equal to the difference, or ſum, as the caſe requires, of the ſemidiameters

of the 1ſt and 3d circles, let another auxiliary circle be deſcribed;

~~and laſtly by ~~
the preceding Problem draw a circle which ſhall touch the two auxiliary ones,

and likewiſe paſs through the point A which is the center of the firſt circle.

~~Let the center of this laſt deſcribed circle be E, and the ſame point E will like-~~
wiſe be the center of the circle required;

~~as will appear by adding equals to ~~
equals, or taking equals from equals, as the caſe requires.

~~
~~

The Caſes are theſe.~~
~~

Case 1ſt. ~~Iſ it be required that the circle ſhould touch and include all the other ~~
three;

~~then let A be the center of the greateſt given circle, B of the next, and D ~~
of the leaſt:

~~and let BG = the difference of the ſemidiameters of the 1ſt and 2d,
~~
~~
and DF = the Difference of the ſemidiameters of the 1ſt and 3d, and through A ~~
deſcribe a circle which ſhall touch the two auxiliary ones externally.

~~
~~

Case 2d. ~~If it be required that the circle ſhould touch all the other three ~~
externally;

~~then the circles being the ſame as before in reſpect to their magnitude, ~~
let BG and DF be alſo the ſame as in the 1ſt Caſe, but through A deſcribe a circle

which ſhall touch and include the two auxiliary ones.

~~
~~

Case 3d. ~~If it be required that the circle ſhould touch and include one of the ~~
given circles A, and touch the other two externally;

~~then let BG and DF = the ~~
the ſum of the ſemidiameters of the 1ſt and 2d, and ſum of the ſemidiameters of

of the 1ſt and 3d reſpectively;

~~and through A deſcribe a circle which ſhall touch ~~
the two auxiliary ones externally.

~~
~~

Case 4th. ~~If it be required that the circle ſhould touch externally one of the ~~
given circles A, and ſhould touch and include the other two;

~~then let BG and ~~
DF = the ſums as in Caſe 3d, but through A deſcribe a circle which ſhall touch

and include the two auxiliary ones.

~~
~~

SUPPLEMENT.
PROBLEM I.
~~
~~~~
~~

~~HAVING two points given A and B, to determine the Locus of the cen-~~
ters of all ſuch circles as may be drawn through A and B.

~~
~~

Join AB and biſect it in the point C, and through C, draw a perpendicular to it CE, and continue it both ways in infinitum, and it is evident that this line

CE will be the Locus required.

~~
~~

PROBLEM II.
~~
~~~~
~~~~
~~

Having two right lines given AB and CD, to determine the Locus of the centers of all ſuch circles as may be drawn touching both the ſaid lines.

~~
~~

Case 1ft. ~~Suppoſe AB and CD to be parallel; ~~
~~then drawing GI perpendicu-~~
lar to them both;

~~biſect it in H, and through H draw EHF parallel to the two ~~
given lines, and it will be the Locus required.

~~
~~

Case 2d. ~~Suppoſe the given lines being produced meet each other in E, ~~
then biſecting the angle BED by the line EHF, this line EHF will be the Locus

required.

~~
~~

PROBLEM III.
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~

Having two circles given whoſe centers are A and B; ~~to determine the Locus ~~
of the centers of all ſuch circles as ſhall touch the two given ones.

~~
~~

Cases 1ſt and 2d. ~~Suppoſe it be required that the circles be touched outwardly ~~
by both the given ones;

~~or that they be touched inwardly by both the given ~~
ones.

~~
~~

Then joining the centers A and B, let AB cut the circumferences in C and D and produced in P and O:

~~let CD which is intercepted between the convex ~~
circumſerences be biſected in E:

~~ſet off from E towards B the center of the
~~
~~
greater of the given circles the line EH = the difference of the Radii of the given ~~
circles, and with A and B as Foci and EH Tranverſe Axis, let two oppoſite

Hyperbolas be deſcribed KEK and LHL:

~~then I ſay that KEK will be the ~~
locus of the centers of all the circles which can be drawn ſo as to be touched

outwardly by both the given circles;

~~and LHL will be the locus of the centers ~~
of all the circles which can be drawn ſo as to be touched inwardly by both the

given circles.

~~For taking any point K in the Hyperbola KEK, and drawing KA ~~
and KB, let theſe lines cut the given circumferences in F and G reſpectively:

~~and make KQ = KA: ~~
~~then from the nature of the curve QB = EH, but by con-~~
ſtruction EH = BG - AF, therefore QB = BG - AF and hence QG = AF, and

then taking equals from equals KG = KF, which is a demonſtration of the 1ſt

Caſe.

~~
~~

Then with regard to the 2d Caſe, taking any point L in the Hyperbola LHL, and drawing LB and LA and producing them to meet the concave cir-

cumferences in M and N, let alſo LR be taken equal to LB;

~~then from the pro-~~
perty of the curve AR = EH, but EH (by conſtruction) = BM - NA;

~~there-~~
fore AR = BM - NA, and NR = BM, and then adding equals to equals LN =

LM, which is a demonſtration of the 2d Caſe.

~~
~~

Case 3d. ~~Suppoſe it be required that the circles to be deſcribed be touched ~~
outwardly by one of the given circles, and inwardly by the other.

~~
~~

~~Then drawing AB, let it cut the convex circumferences in C and D, and the ~~
concave ones in P and O, biſect PD in E, and from E towards B ſet off EH =

the ſum of the given radii.

~~Then with A and B foci and EH tranſverſe axis, ~~
let two oppoſite hyperbolas be deſcribed KEK and LHL:

~~and KEK will be the ~~
locus of the centers of the circles which are touched inwardly by the circle A

and outwardly by the circle B;

~~and LHL will be the locus of the centers of ~~
thoſe circles which are touched inwardly by B and outwardly by A.

~~The demon-~~
ſtration mutatis mutandis is the ſame as before.

~~
~~

Case 4th. ~~Suppoſe the given circle A to include B, and it be required that ~~
the circles to be deſcribed be touched outwardly by them both.

~~
~~

~~Let AB cut the circumferences in C and D, P and O: ~~
~~and biſecting CD in ~~
I, and ſetting off from I towards P, IL = the ſum of the ſemidiameters of the

given circles, and with A and B foci, and IL tranſverſe axis, deſcribing an

ellipſe LKI, it will be the locus of the centers of the circles required.

~~For ~~
taking any point K in the ellipſe, and drawing AK and BK, let AK be con-

tinued to meet one of the given circumferences in G, and let BK meet the other

F.

~~Then from the property of the curve AK + BK = IL = AG + BF (by con-
~~
~~
ſtruction:) ~~
~~hence by ſubſtraction BK = KG + BF, and by ſubtraction again ~~
FK = KG.

~~
~~

Case 5th. ~~Suppoſe the given circle A to include B, and it be required that ~~
the circles to be deſcribed be touched outwardly by A and inwardly by B.

~~
~~

Then let AB cut the circumferences in C and D, P and O: ~~and biſecting ~~
CO in I, and ſetting off from I towards P, IL = the difference of the ſemidia-

meters of the given circles, and with A and B foci and IL tranſverſe axis de-

ſcribing an ellipſe LKI, it will be the locus of the centers of the circles deſcribed,

and the demonſtration, mutatis mutandis, is the ſame as in the laſt caſe.

~~
~~

Cases 6th and 7th. ~~Suppoſe the two given circles cut each other, and it be ~~
required that the circles to be deſcribed either be touched and included in them

both, or be touched by them both and at the ſame time include them both.

~~
~~

These two caſes are ſimilar to caſes 1ſt and 2d, and as there, ſo alſo here, the tranſverſe axis of the two oppoſite hyperbolas, which are the loci required,

muſt be taken = the difference of the ſemidiameters of the given circles.

~~The ~~
demonſtration is ſo alike, it need not be repeated.

~~
~~

PROBLEM IV.
~~
~~~~
~~

Having a given point A, and a given right line BC, to determine the locus of the centers of thoſe circles which ſhall paſs through A and touch BC.

~~
~~

From A draw AG perpendicular to BC, then with focus A and directrix BC let a parabola be deſcribed, and it will be the locus required;

~~for by the propert ~~
of the curve FA always equals FG drawn perpendicular to the directrix.

~~
~~

PROBLEM V.
~~
~~~~
~~~~
~~
~~
~~~~
~~

Having a given point A, and a given circle whoſe center is B, to determine the locus of the centers of all thoſe circles, which paſs through A, and at the

ſame time are touched by the given circle.

~~
~~

Cases 1ſt and 2d. ~~Suppoſe the point A to lie out of the given circle, and ~~
it be required that the circles to be deſcribed be either touched outwardly by the

given circle, or inwardly by it.

~~
~~

Let AB be drawn, and let it cut the given circumference where it is convex towards A in the point C, and where it is concave in the point O:

~~then biſecting ~~
AC in E, and ſetting off from E towards B, EH = BC the given radius, and

with A and B foci and EH tranſverſe Axis deſcribing two oppoſite Hyperbolas

KEK and LHL, it is evident that KEK will be the locus of the centers of thoſe

circles which paſs thrugh A and are touched outwardly by the given circle, and

LHL will be the locus of the centers of thoſe circles which paſs through A and

are touched inwardly by the given circle.

~~
~~

Case 3d. ~~Suppoſe the given point A to lie in the given circle, whoſe center ~~
is B.

~~
~~

Then joining AB and continuing it to meet the given circumference in C and O, biſect AC in E, and from E towards O ſetting off EH = BC the given

Radius, and with A and B Foci and EH tranſverſe Axis deſcribing an Ellipſe

EKH, it will evidently be the Locus required.

~~
~~

PROBLEM VI.
~~
~~~~
~~~~
~~~~
~~~~
~~

Having a right line BC given, and alſo a circle whoſe center is A, to deter-mine the Locus of the centers of the circles which ſhall be touched both by the

given right line and alſo by the given circle.

~~
~~

There are three Caſes, but they are all comprchended under one general ſolution.

~~
~~

Case 1ſt. ~~Let the given right line be without the given circle, and let it be ~~
required that the circles to be deſcribed be touched outwardly by the given circle.

~~
~~

Case 2d. ~~Let the given right line be without the given circle, and let it be ~~
required that the circles to be deſcribed, be touched inwardly by the given

circle.

~~
~~

Case 3d. ~~Let the given right line be within the given circle, and then the ~~
circles to be deſcribed muſt be touched outwardly by the given circle.

~~
~~

From the given center A let fall a perpendicular AG to the given line BC, which meets the given circumference in D [or in Caſes 2d and 3d is produced

to meet it in D] and biſecting DG in F, and ſetting off FM = FA (which is the

ſame thing as making GM = AD the given Radius) and through M drawing

MLK parallel to the given line BC, with A Focus and LK Directrix deſcribe a

Parabola, and it will be the Locus of the centers of the circles required;

~~for ~~
from the property of the Curve FA = FM, and adding equals to equals, or

ſubtracting equals from equals, as the Caſe requires, FD = FG.

~~
~~

A SECOND SUPPLEMENT, BEING Monſ. DE FERMAT’S Treatiſe on Spherical Tangencies.
PROBLEM I.
~~
~~~~
~~~~
~~

~~HAVING four points N, O, M, F, given, to deſcribe a ſphere which ~~
ſhall paſs through them all.

~~
~~

Taking any three of them N, O, M, ad libitum, they will form a triangle, about which a circle ANOM may be circumſcribed, which will be given in

magnitude and poſition.

~~That this circle is in the ſurface of the ſphere ~~
ſought appears from hence;

~~becauſe if a ſphere be cut by any plane, the ~~
ſection will be a circle;

~~but only one circle can be drawn to paſs through the ~~
three given points N, O, M;

~~therefore this circle muſt be in the ſurface of ~~
the ſphere.

~~Let the center of this circle be C, from whence let CB be ~~
erected perpendicular to it’s plane;

~~it is evident that the center of the ſphere ~~
ſought will be in this line CB.

~~From the fourth given point F let FB be ~~
drawn perpendicular to CB, which FB will be alſo given in magnitude and

poſition.

~~Through C draw ACD parallel to FB, and this line will be a
~~
~~
diameter of the circle given in poſition, and therefore the points A and D ~~
will alſo be given.

~~
~~

Letus now ſuppoſe the thing done, and that the center of the ſphere ſought is E, which, as obſerved before, muſt be in the line CB.

~~Drawing ~~
EF, EA, ED, theſe lines muſt be equal, ſince the points F, A, D, have

been ſhewn to be in the ſurface of the ſphere.

~~But theſe lines EF, EA, ED, ~~
are in the ſame plane, ſince FB and AD are parallel, and BC perpendicular

to each of them.

~~If therefore a circle be deſcribed to paſs through the three ~~
points F, A, D, whoſe center is E, it will be in the line CB, and will be

the center of the ſphere required.

~~
~~

PROBLEM II.
~~
~~~~
~~~~
~~

Having three points N, O, M, given, and a plane AD, to deſcribe a-ſphere which ſhall paſs through the three given points;

~~and alſo touch the ~~
given plane.

~~
~~

Let a circle ENOM be deſcribed to paſs through the three given points, it will be in the ſurface of the ſphere ſought, from what was ſaid under the

former Problem.

~~From it’s center I let a perpendicular to it’s plane IA be ~~
erected;

~~the center of the ſphere ſought will be in this line IA; ~~
~~let the line ~~
IA meet the given plane in the point A, which point will be therefore given.

~~From the center of the given circle ENOM, let a perpendicular to the given ~~
plane ID be drawn, the point D will then be given, and therefore the line

AD both in poſition and magnitude, as likewiſe the lines ID, IA, and the

plane of the triangle ADI.

~~But the plane of the circle NOM is alſo given ~~
in poſition, and therefore alſo their common ſection EIF, and hence the

points E and F.

~~
~~

~~Suppoſe now the thing done, and that the center of the ſphere ſought is B. ~~
~~Draw BE, BF, and BC parallel to ID. ~~
~~Since the triangle ADI, and the ~~
line EIF are in the ſame plane, therefore BE, BF, BC, will be in the ſame

plane.

~~But the line ID is perpendicular to the given plane, therefore the ~~
line BC parallel to it, will alſo be perpendicular to the given plane.

~~Since ~~
then a ſphere is to be deſcribed to touch the plane AD, a perpendicular BC

from it’s center B will give the point of contact C;

~~and BC, BE, BF will be ~~
equal, and it has been proved that they are in a plane given in poſition, in

which plane is alſo the right line AD.

~~The queſtion is then reduced to this, ~~
Having two points E and F given, as alſo a right line AD in the ſame plane,

~~
to find the center of a circle which will paſs through the two points, and like-~~
wiſe touch the right line, which is the VIIth of the preceeding Problems.

~~
~~

PROBLEM III.
~~
~~~~
~~~~
~~

Having three points N, O, M, given, as likewiſe a ſphere IG, to de-ſcribe a ſphere which will paſs through the three given points, and likewiſe

touch the given ſphere.

~~
~~

~~The circle NOM in the ſurface of the ſphere ſought is given, and a per-~~
pendicular to its plane from it’s center FA being drawn, the center of the

ſphere required will be in this line.

~~From I the center of the given ſphere ~~
let IB be drawn perpendicular to FA, and through F, ED parallel to IB,

which, from what has been before proved, will be in the plane of the circle

NOM, and the points E and D will be given.

~~
~~

~~Suppoſe now the thing done, and that the center of the ſphere required is ~~
C.

~~Then the lines CI, CE, CD, will be in the ſame plane, which is given, as ~~
the points I, E, and D are given.

~~But the point of contact of two ſpheres is ~~
in the line joining their centers;

~~therefore the ſphere ſought will touch the ~~
ſphere given in the point G, and the line IC will exceed the lines EC, ED, by

IG the radius of the given ſphere:

~~with center I therefore and this diſtance ~~
IG let a circle be deſcribed in the plane of the lines CI, CE, CD, and it

will paſs through the point G and be given in magnitude and poſition;

~~but ~~
the points D and E are alſo in the ſame plane;

~~and therefore the queſtion is ~~
reduced to this, Having two points E and D given, as likewiſe a circle

IGH, to find the center of a circle which will paſs through the two points

and likewiſe touch the circle, which is the XIIth of the preceeding Problems.

~~
~~

PROBLEM IV.
~~
~~~~
~~

Having four planes AH, AB, BC, HG, given; ~~it is required to de-~~
ſcribe a ſphere which ſhall touch them all four.

~~
~~

If two planes touch a ſphere, the center of that ſphere will be in a plane beſecting the inclination of the other two.

~~And if the planes be parallel, it ~~
will be in a parallel plane beſecting their interval.

~~This being allowed, ~~
which is too evident to need further proof;

~~the center of the ſphere ſought ~~
will be in a plane biſecting the inclination of two planes CB and BA;

~~it will ~~
likewiſe be in another plane biſecting the inclination of the two planes BA and

AH;

~~and therefore in a right line, which is the common ſection of theſe two
~~
~~
biſecting planes; ~~
~~let this right line be EF. ~~
~~Moreover, the center of the ~~
ſphere ſought will alſo be in a plane biſecting the inclination of the two planes

AH and GH, and the interſection of this laſt biſecting plane with the right

line EF will give a point D, which will be the center of the ſphere required.

~~
~~

PROBLEM V.
~~
~~~~
~~

Having three planes AB, BC, CD, given, and alſo a point H; ~~to find ~~
a ſphere which ſhall paſs through the given point, and likewiſe touch the

three given planes.

~~
~~

Suppose it done. ~~The three planes, by what was ſaid under the laſt pro-~~
poſition, will give a right line in poſition, in which will be the center of the

ſphere required.

~~Let this right line be GE, perpendicular to which from H ~~
the given point let HI be drawn, which therefore will be given in magnitude

and poſition.

~~Let HI be produced, and FI taken equal to HI; ~~
~~the point ~~
F will then be given.

~~Now ſince the center of the ſphere required is in the ~~
line GE, and FH is perpendicular thereto and biſected thereby, and one

extreme H is by hypotheſis in the ſurface of the ſaid ſphere, the other extreme

F will be ſo too.

~~Nay even a circle deſcribed with I center and IH radius ~~
in a plane perpendicular to GE will be in the ſaid ſpherical ſurface.

~~Here ~~
then we have a circle given in magnitude and poſition, and taking any one

of the given planes AB, by an evident corollary from Problem II.

~~of this ~~
Supplement, a ſphere may be deſcribed which will touch the given plane,

and likewiſe have the given circle in it’s ſurface;

~~and ſuch a ſphere will ~~
anſwer every thing here required.

~~
~~

PROBLEM VI.
~~
~~~~
~~

Having three planes ED, DB, BC, given, and alſo a ſphere RM, to conſtruct a ſphere which ſhall touch the given one, and likewiſe the three

given planes.

~~
~~

Suppose it done, and that the ſphere ERCA is the required one, viz. ~~touches the ſphere in R, and the planes in E, A, C. ~~
~~Let the center of this ~~
ſphere be O;

~~then drawing RO, EO, AO, CO, they will all be equal, and ~~
RO will paſs through M the center of the given ſphere;

~~and EO, AO, CO, ~~
will be perpendicular to the planes ED, DB, BC.

~~Let OU, OG, OI, be ~~
made each equal to OM;

~~and through the points U, G and I, let the planes ~~
UP, GH, IN, be ſuppoſed drawn parallel to the given ones ED, DB, BC,

reſpectively.

~~Since OR is equal to OE, and OM equal to OU, RM will be
~~
~~
equal to UE: ~~
~~but RM is given in magnitude, being the radius of the given ~~
ſphere, therefore UE is alſo given in magnitude.

~~And ſince OE is perpen-~~
dicular to the plane DE, it will be alſo to the plane PU which is parallel

thereto.

~~UE then being given in magnitude, and being the interval be-~~
tween two parallel planes DE, PU, whereof DE is given in poſition by hypo-

theſis, the other PU will alſo be given in poſition.

~~In the ſame manner it ~~
may be proved that the planes GH, IN, are given in poſition, and that the

lines OG, OI, are perpendicular thereto reſpectively, and each alſo equal to

OM.

~~A ſphere therefore deſcribed with center O and OM radius will touch ~~
the three planes PU, GH, IN, given in poſition:

~~but the point M is given, ~~
being the center of the given ſphere.

~~The queſtion is then reduced to this, ~~
Having three planes given PU, GH, IN, and a point M, to find the radius

of a ſphere which ſhall touch the given planes, and paſs through the given

point;

~~which is the ſame as the preceeding Problem. ~~
~~[And this radius be-~~
ing increaſed or diminiſhed by MR, according as R is taken in the further or

nearer ſurface of the given ſphere, will give the radius of a ſphere which will

touch the three given planes DE, DB, BC, and likewiſe the given

ſphere.

~~]~~

By a like method, when among the Data there are no points, but only planes and ſpheres, we ſhall always be able to ſubſtitute a given point in the

place of a given ſphere.

~~
~~

PROBLEM VII.
~~
~~~~
~~~~
~~

Having two points H, M, as alſo two planes AB, BC, given, to find a ſphere which ſhall paſs through the given points, and touch the given

planes.

~~
~~

Draw HM and biſect it in I, the point I will be given, through the point I let a plane be erected perpendicular to the right line HM, this plane

will be given in poſition, and the center of the ſphere required will be in this

plane.

~~But becauſe it is alſo to touch the planes AB, BC, its center will be ~~
alſo in another plane given in poſition (by what has been proved, Prob.

~~IV.) ~~
~~and therefore in a right line which is their interſection, given in poſition, ~~
which let be GE;

~~to which line GE from one of the given points M demit-~~
ting a perpendicular MF, it will be given in magnitude and poſition, and

being continued to D ſo that FD equals MF, the point D will be given;

and, from what has been proved before, will be in the ſpherical ſurface.

~~
~~
~~
Thercfore there are given three points H, M, D, as likewiſe a plane AB, ~~
or AC, through which points the ſphere is to paſs, and alſo touch the given

plane.

~~Hence it appears that this Problem is reduced to the IId of this ~~
Supplement.

~~
~~

Before we proceed, the following eaſy Lemmas muſt be premiſed.~~
~~

LEMMA I.
~~
~~

Let there be a circle BCD, and a point E taken without it, and iſ from E a line EDOB be drawn to paſs through the center, and another line ECA

to cut it any ways;

~~we know from the Elements that the rectangle AEC is ~~
equal to the rectangle BED.

~~Let us now ſuppoſe a ſphere whoſe center is O, ~~
and one of whoſe great circles is ACDB;

~~if from the ſame point E a line ~~
ECA be any-how drawn to meet the ſpherical ſurface in the points C and A,

I ſay the rectangle AEC will ſtill be equal to the rectangle BED.

~~For if we ~~
ſuppoſe the circle and right line ECA to revolve upon EDB as an immove-

able axis, the lines EC and EA will not be changed, becauſe the points C

and A deſcribe circles whoſe planes are perpendicular to that axis;

~~and ~~
therefore the rectangle AEC will in any plane be ſtill equal to the rectangle

BED.

~~
~~

LEMMA II.
~~
~~

By the ſame method of reaſoning, the Vth Lemma immediately preceed-ing Problem XIII, in the Treatiſe of Circular Tangencies, may be extended

alſo to ſpheres, viz.

~~that in any plane (ſee the Figures belonging to that ~~
Lemma) MG X MB = MH X MA.

~~And alſo that MF X MC = ME X MI.~~
~~
~~

LEMMA III.
~~
~~~~
~~

Let there be two ſpheres YN, XM, through whoſe centers let the right line RYNXMU paſs, and let it be as the radius YN to the radius XM, ſo

YU to XU;

~~and from the point U let a line UTS be drawn in any plane, ~~
and let the rectangle S U T be equal to the rectangle RUM;

~~I fay that if ~~
any ſphere OTS be deſcribed to paſs through the points T and S, and to

touch one of the given ſpheres XM as in O, it will alſo touch the other

given ſphere YN.

~~For joining UO, and producing it to meet the ſurface of ~~
the ſphere OTS in Q;

~~the rectangle QUO = the rectangle SUT, by ~~
Lemma I.

~~but the rectangle SUT = the rectangle RUM. ~~
~~by conſtruction,
~~
~~
which RUM by Lemma II. ~~
~~is equal to a rectangle under UO and a line ~~
drawn through the points U and O to the further ſurface of the ſphere YN.

~~Therefore the point Q is in the ſurface of the ſphere YN; ~~
~~it is therefore ~~
common to the ſpheres YN and OTS;

~~and I ſay that theſe ſpheres touch in ~~
the ſaid point Q.

~~For from the point U let a line UZ be drawn in any ~~
plane of the ſphere OTS, and being produced let it cut the three

ſpheres in the points Z, D, H, K, P, B.

~~The rectangle ZUB in the ~~
ſphere OTS is by Lemma I.

~~and II. ~~
~~equal to the rectangle DUP terminated ~~
by the ſpheres XM and YN.

~~But DU is greater than ZU, becauſe the ~~
ſpheres XM and OTS touch in the point O, and therefore any other line from

U but UO muſt meet the ſurface of OTS before it meets the ſurface XM.

Since then ZUB = DUP, and DU is greater than ZU, UP muſt be leſs

than UB, and the point B will fall without the ſphere YN;

~~and by the ~~
fame reaſon, all other points in the ſurface of the ſphere OTS, except the

point Q.

~~
~~

The Demonſtration is ſimilar and equally eaſy in all caſes, whether the ſpheres touch exlernally or internally.

~~
~~

LEMMA IV.
~~
~~

Let there be a plane AC, and a ſphere FGD through whoſe center O let FODB be drawn perpendicular to the plane, and from F any right line

FGA cutting the ſphere in G and the plane in A;

~~I ſay that the rectangle ~~
AFG = the rectangle BFD.

~~For let the given ſphere and plane be cut by ~~
the plane of the triangle ABF, the ſection of the one will be the circle GDF,

and of the other the right line ABC.

~~Since the line FB is perpendicular ~~
to the plane AC, it will be alſo to the right line AC.

~~Having then a circle ~~
FDB, and a right line AC in the ſame plane;

~~and a line FDB paſſing thro' ~~
the center perpendicular to AC, join D and G, and in the quadrilateral

figure ABDG the angles at B and G being both right ones, it will be in a

circle, and the rectangle AFG = the rectangle BFD;

~~and the ſame may be ~~
proved in any other ſection of the ſphere.

~~
~~

LEMMA V.
~~
~~~~
~~~~
~~

Let there be a plane ABD and a ſphere EGF, through whoſe center O let FOEC be drawn perpendicular to the plane, and in any other plane let

FHI be drawn ſo that the rectangle IFH = the rectangle CFE:

~~if through
~~
~~
the points I and H a ſphere be deſcribed which touches the plane AC, I ſay ~~
it will alſo touch the ſphere EGF.

~~From F draw FB to the point of contact ~~
of the ſphere and plane, and make the rectangle BFN = the rectangle CFE,

and the point N will be in the ſurface of the ſphere EGF, by Lemma IV.

~~But the rectangle CFE, by conſtruction, = the rectangle IFH; ~~
~~therefore ~~
IFH = BFN, and the point N will be alſo in the ſurſace of the ſphere IHB.

It remains then to be proved that theſe ſpheres touch in N, which is very eaſy

to be done.

~~For from the point F through any point R in the ſpherical ſur-~~
face EGF let the line FR be drawn, which may cut the ſpherical ſurface

IBH in L and P, and the plane AC in K.

~~The rectangle KFR = the rect-~~
angle CFE, by Lemma IV.

~~= the rectangle IFH, by conſtruction, = the ~~
rectangle PFL.

~~Since then KFR = PFL, and KF is greater than PF, be-~~
cauſe the ſphere IHB touches the plane AC in B, therefore FR is leſs than

FL, and the point R is without the ſphere IHB, and the ſame may be ſhewn

of every other point in the ſpherical ſurface EGF, except the point N.

~~
~~

~~Theſe Lemmas, though they be very eaſy, are very elegant and valuable, ~~
eſpecially the IIId and Vth.

~~In the IIId. ~~
~~though there be an inſinite num-~~
ber of ſpheres which, paſſing through the points T and S, may touch the

ſphere XM, yet they will all alſo touch the ſphere YN, by what is there

proved.

~~In the Vth, though there be an infinite number of ſpheres which, ~~
paſſing through the points I and H, may touch the plane AC, yet they will

all alſo touch the ſphere EGF, by what is there proved.

~~
~~

~~We ſhall now be able to go through the remaining Problems with eaſe.~~
~~
~~

PROBLEM VIII.
~~
~~~~
~~

~~Let there be given a plane ABC, and two points H and M, and alſo a ~~
ſphere DFE;

~~to find a ſphere which ſhall paſs through the given points, and ~~
touch the given plane, and likewiſe the given ſphere.

~~
~~

~~Through the center O of the given ſphere let EODB be demitted perpen-~~
dicular to the given plane ABC, and let HE be drawn, and make the rect-

angle HEG equal to the rectangle BED, and G will then be given.

~~Find ~~
then a ſphere, by Problem II.

~~which ſhall paſs through the three points M, ~~
H, G, and touch the plane ABC, and it will be the ſphere here required.

~~For it paſſes through the points M and H, and touches the plane ABC, ~~
by conſtruction;

~~it likewiſe touches the ſphere DFE, by Lemma V. ~~
~~For ~~
ſince the rectangle HEG = the rectangle BED, every ſphere which paſſes

~~
through the points H and G, and touches the plane ABC, touches likewiſe ~~
the ſphere DFE.

~~
~~

PROBLEM IX.
~~
~~~~
~~

Let there be given two ſpheres AB, DE, as alſo two points H and M; ~~to find a ſphere which ſhall paſs through the two given points, and likewiſe ~~
touch the two given ſpheres.

~~
~~

Let the right line AF be drawn paſſing through the centers of the ſpheres, and as the radius AB is to the radius DE, ſo make BF to EF, and

the point F will be given.

~~Make the rectangle HFG = the rectangle NFA, ~~
and the point G will be given.

~~Now having given three points M, H, G, ~~
as alſo a ſphere DE;

~~find a ſphere by Problem III, which ſhall paſs through ~~
the given points, and touch the given ſphere;

~~and, by Lemma III, it will be ~~
the ſphere here required.

~~
~~

PROBLEM X.
~~
~~~~
~~

Let there be given two planes AB, BD, a point H, and a ſphere EGF;

~~to find a ſphere which ſhall paſs through the given point, and touch ~~
the given ſphere, as alſo the two given planes.

~~
~~

Through the center O of the given ſphere let a perpendicular to either of the given planes CEOF be demitted, and make the rectangle HFI = the

rectangle CFE.

~~Then having given the two points H and I, as alſo the ~~
two planes AB, BD;

~~find a ſphere, by Problem VII, which ſhall paſs ~~
through the two given points, and likewiſe touch the two given planes;

~~and, ~~
by Lemma V, it will be the ſphere required.

~~
~~

PROBLEM XI.
~~
~~~~
~~

Let there be given a point, a plane, and two ſpheres; ~~to find a ſphere ~~
which ſhall paſs through the point, touch the plane, and alſo the two

ſpheres.

~~
~~

~~This Problem, by a like method of reaſoning, is immediately reduced to ~~
the VIIIth, where two points, a plane, and a ſphere are given, and that by

means of the Vth Lemma.

~~But if you chuſe to uſe the IIId Lemma, it will ~~
be reduced to the ſame Problem by a different method, and a different

conſtruction.

~~
~~

PROBLEM XII.
~~
~~~~
~~

Let there be given a point and three ſpheres, to ſind a ſphere which ſhall paſs through the point, and touch all the three ſpheres.

~~
~~

We aſſign no figure to this Problem alſo, becauſe, by help of Lemma III, it may immediately be reduced to Problem IX, where two points and two

ſpheres are given.

~~
~~

PROBLEM XIII.
~~
~~~~
~~

Let there be two planes, and alſo two ſpheres given; ~~to find a ſphere ~~
which ſhall touch the planes, as alſo the ſpheres.

~~
~~

~~Suppoſe the thing done. ~~
~~If therefore we imagine another ſpherical ſurface ~~
parallel to that which is required, and which we now ſuppoſe found, and

whoſe radius is leſs than it's by the radius of the leſſer of the two given

ſpheres;

~~this new ſpherical ſurface will touch two planes parallel to the two ~~
given ones, and whoſe diſtance therefrom will be equal to the radius of the

leſſer of the given ſpheres;

~~it will alſo touch a ſphere concentric to the ~~
greater given one whoſe radius is leſs than it's by the radius of the leſſer given

one;

~~and it will likewife paſs through the center of the leſſer given one. ~~
~~The Queſtion is then reduced to Problem X, where a point, two planes and ~~
a ſphere are given.

~~
~~

PROBLEM XIV.
~~
~~~~
~~

Having three ſpheres and a plane given; ~~to find a ſphere which ſhall ~~
touch them all.

~~
~~

By a like method to what is uſed in the preceeding, and in the VIth Pro-blem, this is reduced to Problem XI, where a point, a plane, and two

ſpheres are given.

~~
~~

PROBLEM XV.
~~
~~~~
~~
~~
~~

Having four ſpheres given; ~~to ſind a ſphere which ſhall touch them all.~~
~~
~~

Suppose the thing done. ~~As, in the treatiſe of Circular Tangencies, the ~~
laſt Problem, where it is required, having three circles given, to find a fourth

which ſhall touch them all, is reduced to another, where a point and two

circles are given;

~~ſo alſo this, by a like method, and ſimilar to what has been ~~
uſed in the preceding Problems, is reduced to Problem XII, where three

ſpheres and a point are given.

~~
~~

The various Caſes, Determinations and other Minuliæ we have taken no notice of:

~~for if we had, this Treatiſe would have very much exceeded that ~~
to which it was intended as a Supplement.

~~
~~

Synopſis of the PROBLEMS.
~~
~~ 1. .... # 4. 1111 # 15. 0000 2. ...1 # 5. 111. # 12. 000. 3. ...0 # 6. 1110 # 14. 0001 7. ..11 # 13. 1100 # 9. 00.. 8. ..10 # 10. 11.0 # 11. 00.1

~~N. ~~
~~B. ~~
~~A point is repreſented by.~~
~~, a plane by 1, and a ſphere by 0.~~
~~
~~

THE TWO BOOKS OF APOLLONIUS PERGÆUS, CONCERNING DETERMINATE SECTION, As they have been Reſtored by WILLEBRORDUS SNELLIUS.
By JOHN LAWSON, B. D. Rector of Swanſcombe, Kent.
TO WHICH ARE ADDED, THE SAME TWO BOOKS, BY WILLIAM WALES, BEING AN ENTIRE NEW WORK.
LONDON: Printed by G. BIGG, Succeſſor to D. LEACH. And ſold by B. White, in Fleet-Street; L. Davis, in Holborne; J. Nourse, in the Strand; and T. Payne, near the Mews-Gate.
MDCC LXXII.

ADVERTISEMENT.
~~
~~~~
~~

~~SINCE the publication of the preceding Tract on ~~

Tangencies, the Tranſlator thereof has obſerved, that thoſe pieces of

Willebrordus Snellius, which he mentioned in his Preface thereto, are exceeding ſcarce

in England.

~~His Reſuſcitata Geometria de ſectione rationis ~~
&

~~ſpatii, 1607, he has never once had an opportunity ~~
of ſeeing;

~~but ſuppoſing this ſhould in a ſhort time be ~~
loſt, more than ample amends is made by what

Dr.

Halley has done on the ſame ſubject. ~~Leſt the ~~
other Tract, De Sectione Determinatâ, ſhould undergo

the ſame fate with the original

Apollonius, he was determined to reſcue it therefrom, or reſpite it at leaſt

for ſome time, by putting it into an Engliſh dreſs.

~~While he was doing this, he happened to communicate ~~
the piece to ſome friends;

~~one of whom has ventured, ~~
after

Snellius, on this ſubject, and he preſumes with ſome ſucceſs, as every Reader will allow, when he peruſes the

Propoſitions here printed after thoſe of

Snellius. ~~Yet, ~~
notwithſtanding this, the Editor perſiſted in his reſolu-

tion of printing his tranſlation of

Snellius, as the work
~~
has much merit, and was in danger of being loſt, and as he ~~
was the firſt that conſtructed Quadratic Equations after

this particular manner, as Dr.

Simson obſerves in his Note on

Euc. ~~VI. ~~
~~28 and 29.~~
~~
~~

~~The Editor leaves his Friend to ſpeak for himſelf ~~
in relation to what he has done, and truſts that the

candid Reader will not think more meanly of his

performance from the modeſt manner in which he

ſpeaks of it himſelf.

~~
~~

EXTRACT from PAPPUS's Preface to his Seventh Book in Dr. HALLEY's Tranſlation.
DE SECTIONE DETERMINATA II.
~~
~~
From hence it appears that Euclid's were called the firſt Elements, and that the other Analytical Tracts, recited by Pappus, were called the ſecond Elements.

~~HIS ſubjiciuntur libri duo de Sectione Determinatâ, ~~
quas etiam ad modum præcedentium unam pro-

poſitionem dicere liceat, ſed disjunctam:

~~quæ hujuſ-~~
modi eſt.

~~“ Datam rectam infinitam in uno puncto ſe-~~
care, ita, ut è rectis interceptis inter illud &

~~puncta ~~
in illâ data, vel quadratum ex unâ, vel rectangulum

ex duabus interceptis, datam habeat rationem, vel ad

contentum ſub aliâ unâ interceptâ &

~~datá quâdum; ~~
~~vel etiam ad contentum ſub duabus aliis interceptis: ~~
idque ad quam partem velis punctorum datorum.

~~” ~~
Hujus autem, quaſi bis disjunctæ, &

~~intricatos Dio-~~
riſmos habentis, per plura neceſſario facta eſt demon-

ſtratio.

~~Hanc autem dedit Apollonius communi methodo ~~
tentamen faciens, ac ſolis rectis lineis uſus, ad exemplum

ſecundi libri Elementorum

~~ac rur-
ſus idem demonſtravit ingenioſe quidem, & ~~
~~magis ad ~~
inſtitutionem accomodate, per ſemicirculos.

~~Habet ~~

~~
autem primus liber Problemata ſex, Epitagmata, ſive ~~
Diſpoſitiones punctorum, ſedecim;

~~Dioriſmos quinque: ~~
~~quorum quatuor quidem Maximi ſunt, Minimus vero ~~
unus.

~~Sunt autem maximi, ad ſecundum Epitagma ſe-~~
cundi problematis;

~~item ad tertium quarti problematis; ~~
ad tertium quinti &

~~ad tertium ſexti. ~~
~~Minimus vero ~~
eſt ad tertium Epitagma tertii problematis.

~~-Secundus ~~
liber de Sectione Determinatâ tria habet Problemata,

Diſpoſitiones novem, Determinationes tres;

~~e quibus ~~
Minima ſunt ad tertium primi, ut &

~~ad tertium ſecun-~~
di;

~~Maximum autem eſt ad tertium tertii problematis. ~~
-Lemmata habit liber primus XXVII, ſecundus vero

XXIV.

~~Inſunt autem in utroque libro de Sectione ~~
determinatâ Theorenata octoginta tria.

~~
~~

THE PREFACE.
~~
~~~~
~~
~~
~~~~
~~~~
~~
The words which are in Italics were entirely omitted in Snellius's Extract from Pappus, both in the Greek and Latin, and are read with ſome variation in Commandine's tranſlation; but are here printed according to Dr. Halley: and though I know not whether in this particular place he had the Authority of either of the Savilian MSS, yet I hope I run no great riſk in ſubſcribing to the opinion of ſo excellent a Geometer.

~~
~~
~~
~~
So called, I conceive, becauſe in every other Caſe of the third Epitagma, except this extreme, or limiting one, there are two points which will ſatisſy the Problem.

~~
~~
~~
~~

~~HAD not a motive more prevalent than Cuſtom induced me to ~~
ſay ſomething by way of Preface to the Performance which

I herewith offer to the Public, the difficulty I find in doing it with

propriety, would have determined me to remain entirely ſilent.

~~
~~

~~The ſubject has employed the Pen of one of the ableſt Geometers ~~
of the laſt Century;

~~it may therefore ſeem very preſumptuous, in ~~
me at leaſt, who am but young in theſe matters, to attempt it after

him.

~~To obviate, if poſſible, this Cenſure is my only intention ~~
here;

~~and I hope I ſhall not be deemed impertinent, if I attempt to ~~
ſhew wherein I apprehend I have come nearer to the great original

than he hath done.

~~
~~

Pappus, in his preface to the ſeventh Book of Matbematical Collections, tells us that this Tract of

Apollonius was divided into two Books;

~~that the firſt Book contained ſix Problems, and the ~~
ſecond three:

~~now ~~
Snellius' whole work contains but four; ~~and ~~
it ſeemed to me difficult to ſhew how thoſe could contain the ſub-

ſtance of nine, and yet the ſix firſt have ſixteen Epitagmas, or ge-

neral Caſes, and the three laſt nine.

~~I firſt, therefore began with ~~
inquiring whether, or no, other Problems could not be found,

wherein the ſection of an indefinite ſtraight line is propoſed to be

effected, “So, that of the ſegments contained between the point of

ſection ſought, and given points in the ſaid line, either the ſquare

on one of them, or the rectangle contained by two of them, may

have a given ratio to the rectangle contained by one of them and a

given external line, or to the rectangle contained by two of them;

~~“ ~~
as is deſcribed by P

APPUS.

~~In this inquiry it ſoon occurred to me, that the three problems ~~
which make my firſt, ſecond and fourth, come within the account

given by P

APPUS;~~and therefore are properly Problems in Determi-~~
ate Section, to be added to the four given by

Snellius: ~~and it does ~~
not appear to me that more can be found which ſhould.

~~Hence I ~~
concluded, that there were in theſe, ſome, more general than thoſe

of

Apollonius, which ought therefore to be divided.~~
~~

~~My next buſineſs was, if poſſible, to find out the order in which ~~

Apollonius had arranged them: ~~and here, with reſpect to the ~~
firſt Book, I had no other information to guide me, but what is to

~~
be met with in the above mentioned Preface of ~~
Pappus; ~~where he ~~
tells us that in the ſix Problems of Book I.

~~there were “Sixteen ~~
Epitagmas, or general Caſes, five Determinations;

~~and of theſe, ~~
four were Maxima, and one a minimum:

~~That the maxima are at the ~~
ſecond Epitagma of the ſecond Problem, at the third of the fourth,

the third of the fifth, and the third of the ſixth;

~~but that the minimum ~~
was at the third Epitagma of the third problem.

are given, would be antecedent to thoſe wherein there were more;

~~and of theſe wherein the number of given points are the ſame, that ~~
thoſe would be prior to the others, wherein there was a given ex-

ternal line concerned:

~~and laſtly, that when the number of given ~~
points were two, the ſecond Caſe, or Epitagma, would naturally

be when the required point O is ſought between the two given

ones.

~~
~~

~~Now the three new Problems, together with the three firſt of ~~

Snellius, making exactly ſixteen Epitagmas, viz. ~~one in the firſt, ~~
and three in each of the others;

~~it ſeemed highly probable, that ~~
theſe compoſed the firſt book.

~~Alſo that the Problem, wherein ~~
only one point was given, would be the firſt;

~~and it ſeemed eaſy ~~
to aſſign the ſecond, becauſe it is the only one wherein the limita-

~~
tion is at the ſecond Epitagma; ~~
~~and farther, the Limiting Ratio is ~~
therein a maximum, as it ought.

~~Again, the Problem, wherein ~~
it is propoſed to make the ſquare on AO in a given ratio to the rect-

angle contained by EO and P, has its limiting ratio a minimum

when the required point is ſought beyond (E) that of the given ones

which bounds the ſegment concerned in the conſequent term of

the ratio;

~~which, therefore, I apprehend muſt have been the third ~~
Epitagma, and if ſo, this of courſe muſt have been the third Pro-

blem:

~~and as there remains only one wherein the number of given ~~
pointsare two, I make that the fourth.

~~With reſpect to the fifth and ~~
ſixth Problems, in which three points are given, it ſhould ſeem

that that would be the firſt in order, wherein there is a given ex-

ternal line concerned.

~~
~~

~~But it ſhould, by no means, be diſſembled that objections may be ~~
brought againſt the identity, and arrangment of ſome of theſe Pro-

blems.

~~For firſt, ~~
Pappus no where expreſsly ſays that Apollo-

NIUS compared together two ſquares, wherefore, if this cannot be implied, the identity of the fourth Problem is deeply ſtruck at:

~~and moreover, this fourth Problem perhaps cannot with propriety, ~~
be ſaid to have its limiting ratio either maximum or minimum, un-

leſs the ratio of equality, can be admitted as ſuch.

~~Laſtly, in the ~~
fifth Problem, the ſaid limiting ratio is a minimum, and not a maxi-

mum as it is ſaid to have been by

Pappus: ~~either, therefore, a
~~
~~
miſtake muſt be admitted in this Author, or the fifth Problem is ~~
wrong placed.

~~I am not prepared farther to obviate theſe objec-~~
tions, and only mention them to ſhew that although I ſaw them

in their full force, I could by no means agree, that they are pow-

erful enough to overturn thoſe already advanced for what I have

~~
~~

~~I come now to Book II, which if I am not much miſtaken, was ~~
entirely employed about what

Snellius makes his fourth Problem. ~~In this I am confirmed not only by the account which ~~
Pappus gives in his Preface, but much more by the Lemmas of

Apollonius

~~For we there find that ~~
Lemma 21, where-in is aſſigned the leaſt ratio which the rectangle contained by AO

and UO can bear to that contained by EO and IO, when O is ſought

between the two mean points of the four given ones, is ſaid to be

concerned in determining the μοναχὴ, or ſingle Caſe

~~
This Problem therefore of ~~
Apollonius contained only thoſe Caſes of the general one, where O is ſought between the two mean

points.

~~In like manner, we gather from Lemma 22, that his ſe-~~
cond Problem was concerned in determining the point O when ſought

between a mean point, and an extreme one.

~~And laſtly, from ~~
Lemma 24, that the third Problem of Book II.

~~determined ~~
the point O when required without all the given ones.

~~
~~
~~
The Limitations of the two former are ſaid by ~~
Pappus to have been minimums, and that of the third a maximum, in conformity,

to which, I have here made them ſo;

~~although I cannot ſee with ~~
what propriety:

~~each of them admitting, in ſome Caſes, of a maxi-~~
mum and in others of minimum, as I have intimated in a ſcbolium at

the end of each Problem.

~~But notwithſtanding I have conformed to ~~
the manner of

Apollonius in dividing this Problem into three, which it muſt be confeſſed contributes much to order in enumera-

ting ſuch a multitude of Caſes, yet have I previouſly ſhewn how

the whole may be generally conſtructed at once;

~~and that by a me-~~
thod, which I flatter myſelf will not be found inferior to any that

hath heretofore been given of this very intricate and general

Problem.

~~
~~

~~Such are the things that I have attempted, and ſuch the reaſons ~~
for what I have done in the following little Tract.

~~The merit due ~~
to each of them I chearfully ſubmit (where every one ought) to

the deciſion of the impartial Reader.

~~In the Conſtructions, my ~~
chief Aim was novelty and uniformity:

~~I could have given more ~~
ſimple conſtructions to one or two of them;

~~in particular the ſixth ~~
of Book I:

~~but it was not my intention to give any thing that I ~~
knew had been done before.

~~I know of many imperfections, but ~~
no falſe reaſonings, and hope none will be found;

~~but if there ~~
ſhould, I hope the candid Geometer will be more inclined to ex-

cuſe than exult, when I aſſure him the greateſt part of the work

has been executed at different times, amidſt the hurry and perplexi-

ties which it may eaſily be conceived attend the fitting out for a

three years Voyage to the ſouth ſeas.

~~
~~

~~I cannot conclude without acknowledging, in the warmeſt man-~~
ner, the obligations I am under to my truly worthy and ingenious

friend, the Tranſlator of

Snellius; ~~for the great pains and trou-~~
ble he hath taken to furniſh me with tranſlations from various Au-

thors, which my utter want of the Greek, and little acquaintance

with the Latin Language made abſolutely neceſſary to me:

~~And ~~
after all, had it not been for his kindneſs, this attempt might ſtill

have remained in as great obſcurity as its Author.

~~
~~

PROBLEMS CONCERNING DETERMINATE SECTION.
PROBLEM I.
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~

~~TO cut a given indefinite right line in one point, ſo that of the ſegments ~~
intercepted between that point and two other points given in the inde-

finite right line, the ſquare of one of them may be to the rectangle under the

other and a given external right line, in a given ratio.

~~
~~

~~In the given indefinite right line let be aſſigned the points A and E, it is then ~~
required to cut it in the point O, ſo that

AO2 may be to OE into a given line AU in the ratio of R to S;

~~which ratio let be expreſſed by AI to AU, ~~
ſetting off AI from A either way, either towards E or the contrary;

~~and ~~
then from A and I erect two perpendiculars AY equal to AE, and IR

equal to AI, and theſe on the ſame ſide of the given indefinite line, if AI

was ſet off towards E;

~~but on oppoſite ſides, if AI was ſet off the other way. ~~
~~The former conſtruction I will beg leave to call Homotactical, and the latter ~~
Antitactical.

~~Let now the extremities of theſe perpendiculars Y and R be ~~
joined, and upon YR as a diameter let a circle be deſcribed, I ſay that the

interſection of this circle with the given indefinite line ſolves the Problem.

If it interſects the line in two places, the Problem admits of two Solutions;

~~
~~
~~
but if it only touches, then only of one; ~~
~~if it neither touches nor cuts, it is ~~
then impoſſible.

~~
~~

Demonstration. ~~Let a point of interſection then be O, and join O Y ~~
and OR.

~~The angles AYO and IOR are equal, the angle AOY being the ~~
complement of each of them to a right one, and hence the triangles AOY and

IOR are ſimilar.

~~
~~

~~Hence AY = AE: ~~
~~AO:~~
~~: OI: ~~
~~IR = AI ~~
And by div .

~~or comp . ~~
~~EO: ~~
~~AO:~~
~~: AO: ~~
~~AI ~~
And

AO2 = EO x AI Therefore

AO2 (= EO x AI): ~~EO x AU:~~
~~: AI: ~~
~~AU:~~
~~: R: ~~
~~S ~~
Q.

~~E. ~~
~~D.~~
~~
~~

~~This Problem admits of two Caſes. ~~
~~The 1ſt determinate or limited, the 2d ~~
unlimited.

~~
~~

Case I. ~~Is when OE the co-efficient of the given external line AU is part of ~~
AO the ſide of the required ſquare [fig.

~~1. ~~
~~2.~~
~~] and here the ~~
Llmitation is, that AI muſt not be given leſs than four times AE, as appears from fig.

~~2. ~~
~~for AE: ~~
~~AO:~~
~~: OI: ~~
~~AI; ~~
~~and here OI being the half of AI, AE will be ~~
the half of AO, or the fourth part of AI.

~~In this Caſe the Homotactical Con-~~
ſtruction is uſed.

~~
~~

Case II. ~~Is when AO the ſide of the required ſquare is part of OE the co-~~
efficient of the given external line AU, [fig.

~~3.~~
~~] and this is unlimited, for here ~~
the Anlitactical Conſtruction is uſed.

~~Or if O be required between A and E, ~~
this is effected by the ſame Conſtruction.

~~
~~

LEMMA I.
~~
~~~~
~~

~~If from the extremes of any diameter perpendiculars be let fall upon any ~~
Chord, I ſay that the ſegments of theſe perpendiculars intercepted by like Arcs

are equal, and moreover alſo the ſegments of the Chords themſelves.

~~
~~

~~That YO is equal to IU may be thus ſhewn. ~~
~~Having joined YI, the ~~
angle IYE is a right one, being in a ſemicircle, and the angle at O is right by

hypotbeſis;

~~hence YI is parallel to the Chord, and YOUI is a parallelogram, ~~
and the oppoſite ſides YO and IU will be equal.

~~In the ſame manner OE is ~~
proved equal to UL.

~~And as to the ſegments of the Chord, it is thus ſhewn. ~~
~~By Euc. ~~
~~III. ~~
~~35. ~~
~~and 36, the rect. ~~
~~EOY = rect. ~~
~~SOR, and rect. ~~
~~LUI = rect. ~~
SUR.

~~But, by what has been juſt proved, rect. ~~
~~EOY = rect. ~~
~~LUI; ~~
~~hence ~~
rect.

~~SOR = rect. ~~
~~SUR, and the ſegments SO and OR are reſpectively equal ~~
to the ſegments UR and SU.

~~
~~

LEMMA II.
~~
~~~~
~~

~~If of four proportionals the ſum of two, being either extremes or means, be ~~
greater than the ſum of the other two;

~~then I ſay theſe will be greateſt and ~~
leaſt of all.

~~
~~

~~This is the converſe of Euc. ~~
~~V. ~~
~~25. ~~
~~and may be thus demonſtrated. ~~
~~Draw ~~
a circle whoſe diameter may be equal to the greater ſum;

~~and in it inſcribe the ~~
leſſer ſum IO, which will therefore not paſs through the center, and let the

parts be IU and UO;

~~then through U draw a diameter AUE, and the other ~~
two terms will be AU and EU, of which AU is greateſt of all and EU leaſt of

all, and IU and UO of intermediate magnitude, by Euc.

~~III. ~~
~~7.~~
~~
~~

LEMMA III.
~~
~~~~
~~

~~If of four proportionals the difference of two, being either extremes or ~~
means, be greater than the difference of the other two, then I ſay theſe will be

the greateſt and leaſt of all.

~~
~~

~~This is demonſtrated in the ſame manner as the preceding by Euc. ~~
~~III. ~~
~~8.~~
~~
~~

PROBLEM II.
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~
~~
~~~~
~~~~
~~

~~To cut a given indefinite right line in one point, ſo that, of the three ſeg-~~
ments intercepted between the ſaid point and three points given in the ſame in-

definite right line, the rectangle under one of them and a given external right

line may be to the rectangle under the other two in a given ratio.

~~
~~

~~In the given indefinite line let the aſſigned points be A, E, I. ~~
~~It is then ~~
required to cut it again in the point O, ſo that AO into a given external line

R may be to EO x IO as R to S.

~~If A be an extreme point and E the ~~
middle one, then ſet off IU = AE the contrary way from A;

~~but if A be the ~~
middle point, then ſet it off towards A.

~~Then from U ſet off UN = S the ~~
conſequent of the given ratio, either towards A, or the contrary way;

~~for as ~~
the Caſes vary, it’s poſition will vary.

~~From A and N erect perpendiculars ~~
AY and NM to the given indefinite right line equal to AE and AI re-

ſpectively, and theſe bomotactical if A be an extreme point, but antitactical if

A be the middle point of the three given ones.

~~Join the extremes of theſe ~~
perpendiculars Y and M, and upon YM as a Diameter deſcribe a circle.

~~I ſay ~~
that the interſection of this circle with the given indefinite line ſolves the

Problem.

~~If it interſects the line in two points, then the Problem admits of
~~
~~
two ſolutions; ~~
~~if it only touches, then but of one; ~~
~~if it neither cuts nor ~~
touches, it is then impoſſible.

~~
~~

Demonstration. ~~Let the point of interſection then be O or o. ~~
~~We ſhall ~~
have, by Lemma I.

~~AO x ON = MN x NK = MN x AY = AI x AE, by ~~
conſtruction.

~~Let now from N be ſet off NL = AI in the ſame direction as A ~~
is from I;

~~then by what has been demonſtrated NL: ~~
~~ON:~~
~~: AO: ~~
~~AE.~~
~~
~~

~~And by Diviſion or Compoſition OL: ~~
~~ON:~~
~~: OE: ~~
~~AE ~~
And by Permutation OL:

~~OE:~~
~~: ON: ~~
~~AE ~~
But by what has been proved ON:

~~AE:~~
~~: AI: ~~
~~AO ~~
Therefore by Equality OL:

~~OE:~~
~~: AI: ~~
~~AO ~~
And by Diviſion or Compoſition LE:

~~OE:~~
~~: OI: ~~
~~AO ~~
And LE x AO = OE x OI

But LE = NU.

~~for NL was put = AI, and IU = AE ~~
Hence NU x AO = S x AO = OE x OI

And R:

~~S:~~
~~: R x AO: ~~
~~S x AO or OE x OI ~~
Q.

~~E. ~~
~~D.~~
~~
~~

~~This Problem may be conſidered as having two ~~
Epitacmas, the firſt, when the ſegment aſſigned for the coefficient of the given external line R is

terminated by an extreme point of the three given ones and the point ſought;

~~and this again admits of three Caſes. ~~
~~The other is when the aforeſaid ſeg-~~
ment is terminated by the middle point of the three given ones and the

point ſought.

~~
~~

Epitagma I. Case I. ~~Let the aſſigned points be A, E, I. ~~
~~A an extreme ~~
and E the middle one.

~~And let the point O ſought (ſuch that AO x R: ~~
~~OE ~~
x OI:

~~: R: ~~
~~S) be required to lie between A and E, or elſe beyond I, which ~~
will ariſe from the ſame conſtruction.

~~
~~

~~Here the Homotactical conſtruction is uſed, and IU as likewiſe UN is ſet off ~~
in the ſame direction as AI.

~~And ſince AO: ~~
~~AE:~~
~~: AI: ~~
~~ON, and AO + ON ~~
is greater than AE + AI or AU, by

Lemma II. ~~AO and ON will be the ~~
leaſt and greateſt of all;

~~and AO will therefore be leſs than AE, as likewiſe ~~
Ao (being equal ON by

Lemma I.) ~~greater than AI. ~~
~~This Caſe is ~~
unlimited.

~~
~~

Case II. ~~Let the aſſigned points be in the ſame poſition as before, and let ~~
the point O ſought be required between E and I.

~~
~~

~~Here the conſtruction is Homotactical, and UN is ſet off the contraty way, viz. ~~
~~in the direction IA. ~~
~~And ſince AO: ~~
~~AE:~~
~~: AI: ~~
~~ON, and AO + ON is leſs ~~
than AE + AI or AU, by

Lemma II. ~~AE and AI will be the leaſt and
~~
~~
greateſt of all, and AE will therefore be leſs than AO, and AI greater. ~~
~~And ~~
the ſame will hold with regard to Ao.

~~
~~

~~Here is a ~~
Limitation, which is this; ~~that UN or S the conſequent of the ~~
given ratio, ſet off from R, muſt not be given greater than the difference of

the ſum of AE and AI and of a line whoſe ſquare is equal to four times their

rectangle [i.

~~e. ~~
~~to expreſs it in the modern manner, UN muſt not exceed AI + ~~
AE - √4 AI x AE*.

~~] This appears by Fig. ~~
~~2. ~~
~~to this Caſe, the circle there ~~
touching the given indefinite line, and pointing out the Limit.

~~
~~

Case III. ~~Let the aſſigned points be ſtill in the ſame poſition, and let the ~~
point ſought be now required on the contrary ſide of A.

~~
~~

~~Here the conſtruction is ſtill Homotactical, and UN is ſet off the ſame way as ~~
in the laſt Caſe;

~~and the ~~
Limitation is, that UN muſt not be given leſs than the ſum of AI, AE, and a line whoſe ſquare is equal to four times their rect-

angle [or expreſſing it Algebraically, UN muſt not be leſs than AI + AE +

√4 AI x AE*.

~~]~~

Epitagma II. Case IV. ~~Let now A be the middle point of the given ~~
ones, and let O the point ſought be required either between A and one of the

extremes, or beyond either of the extremes.

~~
~~

~~Here having ſet off IU = AE toward A, you may ſet off UN either way, ~~
and uſing the Antitactical conſtruction, the ſolution will be unlimited.

~~The ~~
only difference is, that if UN be in the direction UI, two ſolutions will ariſe,

whereof in one the point O will fall between A and E, and in the other be-

yond I;

~~but if UN be in the direction IU, two ſolutions will ariſe, whereof ~~
in one the point will fall between A and I, and in the other beyond E.

~~In ~~
proof of which

Lemma III. ~~is to be uſed, as ~~
Lemma II. ~~was in Caſe I. ~~
~~II.~~
~~
~~

Corollary I. ~~If then the given ratio be that of AT to TI, or of AE to ~~
EP ſet off from A the other way, ſo that EP be leſs than AE, I ſay then

that O will fall between E and P, as likewiſe ο between T and I, provided o

falls beyond I.

~~
~~

~~For by conſtruction IU = AE, and UN = PE. ~~
~~therefore IN = AP. ~~
~~But by ~~

Lemma I. ~~oN = AO. ~~
~~therefore (o falling beyond I by hypotbeſis) O will fall ~~
beyond P;

~~but by hypotbeſis it falls ſhort of E; ~~
~~therefore O falls between ~~
P and E.

~~
~~

~~Next to ſhew that ο will fall between T and I, we have AT: ~~
~~TI:~~
~~: AE: ~~
~~EP~~

~~And by Diviſion AT: ~~
~~AI:~~
~~: AE: ~~
~~AP~~

~~Hence AT x AP = IAE or o AO~~

~~Therefore AT x AO is greater than o AO~~

~~Or AT greater than Ao.~~
~~
~~

Corollary II. ~~If the three given points be I, A, E; ~~
~~and O falls between ~~
A and I, ſo as to make AO x PE:

~~IOE:~~
~~: AL: ~~
~~LI, I ſay then O will fall ~~
beyond L.

~~
~~

~~For let us ſuppoſe that O and L coincide; ~~
~~then by hypotbeſis AL: ~~
~~LI:~~
~~: ~~
AL x PE:

~~IL x LE~~

~~And by the next following ~~
Lemma IV. ~~AL x IL: ~~
~~IL x PE:~~
~~: AL: ~~
~~LE ~~
i.

~~e. ~~
~~AL: ~~
~~PE:~~
~~: AL: ~~
~~LE~~

~~Hence PE is equal to LE, a part to the whole, which is manifeſtly abſurd.~~
~~
~~

LEMMA IV.
~~
~~~~
~~~~
~~~~
~~

~~If it be as a line to a line ſo a rectangle to a rectangle; ~~
~~then I ſay it will be ~~
as the flrſt line into the breadth of the ſecond rectangle to the ſecond line into

the breadth of the firſt rectangle, ſo the length of the firſt rectangle to the

length of the ſecond.

~~
~~

~~Suppoſition. ~~
~~AE: ~~
~~IO:~~
~~: UYN: ~~
~~SRL.~~
~~
~~

~~Concluſion. ~~
~~AE x RL: ~~
~~IO x YN:~~
~~: UY: ~~
~~SR.~~
~~
~~

Dem. ~~AE: ~~
~~IO:~~
~~: AE x YN: ~~
~~IO x YN:~~
~~: UYN: ~~
~~SRL~~

~~And by Permutation AE x YN: ~~
~~UYN:~~
~~: AE: ~~
~~UY:~~
~~: IO x YN: ~~
~~SRL~~

~~But SR: ~~
~~AE:~~
~~: SRL:~~
~~: AE x RL~~

~~Therefore ex æquo perturbatè SR: ~~
~~UY:~~
~~: IO x YN: ~~
~~AE x RL~~

~~Q. ~~
~~E. ~~
~~D.~~
~~
~~

LEMMA V.
~~
~~~~
~~~~
~~~~
~~~~
~~

~~If a right line be cut in two points, I fay the rectangle under the alternate ~~
ſegments is equal to that under the whole and the middle ſegment, together

with the rectangle under the extremes.

~~
~~

Dem. ~~AI x IE + IO x IE = AO x IE.~~
~~
~~

~~Hence {AI x IE + IO x IE + AE x IO \\ i. ~~
~~e. ~~
~~AI x IE + AI x IO \\ i. ~~
~~e. ~~
~~AI x EO} = AO x IE + AE x IO.~~
~~
~~

~~Q. ~~
~~E. ~~
~~D.~~
~~
~~

~~N. ~~
~~B. ~~
~~Theſe two ~~
Lemmas ſave much Circumlocution and Tautology in the two following Propoſitions, and indeed are highly uſeful in all caſes where

compound ratios are concerned.

~~
~~

PROBLEM III.
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~

~~To cut a given indefinite right line in one point, ſo that of the three ſeg-~~
ments intercepted between the ſame, and three points given, the rectangle

under two of them may be to the ſquare of the remaining one in a given ratio.

~~
~~

~~In the indefinite line let the three points be A, E, I. ~~
~~it is then required to be ~~
cut again in O, ſo that OA x OE may be to

OI2 (let the ſituation of I be what it may) in a given ratio, which ratio let be expreſſed by EL to LI.

~~[And here I cannot but obſerve with ~~
Hugo D'Omerique, page 113. ~~that ~~
this Problem, viz.

~~‘To exhibit two lines in a given ratio whoſe ſum, or whoſe ~~
difference is given,’ ought to have had a place in the Elements as a Propoſition;

or at leaſt to have been annext as a Scholium to the 9th or 10th of the VIth

Book.

~~] And be the ſituation of L alſo what it may, either between A and E, ~~
or between A and I, or between E and I, or beyond either extreme.

~~To the three ~~
points E, L, I, and the right line AI, let be found, by

Problem II, a fourth point O ſuch, that AI x OE:

~~OI x OL:~~
~~: EI: ~~
~~IL. ~~
~~And let ſuch a Caſe be ~~
choſen of

Problem II, that, according as AO is greater or leſs than AI, ſo of the three rectangles, deſcribed in

Lemma V, made by the four points E, O, I, L, that of IO x EL may accordingly be greater or leſs than that of EI x OL.

~~
~~

~~D~~
EMONSTRATION .~~On ſuppoſition then that ſuch a Caſe of ~~
Problem II. ~~is ~~
made uſe of, we have

AI x OE:

~~OI x OL:~~
~~: EI: ~~
~~IL~~

~~And by ~~
Lemma IV, OL x EI: ~~OE x IL:~~
~~: AI: ~~
~~OI~~

~~And by Diviſion or Compoſition EL x OI: ~~
~~OE x IL:~~
~~: AO: ~~
~~OI~~

~~This appears from ~~
Lemma V.~~
~~

~~Then again by ~~
Lemma IV, AO x OE: OI2:~~: EL: ~~
~~IL.~~
~~
~~

~~Q. ~~
~~E. ~~
~~D.~~
~~
~~

~~This Problem has two ~~
Epitagmas. ~~The firſt wherein OI, whoſe ſquare is ~~
ſought, is bounded by I an extreme point of the three given ones.

~~And this ~~
again admits of three Caſes.

~~The ſecond is when the point I is the middle ~~
point.

~~And this again has three caſes. ~~
~~And there remain two Anomalous ~~
Caſes, wherein Problem II.

~~is of no uſe, which muſt therefore be conſtructed ~~
by themſelves.

~~
~~

Epitagma I. Case I. ~~Let the ratio given, EL to LI, be inequalitatis ~~
majoris, i.

~~e. ~~
~~of a greater to a leſs; ~~
~~and the point O ſought be required to lie ~~
between I and the next point to it E, or elſe to lie beyond I the other way;

~~for the ſame conſtruction ſerves for both. ~~
~~Here ~~
Case I. ~~of ~~
Problem II. ~~is to
~~
~~
be uſed, and the point O will fall between E and I, and the point o beyond ~~
L, much more beyond I.

~~
~~

Case II. ~~Let the given ratio, EL to LI, be inæqualitatis minoris, i. ~~
~~e. ~~
~~of a ~~
leſs to a greater, and the point O ſought be required to lie between I and the

next point to it E;

~~or elſe to fall beyond A the other extreme. ~~
~~For the ſame ~~
conſtruction ſerves for both.

~~Here ~~
Case IV. ~~of ~~
Problem II. ~~is to be uſed, and ~~
the point O will fall between E and I, and o beyond A, if we uſe one of the

conſtructions there recited:

~~but if we uſe the other, the points will ſhift places, ~~
as was obſerved under that Caſe, viz.

~~O will fall beyond I the other way, and ~~
o between L and E.

~~
~~

Case III. ~~Let now the point O be ſought between A and E. ~~
~~Here ſet off ~~
the given ratio in ſuch a manner that EI may be the ſum of the terms, and

make uſe of the IIId

Case of Problem II. ~~and the ~~
Limitation here will be evident from the

Limitation there given, viz. ~~making EI: ~~
~~IL:~~
~~: AI: ~~
~~X, ~~
the

Limitation here is that X muſt not be leſs than IE + EL + √4 IEL*.~~
~~

Epitagma II. Case IV. ~~Here OI the line whoſe ſquare is concerned is ~~
to be bounded by I the middle point of the three given ones, and O or o, its

other bound is to be ſought between I and either extreine A or E.

~~the ſame ~~
conſtruction ſerving for both.

~~The given ratio muſt here be ſet off in ſuch a ~~
manner that EI may be the ſum of the terms of it;

~~and make uſe ~~
of Iſt

Case of the IId Problem; ~~with this caution, that of the two ſegments ~~
AI, IE, you choſe the leſſer IE whereon to exhibit the given ratio;

~~for then ~~
it will appear by the work itſelf that O falling between E and L, o will alſo

fall between A and I:

~~otherwiſe, if AI was leſs than IE, there would want ~~
ſome proof of this.

~~Therefore of the two extreme given points call that E ~~
which bounds the leſſer ſegment, and then the general Demonſtration will fit

this Caſe as well as the reſt.

~~
~~

Case V. ~~Let the given ratio of EL to LI be inæqualitatis minoris; ~~
~~and let ~~
the point ſought be required to lie beyond either extreme.

~~The ſame con-~~
ſtruction ſerves for both.

~~Here we muſt uſe the IVth ~~
Case of the IId Pro-

BLEM, and O being made to fall between E and L, o will fall always beyond A, provided we call that point E which bounds the bigger ſegment.

~~I have ~~
in the Figure made AI = IE on purpoſe to ſhew that in that caſe the point N

will coincide with A.

~~But if IE be greater than AI, the point N will ~~
always fall beyond A, and conſequently the point o more ſo.

~~
~~

Case VI. ~~Let the given ratio of EL to LI be inæqualitatis majoris, and ~~
let the point ſought be required to lie beyond either extreme.

~~Here we muſt ~~
uſe the IIId

Case of the IId Problem; ~~and the ~~
Determination is that UN (found in the ſame ratio to AI as IL is to IE) muſt not be leſs than

IE + EL + √4 IEL*.

~~
~~

Case VII. ~~Let the ſituation of O be required the ſame as in the two laſt ~~
Caſes, but let the given ratio be that of equality, which was there ſuppoſed

of inequality.

~~Here the IId ~~
Problem will be of no uſe, and this Caſe requires a particular conſtruction.

~~
~~

~~Let then the three Points be A, I, E, and I the middle one; ~~
~~and let it ~~
be required to find a fourth O beyond E, ſuch that AO x OE may equal

OI2.~~
~~

Construction. ~~Upon AE diameter deſcribe a circle,, and let another YS ~~
cut the former at right angles.

~~Join SI, and continue it to meet the circum-~~
ference in R.

~~From R draw a tangent to meet the given line in O, and I ~~
ſay O is the point required.

~~
~~

Demonstration. ~~Joining YR, the triangles SUI and SYR will be ſimi-~~
lar, and the angle UIS or RIO = SYR.

~~But the angle IRO made by the ~~
tangent and ſecant = SYR in the alternate ſegment.

~~Therefore RIO = IRO, ~~
and OR = OI.

~~But by the property of the circle AOE = ~~
OR2. ~~And ~~
therefore AOE =

OI2.~~
~~

~~Q. ~~
~~E. ~~
~~D.~~
~~
~~

~~The ~~
Determination is that AI muſt be greater than IE.~~
~~

Case VIII. ~~Whereas in the Iſt and IId ~~
Cases the given ratio was that of inequality, let us now ſuppoſe it that of equality;

~~and let the three points be ~~
A, E, I, and E the middle one;

~~and let a fourth O be ſought between E and ~~
I, ſuch that AOE may equal

OI2.~~
~~

~~The ~~
Construction and Demonstration of this Caſe is in every reſpect the ſame as that of the preceeding, as will appear by comparing the figures.

~~
~~

PROBLEM IV.
~~
~~
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~
~~
~~~~
~~~~
~~~~
~~

~~To cut a given indefinite right line ſo in one point that, of the four ſeg-~~
ments intercepted between the ſame and four points given in the indefinite

line, the rectangle under any two aſſigned ones may be to the rectangle under

the two remaining ones in a given ratio.

~~
~~

~~In the indefinite line let the four points be A, E, I, U. ~~
~~It is then required ~~
to be cut again in O ſo that OA x OU may be to OE x OI (be the Poſition

of the four given points what they may) in the ratio of AL to LE, let the

point L fall alſo as it may.

~~
~~

Construction. ~~To the three points E, L, U, and the right line UI, let ~~
be found by the IId

Problem a fourth point O, ſo that UI x OE may be to OU x OL as AE to AL.

~~And let ſuch a ~~
Case be choſen of the IId Problem that, according as UO is required greater or leſs than UI, or their ſum ſhall

conſtitute OI, ſo of the three rectangles deſcribed in the Vth

Lemma made by the four points E, O, A, L, that of OE x AL may accordingly be greater

or leſs than OL x AE, or their ſum conſtitute that of OA x EL.

~~
~~

~~N. ~~
~~B. ~~
~~U being now uſed to repreſent one of the given points, in all the ~~
following Diagrams I have ſubſtituted V in the place where U was uſed

before.

~~
~~

Demonstration. ~~On ſuppoſition therefore that ſuch a ~~
Case of the IId

Problem is made uſe of, We have UI x OE:

~~OU x OL:~~
~~: AE: ~~
~~AL~~

~~And by Inverſion OU x OL: ~~
~~UI x OE:~~
~~: AL: ~~
~~AE~~

~~And by ~~
Lemma IV. ~~AL x OE: ~~
~~AE x OL:~~
~~: OU: ~~
~~UI~~

~~Hence by Compoſition or Diviſion, &~~
~~c. ~~
~~AL x OE: ~~
~~OA x LE:~~
~~: OU ~~
:

~~OI as appears by ~~
Lemma V.~~
~~

~~Then again by ~~
Lemma IV. ~~OU x OA: ~~
~~OI x OE:~~
~~: AL: ~~
~~LE~~

~~Q. ~~
~~E. ~~
~~D.~~
~~
~~

~~This ~~
Problem has three Epitagmas. ~~The Iſt whereof is when of the ~~
two aſſigned points A and U, the one of them is an extreme, and the other an

alternate mean;

~~and this admits of three ~~
Cases. ~~The IId is when A and U ~~
are both of them extremes;

~~and this has four ~~
Cases. ~~The IIId is when of ~~
A and U one of them is an extreme, and the other is the point next to it;

~~and ~~
this has three Caſes.

~~And there remain three more Anomalous Caſes, wherein ~~
the IId

Problem is of no uſe, but which may be reduced to one, as ſhall be ſhewn in it's proper place.

~~
~~

Epitagma I. Case I. ~~Let A the firſt aſſigned point be an extreme, and ~~
U the ſecond aſſigned point be an alternate mean;

~~and let the point O be ~~
ſought between the firſt aſſigned A and the next point to it E;

~~or between ~~
the ſecond aſſigned U and the laſt I.

~~For the ſame Conſtruction ſerves ~~
for both.

~~
~~

~~Here AE is to be made the ſum of the terms of the given ratio, and we are ~~
to uſe the IVth

Case of the IId Problem, whereby O falling between L and E, o will fall beyond U;

~~and that it will fall ſhort of I appears from ~~
the Iſt

Corollary from the IVth Case of the IId Problem.~~
~~

Case II. ~~The given ratio being inæqualitatis minoris, let the point ſought ~~
be required between the ſecond aſſigned U and the ſecond in order E, or be-

yond the firſt A, which ariſes from the ſame Conſtruction.

~~Here AE is to be ~~
made the difference of the terms of the given ratio, and we are to uſe the

IVth

Case of the IId Problem, where O being made to fall between U and E, o will fall beyond L, much more beyond A.

~~
~~

Case III. ~~The given ratio being inæqualitatis majoris, let the point ſought ~~
be required between the ſecond aſſigned U and the ſecond in order E, or be-

yond the laſt I, which ariſes from the ſame Conſtruction.

~~Here AE is to be ~~
made the difference of the terms of the given ratio, and L is to be ſet off the

contrary way to what it was in the laſt

Case; ~~and we are to uſe the Iſt ~~
Case of the IId

Problem, whereby O being made to fall between E and L, or between E and U, according as L or U is neareſt to the point E, o will fall

beyond I, as any one will ſee who conſiders the Conſtruction of that

Case with due attention.

~~
~~

Epitagma II. Case IV. ~~Let the aſſigned points now be the extremes ~~
A and U, and let O the point ſought be required now between the firſt

aſſigned A and the next to it E, or, which is effected by the ſame Conſtruction,

between the ſecond aſſigned U and the next to it I.

~~Here AE is to be made ~~
the ſum of the terms of the given ratio, and the IVth

Case of the IId Pro-

BLEM is to be uſed, ſo that of the three points L, E, U, O being made to fall beyond L one of the extremes, and o within U the other extreme, it will appear

from the Iſt

Corollary from the IVth Case of the ſaid Problem that O will fall between A and E, and o between U and I.

~~
~~

Case V. ~~The given ratio being inæqualitatis majoris, let the point ſought be ~~
required between the ſecond and third in order, viz.

~~between E and I. ~~
~~Here ~~
AE muſt be the difference of the terms of the given ratio, and L ſet off to-

wards I, and the IId

Case of the IId Problem uſed, and then O, as like-wiſe o, will fall between E and I, if the

Problem be poſſible.~~
~~

~~As to the ~~
Determination, ſee Lemma VII. ~~following.~~
~~
~~

Case VI. ~~The given ratio being inæqualitatis majoris, let the point ſought ~~
be required beyond the laſt aſſigned, that is the laſt in order, U.

~~Here AE ~~
muſt be the difference of the terms of the given ratio, [and L muſt evidently

~~
fall beyond U, but for a more particular ~~
Determination ſee Lemma VII. ~~following] and the IId ~~
Case of the IId Problem is to be uſed, and then O, as likewiſe o, will fall beyond U.

~~
~~

Case VII. ~~The given ratio being inæqualitatis minoris, let the point ſought ~~
be required to lie beyond either of the aſſigned ones, i.

~~e. ~~
~~beyond either ex-~~
treme, the ſame Conſtruction ſerving for both.

~~Here AE is to be the diſ-~~
ference of the terms of the given ratio, and L to be ſet off backwards beyond

A;

~~and the IVth ~~
Case of the IId Problem uſed, that ſo O being made to fall beyond U, it will appear, by the IId

Corollary from the IVth Case of the ſaid

Problem, that o will alſo fall beyond A.~~
~~

Epitagma III. Case VIII. ~~Let the aſſigned points A and U be now ~~
one an extreme, and the other the point next it:

~~and let the point ſought be re-~~
quired to fall between the two aſſigned ones.

~~Here AE muſt be the ſum of ~~
the terms of the given ratio, and the IId

Case of the IId Problem uſed. ~~And ſo O, as likewiſe o, being made to fall between L and U, they will ~~
much more fall between A and U.

~~
~~

~~The ~~
Limitation is, that VN (found in the ſame ratio to UI as AL to AE) muſt not exceed LE + EU - √4 LEU*.

~~
~~

Case IX. ~~The given ratio being inæqualitatis minoris, let the point ſought ~~
be required between the ſecond aſſigned U and the third in order E, or elſe

beyond the firſt aſſigned A, the ſame Conſtruction ſerving for both.

~~Here AE ~~
is to be the difference of the terms of the given ratio, and L to be ſet off be-

yond A, and the Iſt

Case of the IId Problem uſed: ~~and ſo O being made ~~
to fall between E and U, o will fall beyond L, and much more be-

yond A.

~~
~~

Case X. ~~The given ratio being inæqualitatis majoris, let the point ſought ~~
be required between the ſecond aſſigned U and the third in order E, or elſe

beyond the laſt in order I, the ſame conſtruction ſerving for both.

~~Here AE ~~
is to be the difference of the terms of the given ratio, and L to be ſet off be-

yond E, and the IVth

Case of the IId Problem uſed, ſo that O being made to fall between U and E, o will fall beyond I, as any one will ſee who

conſiders the conſtruction of that

Case with due attention.~~
~~

Case XI. ~~As to the three Anomalous Caſes, in which the IId ~~
Problem is of no uſe, and which I ſaid before might be reduced to one, they are theſe:

~~whereas in the IId and IIId ~~
Cases, as alſo in the VIth and VIIth, and like-wiſe in the IXth and Xth the ratio given was that of inequality, let us now

~~
ſuppoſe it that of equality, and they may all be ſolved by one conſtruction, viz. ~~
~~the rectangle AOU made equal to the rectangle EOI.~~
~~
~~

~~Let it be made as UI: ~~
~~AE:~~
~~: UO: ~~
~~EO~~

~~Then by permutation UO: ~~
~~UI:~~
~~: EO: ~~
~~AE~~

~~And by comp. ~~
~~or diviſ. ~~
~~UO: ~~
~~OI:~~
~~: EO: ~~
~~AO~~

~~Hence AOU = EOI.~~
~~
~~

Lemma VI. ~~Let there be two ſimilar triangles IAE, UAO, having their ~~
baſes IE and UO parallel;

~~I ſay Iſt when they are right-angled, that the ex-~~
ceſs of the rectangle EAO, under the greater ſides of each, above the rect-

angle IAU, under the leſſer ſides of each, will be equal to the rectangle

IE x OU, under their baſes.

~~IIdly, When they are obtuſe-angled, that the ~~
ſaid exceſs will be equal to the rectangle under the baſe of one and the ſum

of the diſtances of the angles at the baſe of the other from the perpendicular,

viz.

~~EI x ~~
OS + US. ~~IIIdly, When they are acute-angled, that then the ſaid ~~
exceſs will be equal to the rectangle under the baſe of one and the difference

of the ſegments of the baſe of the other made by the perpendicular, viz.

~~OU x EL.~~
~~
~~

Demonstration. ~~Since EA: ~~
~~AO:~~
~~: IA: ~~
~~AU:~~
~~: EI: ~~
~~OU, the rect-~~
angles EAO, IAU, and EI x OU will be ſimilar, and when Iſt the triangles

are right-angled EAO = IAU + EI x OU by Euc.

~~VI. ~~
~~19. ~~
~~and I. ~~
~~47. ~~
~~But if ~~
they be oblique-angled, draw the perpendicular YAS.

~~Then IIdly, in caſe ~~
they be obtuſe-angled, EAO = YAS + EY x OS by part Iſt;

~~and IAU = ~~
YAS + IY x US by the ſame.

~~And therefore EAO - IAU = EY x OS -~~
IY x US =

EY - IY or EI x OS + US. ~~But if IIIdly they be acute-angled, ~~
and EY be greater than IY, then from Y ſet off YL = YI, and draw LAR

which will be equal and ſimilarly divided to IAU.

~~Then by part IId EAO ~~
- LAR, i.

~~e. ~~
~~EAO - IAU = EL x ~~
OS + RS = EL x OU.~~
~~

~~Q. ~~
~~E. ~~
~~D.~~
~~
~~

Lemma VII. ~~If a right line VY, joining the tops of two perpendiculars ~~
drawn from two points of the diameter of a circle E and I to the circum-

ference on oppoſite ſides of the diameter, cut the ſaid diameter in O, and

A and U be the extremes of the ſaid diameter, I ſay that the ratio of the

rectangle AOU to the rectangle EOI is a Minimum.

~~
~~

~~But if VY joins the tops of two perpendiculars from E and I drawn on ~~
the ſame ſide of the diameter, and conſequently meets the diameter produced

in O, that then the ratio of AOU to EOI is a Maximum.

~~
~~

Demonstration. ~~Through S, any other point taken at pleaſure, draw ~~
LSM parallel to VOY, and join VS and produce it to meet the perpendi-

cular in N and the circumference in R.

~~Produce alfo the perpendicular YI ~~
to meet the circumference again in F, and join RF;

~~then from ſrmilar tri-~~
angles it appears that the rectangle LSM:

~~ESI:~~
~~: VOY i. ~~
~~e. ~~
~~AOU: ~~
~~EOI. ~~
~~But the rectangle ASU i. ~~
~~e. ~~
~~VSR is greater than LSM. ~~
~~(for LV i. ~~
~~e. ~~
~~MY x ~~

NI + MI together with VS x SN is by the preceeding Lemma equal to LSM. But ASU or VSR is equal to VS x SN, together with VS x NR;

~~for which ~~
laſt rectangle we may ſubſtitute MY x NF:

~~for the triangles VLS and NRF ~~
are ſimilar, being each of them ſimilar to VNY;

~~therefore VL or MY: ~~
~~VS ~~
:

~~: NR: ~~
~~NF and MY x NF = VS x NR. ~~
~~Now NF being always greater ~~
than

NI + MI, it appears from thence that ASU is greater than LSM.) Therefore the ratio of ASU to ESI is greater than that of AOU to EOI.

And the ſame holds good with regard to any other point S taken between E

and I, ſo that the ratio of AOU to EOI is a Minimum, and ſingular, or

what the Antients called μοναχ@.

~~
~~

~~Let now VY join the tops of two perpendiculars drawn on the ſame ſide of ~~
the diameter, and meet the diameter produced in O;

~~I ſay that the ratio of ~~
AOU to EOI is a Maximum.

~~
~~

~~For uſing the ſame conſtruction as before, it will appear that the rectangle ~~
LSM:

~~ESI:~~
~~: VOY or AOU: ~~
~~EOI. ~~
~~And it may be proved in the ſame ~~
manner that VSR or ASU is leſs than LSM.

~~(LV i. ~~
~~e. ~~
~~MY x ~~
NI + MI or (making IK = MI) MY x NK, together with LSM is by the preceding

Lem-

MA equal to VS x SN. ~~But VS x NR, together with VSR, is alſo equal to ~~
VS x SN.

~~Now VS x NR is equal to MY x NF or LV x NF from the ſimi-~~
larity of the triangles LVS, NRF.

~~Therefore now alſo MY x NF together ~~
with VSR is proved equal to VS x SN.

~~But as NK is leſs than NF, VSR ~~
will be a leſs rectangle than LSM) Hence the ratio of LSM to ESI or it's

equal AOU to EOI is greater than the ratio of VSR or ASU to ESI.

~~And ~~
the ſame holds with regard to any other point taken in the diameter pro-

duced.

~~Therefore the ratio of AOU to EOI is a Maximum.~~
~~
~~

~~Q. ~~
~~E. ~~
~~D.~~
~~
~~

DETERMINATE SECTION.
BOOK I.
PROBLEM I. (Fig. 1.)
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~

~~In any indefinite ſtraight line, let the Point A be aſſigned; ~~
~~it is required ~~
to cut it in ſome other point O, ſo that the ſquare on the ſegment AO

may be to the ſquare on a given line, P, in the ratio of two given ſtraight

lines R and S.

~~
~~

Analysis. ~~Since, by Hypotheſis, the ſquare on AO muſt be to the ~~
ſquare on P as R is to S, the ſquare on AO will be to the Square on P as

the ſquare on R is to the rectangle contained by R and S (

Eu. ~~V. ~~
~~15.) ~~
~~Let there be taken AD, a mean proportional between AB (R) and AC ~~
(S);

~~then the Square on AO is to the ſquare on P as the ſquare on R is ~~
to the ſquare on AD, or (

Eu. ~~VI. ~~
~~22) AO is to P as R to AD; ~~
~~conſe-~~
quently, AO is given by

Eu. ~~VI. ~~
~~12.~~
~~
~~

Synthesis. ~~Make AB equal to R, AC equal to S, and deſcribe on ~~
BC a ſemi-circle;

~~erect at A the indefinite perpendicular AF, meeting the ~~
circle in D, and take AF equal to P;

~~draw DB, and parallel thereto FO, ~~
meeting the indefinite line in O, the point required.

~~
~~

~~For, by reaſon of the ſimilar triangles ADB, AFO, AO is to AF (P) as ~~
AB (R) is to AD;

~~therefore (~~
Eu. ~~VI. ~~
~~22.) ~~
~~the ſquare on AO is to the ~~
ſquare on P as the ſquare on R is to the ſquare on AD;

~~but the ſquare on ~~
AD is equal to the rectangle contained by AB (R) and AC (S) by

Eu. ~~VI. ~~
~~13. ~~
~~17; ~~
~~and ſo the ſquare on AO is to the ſquare on P as the ſquare on R ~~
is to the rectangle contained by R and S;

~~that is (~~
Eu. ~~V. ~~
~~15.) ~~
~~as R is to S.~~
~~
~~

~~Q. ~~
~~E. ~~
~~D.~~
~~
~~

Scholium. ~~Here are no limitations, nor any precautions whatever to be ~~
obſerved, except that AB (R) muſt be ſet off from A that way which O

is required to fall.

~~
~~

PROBLEM II. (Fig. 2 and 3.)
~~
~~~~
~~~~
~~~~
~~~~
~~

~~In any indefinite ſtraight line, let there be aſſigned the points A and E; ~~
~~it is required to cut it in another point O, ſo that the rectangle contained ~~
by the ſegments AO, EO may be to the ſquare on a given line P, in the

ratio of two given ſtraight lines R and S.

~~
~~

Analysis. ~~Conceive the thing done, and O the point ſought: ~~
~~then ~~
would the rectangle AO, EO be to the ſquare on the given line P as R is

to S, by hypotheſis.

~~Make AQ to P as R is to S: ~~
~~then the rectangle AO, ~~
EO will be to the ſquare on P as AQ to P;

~~or (~~
Eu. ~~V. ~~
~~15. ~~
~~and 16) the ~~
rectangle AO, EO will be to the rectangle AQ, EO as the ſquare on P is

to the rectangle EO, P;

~~and therefore AO is to AQ as P is to EO; ~~
~~con-~~
ſequently (

Eu. ~~VI. ~~
~~16.) ~~
~~the rectangle AO, EO is equal to the rectangle ~~
AQ, P;

~~and hence, as the ſum, or difference of AO and EO is alſo given, ~~
theſe lines themſelves are given by the 85th or 86th of the Data.

~~
~~

Synthesis. ~~On AE deſcribe a circle, and erect at A the indefinite ~~
perpendicular AK;

~~and, having taken AQ a fourth proportional to S, R ~~
and P, take AD a mean proportional between AQ and P;

~~from D draw ~~
DH, parallel to AE if O be required to fall between A and E;

~~but through ~~
F, the center of the Circle on AE, if it be required beyond A or E, cutting

the circle in H;

~~laſtly, draw HO perpendicular to DH, meeting the inde-~~
finite line in O, the point required.

~~
~~

~~For it is manifeſt from the conſtruction that AD and HO are equal; ~~
~~hence, and ~~
Eu. ~~VI. ~~
~~17, the rectangle AQ, P is equal to the ſquare on HO; ~~
conſequently equal to the rectangle AO, EO (

Eu. ~~III. ~~
~~35. ~~
~~36) and ſo ~~
(

Eu. ~~VI. ~~
~~16.) ~~
~~AO is to P as AQ is to EO; ~~
~~but by conſtruction AQ is to ~~
P as R to S, therefore by compound ratio, the rectangle AO, AQ is to the

ſquare on P as the rectangle AQ, R is to the rectangle EO, S:

~~hence ~~
(

Eu. ~~V. ~~
~~15. ~~
~~16.) ~~
~~the rectangle AO, EO is to the ſquare on P as the rect-~~
angle EO, R is to the rectangle EO, S, that is, as R is to S.

~~Q. ~~
~~E. ~~
~~D.~~
~~
~~

Scholium. ~~This Problem has three Epitagmas. ~~
~~Firſt, when O is ſought ~~
beyond A;

~~ſecondly, when it is ſought between A and E, and laſtly, when ~~
it is ſought beyond E.

~~The firſt and laſt of theſe are conſtructed by ~~
Fig.

~~2, and have no limitations; ~~
~~but in the ſecond, (Fig. ~~
~~3.) ~~
~~the given ~~
ratio of R to S muſt not be greater than that which the ſquare on half

AE bears to the ſquare on P:

~~ſince if it be, a third proportional to S
~~
~~
and P will be greater than one to R and half AE, and of courſe, AQ ~~
(a fourth proportional to S, R and P) greater than a third proportional

to P and half AE;

~~in which caſe the rectangle AQ, P will be greater ~~
than the ſquare on half AE, and ſo AD (a mean proportional between

AQ and P) greater than half AE;

~~but when this happens, it is plain that ~~
DH can neither cut nor touch the circle on AE, and therefore, the

problem becomes impoſſible.

~~
~~

PROBLEM III. (Fig. 4. and 5.)
~~
~~~~
~~~~
~~~~
~~~~
~~

~~In any indefinite ſtraight line let there be aſſigned the points A and E; ~~
~~it is required to cut it in another point O, ſo that the ſquare on the ſegment ~~
AO may be to the rectangle contained by the ſegment EO and a given line

P, in the ratio of two given ſtraight lines R and S.

~~
~~

Analysis. ~~Suppoſe the thing done, and that O is the point ſought: ~~
~~then will the ſquare on AO be to the rectangle EO, P as R to S. ~~
~~Make ~~
AQ to P as R is to S;

~~then will the ſquare on AO be to the rectangle EO, ~~
P as AQ is to P;

~~or (~~
Eu. ~~V. ~~
~~15.) ~~
~~the ſquare on AO is to the rectangle EO, ~~
P as the rectangle AQ, AO is to the rectangle P, AO;

~~wherefore AO is to ~~
EO as AQ to AO;

~~conſequently by compoſition, or diviſion, AO is to AE ~~
as AQ is to OQ, and ſo (

Eu. ~~VI. ~~
~~16.) ~~
~~the rectangle AO, OQ is equal to ~~
the rectangle AE, AQ;

~~and hence, as the ſum or difference of AO and OQ ~~
is alſo given, theſe lines themſelves are given by the 85th or 86th of the

Data.

~~
~~

Synthesis. ~~Take AQ a fourth proportional to S, P and R, and ~~
deſcribe thereon a circle;

~~erect at A, the indefinite perpendicular AK, and ~~
take therein AD, a mean proportional between AE and AQ;

~~from D, ~~
draw DH, parallel to AE, if O be required beyond E;

~~but through F the ~~
center of the circle on AQ, if it be ſought beyond A, or between A and

E, cutting the ſaid circle in H:

~~Laſtly, from H draw HO perpendicular ~~
to DH, which will cut the indefinite line in O, the point required.

~~
~~

~~For it is plain from the Conſtruction, that AD and HO are equal; ~~
~~and ~~
(

Eu. ~~VI. ~~
~~17) the rectangle AE, AQ is equal to the ſquare on AD, and ~~
therefore equal to the ſquare on HO;

~~but the ſquare on HO is equal to the ~~
rectangle AO, OQ, (

Eu. ~~III. ~~
~~35. ~~
~~36) conſequently the rectangle AO, OQ
~~
~~
is equal to the rectangle AE, AQ; ~~
~~and hence (~~
Eu. ~~VI. ~~
~~16.) ~~
~~OQ is to AQ ~~
as AE is to AO;

~~therefore, by compoſition or diviſion, AO is to AQ as ~~
EO is to AO;

~~but by conſtruction, AQ is to R as P is to S, and ſo, by ~~
compound ratio, the rectangle AO, AQ is to the rectangle R, AQ as the

rectangle EO, P is to the rectangle AO, S;

~~or (~~
Eu. ~~V. ~~
~~15. ~~
~~and 16.) ~~
~~the ſquare ~~
on AO is to the rectangle EO, P as the rectangle AO, R is to the rectangle

AO, S;

~~that is, as R to S.~~
~~
~~

Scholium. ~~This Problem hath three Epitagmas alſo, which I ſtill enu-~~
merate as before.

~~The firſt and ſecond are conſtructed by Fig. ~~
~~4, where DH ~~
is drawn through F, the center of the circle on AQ:

~~and theſe have no limi-~~
tations.

~~The third is conſtructed as in Fig. ~~
~~5, where DH is drawn parallel ~~
to AQ;

~~and here the given ratio of R to S muſt not be leſs than the ratio ~~
which four times AE bears to P:

~~for if it be, AE will be greater than ~~
one-fourth Part of AQ (a fourth proportional to S, R and P) in which

caſe the rectangle contained by AE and AQ will be greater than the ſquare

on half AQ, and conſequently AD (a mean proportional between AE and

AQ) greater than half AQ;

~~but it is plain when this is the caſe, that DH ~~
will neither cut nor touch the circle on AQ, and therefore the problem is

impoſſible.

~~
~~

PROBLEM IV. (Fig. 6. 7. and 8.)
~~
~~~~
~~
~~
~~~~
~~~~
~~~~
~~

~~In any indefinite ſtraight line, let there be aſſigned the points A and E; ~~
~~it is required to cut it in another point O, ſo that the two ſquares on the ~~
ſegments AO, EO, may obtain the Ratio of two given ſtraight lines,

R and S.

~~
~~

Analysis. ~~Imagine the thing to be effected, and that O is really the ~~
point required:

~~then will the ſquare on AO be to the ſquare on EO as ~~
R to S;

~~or (~~
Eu. ~~V. ~~
~~15.) ~~
~~the ſquare on AO will be to the ſquare on EO ~~
as the ſquare on R is to the rectangle contained by R and S.

~~Let DE be ~~
made a mean proportional between EB (R) and EC (S).

~~Then ~~
(

Eu. ~~VI. ~~
~~17.) ~~
~~the ſquare on AO will be to the ſquare on EO as the ~~
ſquare on R to the ſquare on DE;

~~and ſo (~~
Eu. ~~VI. ~~
~~22.) ~~
~~AO to EO as R to ~~
DE;

~~and hence both AO and EO will be given by the converſe of Prop. ~~
~~38. ~~
~~of Eu. ~~
~~Data.~~
~~
~~

Synthesis. ~~Make EB equal to R, EC equal to S, and deſcribe on ~~
BC a circle;

~~erect at E the perpendicular ED, meeting the periphery of the ~~
circle in D;

~~alſo at A erect the perpendicular AF equal to R; ~~
~~draw AD, ~~
which produce, if neceſſary, to cut the indefinite line, as in O, which will

be the point required.

~~
~~

~~For becauſe of the ſimilar triangles AOF, EOD, AO is to EO as AF ~~
(R) is to DE;

~~therefore the ſquare on AO is to the ſquare on EO as the ~~
ſquare on R is to the ſquare on DE (

Eu. ~~VI. ~~
~~22); ~~
~~but the ſquare on DE ~~
is equal to the rectangle contained by R and S;

~~therefore the ſquare on AO is ~~
to the ſquare on EO as the ſquare on R is to the rectangle R, S;

~~that is as ~~
R is to S, by

Eu. ~~V. ~~
~~15.~~
~~
~~

~~Q. ~~
~~E. ~~
~~D.~~
~~
~~

Scholium. ~~This Problem alſo hath three Epitagmas, which I enumerate ~~
as in the laſt.

~~The firſt is conſtructed by Fig. ~~
~~6, wherein the perpendicu-~~
lars DE and AF are ſet off on the ſame ſide of the indefinite line;

~~the ſecond ~~
by Fig.

~~7, where they are ſet off on contrary ſides, and the third by Fig. ~~
~~8, ~~
in which they are again ſet off on the ſame ſide.

~~The ſecond has no limits; ~~
~~but in the firſt R muſt be leſs, and in the third greater than S, for reaſons ~~
too obvious to be inſiſted on;

~~and hence, both theſe caſes are impoſſible ~~
when the given ratio is that of equality.

~~
~~

PROBLEM V. (Fig. 9. 10. 11. 12. 13. 14. 15. 16.)
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~

~~In any indefinite ſtraight line let there be aſſigned the points A, E and I; ~~
~~it is required to cut it in another point, O, ſo that the rectangle contained ~~
by the ſegment AO and a given ſtraight line P may be to the rectangle

contained by the ſegments EO, IO in the ratio of two given ſtraight lines

R and S.

~~
~~

Analysis. ~~Conceive the thing done, and O the point ſought: ~~
~~then ~~
would the rectangle AO, P be to the rectangle EO, IO as R to S.

~~Make ~~
IQ to P as S is to R;

~~then will the rectangle AO, P be to the rectangle ~~
EO, IO as P is to IQ;

~~or (~~
Eu. ~~V. ~~
~~15.) ~~
~~the rectangle AO, P be to the ~~
rectangle EO, IO as the rectangle IO, P is to the rectangle IO, IQ;

~~and ~~
hence (

Eu. ~~V. ~~
~~15. ~~
~~16) AO is to EO as IO is to IQ; ~~
~~whence, by compoſition ~~
or diviſion, AE is to EO as OQ is to IQ:

~~therefore (~~
Eu. ~~VI. ~~
~~16.) ~~
~~the
~~
~~
rectangle EO, OQ is equal to the rectangle AE, IQ; ~~
~~conſequently, as the ~~
ſum or difference of EO and OQ is alſo given, thoſe lines themſelves are

given by the 85th or 86th of the Data.

~~
~~

Synthesis. ~~Take IQ a fourth proportional to R, S and P, and ~~
deſcribe on EQ a circle;

~~erect at E the indefinite perpendicular EK, and ~~
take therein ED a mean proportional between AE and IQ;

~~from D draw ~~
DH, parallel to EQ, if O muſt lie any where between the points E and Q;

~~but through F, the center of the circle on EQ if it muſt fall without them, ~~
cutting the ſaid circle in H:

~~Laſtly, draw HO perpendicular to DH, which ~~
will meet the indeſinite line in O, the point required.

~~
~~

~~For it is manifeſt from the conſtruction that ED and HO are equal; ~~
~~and ~~
(

Eu. ~~VI. ~~
~~17.) ~~
~~the rectangle AE, IQ is equal to the ſquare on ED, and ~~
therefore equal to the ſquare on HO;

~~but the ſquare on HO is equal to the ~~
rectangle EO, OQ (

Eu. ~~III. ~~
~~35. ~~
~~36.)~~
~~: ~~
~~therefore the rectangle AE, IQ is ~~
equal to the rectangle EO, OQ;

~~and hence (~~
Eu. ~~VI. ~~
~~16.) ~~
~~AE is to OE as OQ ~~
to IQ, whence, by compoſition or diviſion, AO is to EO as OI to IQ;

~~but ~~
IQ is to P as S to R, or inverſely, P is to IQ as R to S;

~~and ſo, by compound ~~
ratio, the rectangle AO, P is to the rectangle EO, IQ as the rectangle IO,

R is to the rectangle IQ, S;

~~that is (~~
Eu. ~~V. ~~
~~15 and 16.) ~~
~~the rectangle AO, ~~
P is to the rectangle IO, R as EO is to S;

~~or the rectangle AO, P is to the ~~
rectangle IO, R as the rectangle IO, EO is to the rectangle IO, S

(

Eu. ~~V. ~~
~~16.) ~~
~~the rectangle AO, P is to the rectangle EO, IO as the rectangle ~~
IO, R is to the rectangle IO, S;

~~that is (~~
Eu. ~~V. ~~
~~15.) ~~
~~as R is to S. ~~
~~Q. ~~
~~E. ~~
~~D.~~
~~
~~

Scholium. ~~This Problem may be conſidered as having three Epitagmas, ~~
or general Caſes, viz.

~~when A, the point which bounds the ſegment aſſigned ~~
for the co efficient of the given line P being an extreme, O is ſought be-

tween it and the next thereto, or beyond all the points with reſpect to A;

~~ſecondly, where A is the middle point; ~~
~~and thirdly, when A being again ~~
an extreme, O is ſought beyond it, or between the other two points E and

I:

~~and each of theſe is ſubdiviſible into four more particular ones.~~
~~
~~

Epitagma I. ~~Here the four Caſes are when E being the middle point, ~~
O is required between A and E, or beyond I;

~~and theſe are both con-~~
ſtructed at once by Fig.

~~9: ~~
~~when I is the middle point and O ſought between ~~
A and I or beyond E;

~~and theſe are both conſtructed at once by Fig. ~~
~~10. ~~
~~and in both of theſe IQ is ſet off from I contrary to that direction which A
~~
~~
bears therefrom, and DH drawn through F, the center of the circle on EQ: ~~
~~none of theſe Caſes are ſubject to any Limitations.~~
~~
~~

Epitagma II. ~~Wherein A is the middle point, and the Caſes, when O ~~
is ſought beyond E, between E and A, between A and I or beyond I.

~~The ~~
firſt and third of which are conſtructed at once by Fig.

~~11, wherein IQ is ~~
ſet off from I towards A and DH drawn through F, the center of the circle

on EQ.

~~The ſecond and fourth are conſtructed at once, alſo, by Fig. ~~
~~12. ~~
~~where IQ is ſet off from I the contrary way to that which A lies, and DH ~~
drawn parallel to EQ.

~~There are no Limitations to any of theſe Caſes.~~
~~
~~

Epitagma III. ~~Here, E being the middle point, the Caſes are, when O ~~
muſt lie beyond A, or between E and I;

~~and the ſame Caſes occur when ~~
I is made the middle point.

~~The firſt is conſtructed by Fig. ~~
~~13, the ſecond ~~
by Fig.

~~14, the third by Fig. ~~
~~15, and the fourth by Fig. ~~
~~16: ~~
~~in every one ~~
of which IQ is ſet off from I towards A, and DH drawn parallel to EQ.

~~The Limits are that the given ratio of R to S, muſt not be leſs than the ratio ~~
which the rectangle AE, P bears to the ſquare on half the Sum, or half the

difference of AE, and a fourth propor tional to R, S and P;

~~that is, to the ~~
ſquare on half EQ:

~~ſince if it ſhould, the rectangle contained by AE and ~~
the ſaid fourth proportional will be greater than the ſquare on half EQ;

and of courſe ED (a mean proportional between them) greater than half

EQ, in which Caſe DH can neither cut nor touch the circle on EQ, and

ſo the problem be impoſſible.

~~It is farther obſervable in the two laſt caſes, ~~
that to have the former of them poſſible, AE muſt be leſs, and to have

the latter poſſible, EI muſt be greater than the above-mentioned half

ſum;

~~for if this latter part of the Limitation be not obſerved, theſe caſes ~~
are changed into one another.

~~
~~

PROBLEM VI. (Fig. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.)
~~
~~
~~
~~~~
~~~~
~~~~
~~
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~See Prop. A. Book V. of Dr. Simſon’s Euclid.

~~In any indefinite ſtraight line let there be aſſigned the points A, E and I; ~~
~~it is required to cut it in another point O, ſo that the rectangle contained ~~
by the ſegments AO, EO may be to the ſquare on IO in the ratio of two

given ſtraight lines, R and S.

~~
~~

Analysis. ~~Let us conceive the thing effected, and that O is really the ~~
point ſought.

~~Then, by ſuppoſition, the rectangle AO, EO is to the ~~
ſquare on IO as R to S.

~~Make EC to IC as R is to S; ~~
~~and the rectangle ~~
AO, EO is to the ſquare on IO as EC to IC.

~~Let now OB be taken a ~~
fourth proportional to EO, EC and IO;

~~then (~~
Eu. ~~V. ~~
~~15.) ~~
~~the rectangle ~~
AO, EO is to the ſquare on IO as the rectangle EC, OB is to the rectangle

IC, OB;

~~and ſo by permutation, the rectangle AO, EO is to the rectangle ~~
EC, OB as the ſquare on IO is to the rectangle IC, OB;

~~and becauſe EO is ~~
to EC as IO to BO, AO will be to OB as IO to IC, and ſo by compoſition,

or diviſion CO is to EC as IB to OB, and AB is to OB as CO to IC;

~~whence, ex æquo perturb. ~~
~~et permut. ~~
~~AB is to IB as EC to IC; ~~
~~that is in the ~~
given ratio, and hence is given BC, the ſum or difference of CO and BO,

as alſo the rectangle contained by them, equal to the rectangle AB, IC,

wherefore theſe lines themſelves are given by the 85th or 86th of the Data.

~~
~~

Synthesis. ~~Make AB to IB and EC to IC in the given ratio, and ~~
deſcribe on BC a circle;

~~erect, at B, the indefinite perpendicular BK, and ~~
take therein BD a mean proportional between AB and IC, or between IB

and EC:

~~from D draw DH parallel to CB, if O muſt fall between B and C; ~~
~~but through F, the center of the circle on BC, if it muſt fall without them, ~~
cutting the ſ@id circle in H;

~~then draw HO perpendicular to DH, which ~~
will cut the indefinite line in O, the point required.

~~
~~

~~For it is plain from the conſtruction that BD and HO are equal, and ~~
(

Eu. ~~IV. ~~
~~17.) ~~
~~the rectangle AB, IC, or the rectangle IB, EC is equal to the ~~
ſquare on BD, and therefore equal to the ſquare on HO, which (

Eu. ~~III. ~~
~~35. ~~
~~36.) ~~
~~is equal to the rectangle BO, CO: ~~
~~conſequently (~~
Eu. ~~VI. ~~
~~16.) ~~
~~AB ~~
is to BO as CO to IC;

~~alſo EC is to CO as BO is to IB; ~~
~~wherefore, by ~~
compoſition or diviſion, AO is to BO as IO to IC, and EO to EC as IO

to BO:

~~conſequently by compound ratio, the rectangle contained by AO and ~~
EO is to the rectangle contained by BO and EC, as the ſquare on IO is to

the rectangle contained by BO and IC;

~~by permutation, the rectangle ~~
contained by AO and EO is to the ſquare on IO as the rectangle contained

by BO and EC is to the rectangle contained by BO and IC, that is (Euc.

v.

~~15.) ~~
~~as EC is to IC, or as R to S.~~
~~
~~

~~Q. ~~
~~E. ~~
~~D.~~
~~
~~

Scholium. ~~This Problem alſo hath three Epitagmas; ~~
~~ſirſt, when O is ~~
ſought between I, the point which bounds the ſegment whoſe ſquare is

concerned, and either of the other given ones;

~~ſecondly, the ſaid point ~~
being an extreme one, when O is ſought beyond it, or beyond both the

other given points with reſpect to it;

~~and thirdly, when O is required be-~~
yond the next in order to the abovementioned point I:

~~theſe are each of ~~
them ſubdiviſible into other more particular caſes.

~~
~~

Epitagma I. ~~Here O is ſought between I, the point which bounds the ~~
ſegment whoſe ſquare is concerned, and the next in order to it:

~~and there ~~
are four caſes, viz.

~~when I is an extreme point; ~~
~~and the given ratio of R ~~
to S the ratio of a greater to a leſs;

~~when I, remaining as before, the given ~~
ratio is of a leſs to a greater;

~~again when I is the middle point, and O ~~
ſought between it and either of the other given ones.

~~
~~

Case I. ~~Here the points B and C are both made to fall beyond I ~~
(Fig.

~~17.) ~~
~~and DH is drawn through the center of the circle on BC, and O ~~
will fall between the point I, and the next in order thereto;

~~becauſe by ~~
conſtruction, EC is to CO as BO is to IB, and therefore when EC is greater

than CO, BO will be greater than IB, and when leſs, leſs;

~~but it is plain ~~
that if O ſhould fall either beyond E or I, this could not be the Caſe.

~~It ~~
is farther manifeſt that ſhould the points A and E change places, the con-

ſtruction would be no otherwiſe altered than that AB would then be greater

than IC.

~~
~~

Case II. ~~If the given points retain their poſition, but the ratio be made ~~
of a leſs to a greater, the conſtruction will then be by Fig.

~~18, where B ~~
muſt be made to fall beyond A, and C beyond E with reſpect to I;

~~but ~~
DH is ſtill drawn through the center of the circle on BC:

~~and that O will ~~
fall as required may be made appear by reaſonings ſimilar to thoſe uſed in

Caſe I.

~~Moreover no change will enſue in the conſtruction when the Points ~~
A and E change places, except that B and C will change ſituations alſo.

~~
~~

Cases III and IV, Are conſtructed at once by Fig. ~~19, when B muſt ~~
fall between A and I, C between I and E, and DH be drawn as before:

~~and it is here evident that the conſtruction will be the ſame let the given ~~
ratio be what it will.

~~None of thoſe Caſes admit of any Limitations.~~
~~
~~

Epitagma II. ~~There are here only two Caſes, viz. ~~
~~when O is required ~~
beyond I, the point which bounds the ſegment whoſe ſquare is concerned;

~~
~~
~~
and ſecondly when it is ſought beyond both the other given points: ~~
~~in the ~~
firſt, the ratio of R to S muſt be of a greater to a leſs, and in the latter of

a leſs to a greater.

~~The former is conſtructed in Fig. ~~
~~17. ~~
~~at the ſame time ~~
with Caſe I of Epitagma I.

~~and is repreſented by the ſmall letters h and o: ~~
~~the latter in Fig. ~~
~~18, and pointed out by the ſame letters. ~~
~~That O will ~~
fall in both as is required needs not inſiſting on.

~~
~~

Epitagma III. ~~In which there are ſix Caſes, viz. ~~
~~I being the middle ~~
point, when O is ſought beyond A, or beyond E;

~~and that whether the ~~
given ratio be of a leſs to a greater, or of a greater to a leſs;

~~and again, I ~~
being an extreme point, when O is ſought between A and E, and that let

the order of the points A and E be what it will.

~~
~~

Cases I and II. ~~Are when I is a mean point and the given ratio of a ~~
leſs to a greater;

~~and theſe are both conſtructed at once by Fig. ~~
~~20, ~~
wherein B is made to fall beyond A, and C beyond E with reſpect to the

middle point I, and DH is drawn through the center of the circle on BC.

~~
~~

Cases III and IV. ~~Here, the points remaining as before, the given ~~
ratio is of a greater to a leſs;

~~and the conſtruction will be effected by ~~
making B fall beyond E, and C beyond A, and drawing DH parallel to

BC, as in Fig.

~~21 and 22.~~
~~
~~

Case V. ~~Wherein I is one extreme point and A the other, and O is ~~
ſought between A and E:

~~in conſtructing this Caſe, B muſt be made to ~~
fall between A and I, C between E and I, and DH drawn parallel to BC,

as is done in Fig.

~~23. ~~
~~The directions for Conſtructing Caſe VI. ~~
~~are exactly ~~
the ſame, as will appear by barely inſpecting Fig.

~~24.~~
~~
~~

Limitation. ~~It is plain that in the four laſt Caſes, the ratio which the ~~
rectangle contained by AO and EO bears to the ſquare on IO, or which is

the ſame thing, the given ratio of R to S cannot exceed a certain limit;

~~and it is farther obvious that the ſaid limit will be when the ſtraight line ~~
DH becomes a tangent to the circle on BC, as in Fig.

~~25. ~~
~~26, for after ~~
that the problem is manifeſtly impoſſible.

~~Now when DH is a tangent to ~~
the circle on BC, HO will be equal to half BC;

~~but the ſquare on HO ~~
is equal to the rectangle contained by IB and EC, wherefore the ſquare on

half BC will then be equal to the rectangle contained by IB and EC.

Moreover, by the conſtruction, R is to S as AB is to IB, and as EC is to

IC;

~~therefore by compoſition or diviſion, the ſum or difference of R and ~~
S is to R as EI to EC, and the ſaid ſum or difference is to S as AI is to

~~
IB, as EI is to IC: ~~
~~and hence, by compound ratio, the ſquare on the ~~
abovementioned ſum or difference is to the rectangle contained by R and S

as the rectangle contained by AI and EI is to the rectangle contained by

IB and EC, alſo by permutation, AI is to EI as IB is to IC;

~~wherefore, ~~
by compoſition or diviſion, AE is to AI as BC is to IB, by permutation,

AE is to BC as AI is to IB, therefore by equality, the ſum or difference

of R and S is to S as AE is to BC;

~~or (~~
Eu. ~~V. ~~
~~15.) ~~
~~as half AE is to ~~
half BC;

~~conſequently (~~
Eu. ~~VI. ~~
~~22.) ~~
~~the ſquare on the above mentioned ~~
ſum or diſſerence is to the ſquare on S as the ſquare on half AE is to the

ſquare on half BC, or to the rectangle contained by IB and EC.

~~Hence ~~
exæquo perturbaté, the rectangle contained by R and S is to the ſquare on

S as the ſquare on half AE is to the rectangle contained by AI and EI,

or (

Eu. ~~V. ~~
~~15.) ~~
~~R is to S as the ſquare on half AE is to the rectangle ~~
contained by AI and EI;

~~and which is therefore the greateſt ratio which ~~
R can have to S in thoſe Caſes.

~~
~~

~~It ought farther to be remarked, that to have Caſe III poſſible, where ~~
O is ſought beyond A, and the ratio of a greater to a leſs, it is neceſſary

that AI be leſs than IE, and to have Caſe IV.

~~poſſible, that it be greater. ~~
~~For it is plain from the Conſtruction, that IB muſt in the former caſe be ~~
leſs, and in the latter greater than I C;

~~but as R is to S ſo is AB to ~~
IB, and ſo is EC to IC, wherefore by diviſion, the exceſs of R above S

is to S as AI is to IB, and as EI is to IC;

~~and ſo by permutation AI is ~~
to EI as IB is to IC:

~~conſequently when IB is greater than IC, AI will ~~
be greater than EI;

~~and when leſs, leſs.~~

~~With reſpect to thoſe caſes wherein the given ratio is that of equality, ~~
it may be ſufficient to remark, that none of the Caſes of Epitagma II.

~~are poſſible under that ratio: ~~
~~that one of Caſes III. ~~
~~and IV. ~~
~~Epitagma III. ~~
is always impoſſible when the given ratio of R to S is the ratio of equality;

and both are ſo if AI be at the ſame time equal to IE.

~~Laſtly Caſes V. ~~
and VI.

~~are never poſſible under the ratio of equality, unleſs the ſquare on ~~
half AE be equal to, or exceed the rectangle contained by AI and EI;

all which naturally follows from what has been delivered above.

~~
~~

THE END OF BOOK I.

DETERMINATE SECTION. BOOK II. LEMMA I.
~~
~~~~
~~

~~If from two points E and I in the diameter AU of a circle AYUV (Fig. ~~
~~27.) ~~
~~two perpendiculars EV, IY be drawn contrary ways to terminate in ~~
the Circumference;

~~and if their extremes V and Y be joined by a ſtraight ~~
line VY, cutting the faid diameter in O;

~~then will the ratio which the ~~
rectangle contained by AO and UO bears to the rectangle contained by

EO and IO be the leaſt poſſible.

~~
~~

~~*** This being demonſtrated in the preceeding Tract of ~~
Snellius, I ſhall not attempt it here.

~~
~~

LEMMA II.
~~
~~~~
~~~~
~~~~
~~

~~If to a circle deſcribed on AU, tangents EV, IY (Fig. ~~
~~28. ~~
~~29.) ~~
~~be drawn ~~
from E and I, two points in the diameter AU produced, and through the

points of contact V, and Y, a ſtraight line YVO be drawn to cut the line

AI in O;

~~then will the ratio which the rectangle contained by AO and ~~
UO bears to that contained by EO and IO be the leaſt poſſible:

~~and ~~
moreover, the ſquare on EO will be the ſquare on IO as the rectangle con-

tained by AE and UE is to the rectangle contained by AI and UI.

~~
~~

Demonstration. ~~If the ſaid ratio be not then a minimum, let it be ~~
when the ſegments are bounded by ſome other point S, through which and

the point V, let the ſtraight line SV be drawn, meeting the circle again in

R;

~~draw SM parallel to OY, meeting the tangents EV and IY in L and ~~
M, and through R and Y draw the ſtraight line RY meeting SM produced

in N:

~~the triangles ESL and EOV, ISM and IOY are ſimilar; ~~
~~wherefore ~~
LS is to SE as VO is to EO, and SM is to SI as YO is to OI;

~~conſe-
~~
~~
quently, by compound ratio, the rectangle contained by LS and SM is to ~~
that contained by SE and SI as the rectangle contained by VO and OY,

or its equal, the rectangle contained by AO and OU is to that contained

by EO and IO.

~~Now the triangles VSL and NSR having the angles at ~~
R and L equal, and the angle at S common, are ſimilar;

~~and therefore SR ~~
is to SN as SL is to SV;

~~conſequently, the rectangle contained by SR and ~~
SV, or its equal, the rectangle contained by SA and SU is equal to that

contained by SN and SL:

~~but SN is neceſſarily greater than SM, in con-~~
ſequence whereof the rectangle contained by SN and SL, or its equal, the

rectangle contained by AS and SU is greater than that contained by SM

and SL;

~~wherefore the ratio which the rectangle AS, SU bears to the rect-~~
angle ES, SI is greater than that which the rectangle SM, SL bears to it,

and of courſe, greater than the ratio which the rectangle AO, UO bears to

the rectangle EO, IO;

~~and that, on which ſide ſoever of the point O, S is ~~
taken.

~~
~~

~~Again, on YO produced, let fall the perpendiculars EB and IC: ~~
~~the ~~
triangles EBV and ICY, EBO and ICO are ſimilar, becauſe the angles EVO

and IYO are equal, and ſo EO is to IO as EB is to IC, alſo EV is to IY

as EB is to IC;

~~therefore by equality of ratios EO is to IO as EV is to IY, ~~
and (

Eu. ~~VI. ~~
~~22.) ~~
~~the ſquare on EO is to the ſquare on IO as the ſquare ~~
on EV is to the ſquare on IY;

~~that is (~~
Eu. ~~III. ~~
~~36.) ~~
~~as the rectangle con-~~
tained by AE and UE is to that contained by AI and UI.

~~
~~

~~Q. ~~
~~E. ~~
~~D.~~
~~
~~

LEMMA III.
~~
~~
~~
~~

~~If from two points E and I, in the diameter AU, of a circle, AVYU ~~
(Fig.

~~30.) ~~
~~two perpendiculars EV, IY be drawn on the ſame ſide thereof to ~~

ſtraight line VY, cutting the ſaid diameter, produced, in O;

~~then will the ~~
ratio which the rectangle contained by AO and UO bears to the rectangle

contained by EO and IO be the greateſt poſſible.

~~
~~

~~*** This, like the ſirſt, is demonſtrated by ~~
Snellius, and needs not be repeated.

~~
~~

LEMMA IV.
~~
~~~~
~~
~~
~~~~
~~

~~If EK and IY (Fig. ~~
~~27.) ~~
~~be any perpendiculars to the diameter AU of ~~
a circle AYUV, terminating in the circumference, and if KY be drawn,

on which, from U, the perpendicular UF is demitted;

~~then will KF be a ~~
mean proportional between AI and EU, alſo YF a mean proportional

between AE and IU.

~~
~~

Demonstration. ~~Draw UY, UK, KA and AY, the angles I and F ~~

and conſequently IY is to UY as KF is to UK;

~~or (~~
Eu. ~~VI. ~~
~~22.) ~~
~~the ~~
ſquare on IY is to the ſquare on UY as the ſquare on KF is to the ſquare

on UK:

~~now the ſquare on IY is (~~
Eu. ~~VI. ~~
~~8. ~~
~~17.) ~~
~~equal to the rectangle ~~
contained by AI and IU, the ſquare on UY to the rectangle contained by

AU and IU, and the ſquare on UK to the rectangle contained by AU and

EU;

~~wherefore the rectangle contained by AI and IU is to that contained ~~
by AU and IU as the ſquare on KF is to the rectangle contained by AU

and UE, whence (

Eu. ~~V. ~~
~~15.) ~~
~~AI is to AU as the ſquare on KF is to the ~~
rectangle contained by AU and UE, or the rectangle contained by AI and

UE is to that contained by AU and UE as the ſquare on KF is to the

rectangle contained by AU and UE;

~~ſeeing then that the conſequents are ~~
here the ſame, the antecedents muſt be equal, and therefore AI is to KF

as KF is to UE.

~~
~~

~~Again, the angle AKE is equal to AUK, which is equal to the angle ~~
AYK, of which the angle UYF is the complement, becauſe AYU is a

right angle;

~~and therefore as the angles F and E are both right, the tri-~~
angles AKE and YUF are ſimilar, and AK is to AE as YU is to YF,

wherefore the ſquare on AK is to the ſquare on AE as the ſquare on YU

is to the ſquare on YF:

~~but the ſquare on AK is equal to the rectangle ~~
contained by AU and AE, and the ſquare on YU is equal to the rectangle

contained by AU and IU, conſequently the rectangle contained by AU

and AE is to the ſquare on AE as the rectangle contained by AU and IU

is to the ſquare on YF;

~~whence (~~
Eu. ~~V. ~~
~~15.) ~~
~~AU is to AE as the rect-~~
angle contained by AU and IU is to the ſquare on YF, or the rectangle

contained by AU and IU is to that contained by AE and IU as the rect-

~~
angle contained by AU and IU is to the ſquare on YF; ~~
~~hence the rect-~~
angle AE, IU is equal to the ſquare on YF, and AE is to YF as YF is

to IU.

~~
~~

~~Q. ~~
~~E. ~~
~~D.~~
~~
~~

LEMMA V.
~~
~~~~
~~
See Pap. Math. Collect. B. vii. prop. 60.
~~
~~~~
~~

~~If in any ſtraight line four points A, U, E and I (Fig. ~~
~~31.) ~~
~~be aſſigned, ~~
and if the point O be ſo taken by

Lemma II, that the ratio of the rect-angle contained by AO and UO to that contained by EO and IO may be

the leaſt poſſible;

~~alſo if through O the indefinite perpendicular FG be ~~
drawn;

~~and laſtly, if from E and I, EG and IF be applied to FG, the ~~
former equal to a mean proportional between AE and UE, and the latter

to one between AI and UI:

~~then ſhall FG be equal to the ſum of two ~~
mean proportionals between AE and UI, AI and UE.

~~
~~

Demonstration. ~~Draw AF and AG, and, through U, FV and GY, ~~
produce GE to meet FV in H, and let fall on FV the perpendicular XI,

cutting FG in N;

~~moreover draw UM through N, and NP through E, ~~
and theſe two laſt will be reſpectively perpendiculars to IF and UG, be-

cauſe the three perpendiculars of every plane triangle meet in a point.

~~Since by conſtruction and ~~
Eu. ~~VI. ~~
~~17, the ſquare on EG is equal to the ~~
rectangle contained by AE and UE, and the ſquare on IF to that con-

tained by AI and UI, and becauſe (

Lem. ~~II.) ~~
~~the ſquare on EO is to the ~~
ſquare on IO as the rectangle AE, UE is to the rectangle AI, UI;

~~the ~~
ſquare on EO is to the ſquare on IO as the ſquare on EG is to the ſquare

on IF, and (

Eu. ~~VI. ~~
~~22.) ~~
~~EO is to IO as EG is to IF; ~~
~~from whence it ~~
appears that the triangles EOG and IOF are ſimilar, and HG parallel to

IF, and the angle UHE equal to the angle UFI.

~~Again, becauſe AI is ~~
to IF as IF is to UI, the triangles AIF, and UFI are ſimilar, and ſo, for

like reaſons, are the triangles AEG and GEU, wherefore the angles UFI

and FAE are equal, and alſo the angles UGE and UAG;

~~hence there-~~
fore (

Eu. ~~I 32.) ~~
~~the angle YUF is equal to the angle UAF (UHE) toge-~~
ther with the angle UAG (UGE) and conſequently, the angles VAY and

YUV are together equal to two right angles;

~~wherefore the points AYUV ~~
are in a circle:

~~hence, and becauſe AO is perpendicular to FG, ~~
~~
be perpendicular to AF, and FV to AG; ~~
~~whence it follows that GA is ~~
parallel to XI, and the angle NIU equal to the angle UAG;

~~but UAG is ~~
equal to UGE, which is equal to CNE;

~~wherefore, in the triangles UNI, ~~
UNE, the angles at I and N being equal, and that at U common, they

are ſimilar, and UN is to UI as UE is to UN, conſequently the ſquare on

UN is equal to the rectangle contained by UI and UE.

~~Moreover, ſince ~~
IX paſſes through N, and is perpendicular to FU, by

Eu. ~~I. ~~
~~47, the dif-~~
ference of the ſquares on IF and IU is equal to the difference of the

ſquares on NF and NU:

~~now the ſquare on IF being equal to the rect-~~
angle contained by AI and UI, that is (

Eu. ~~II. ~~
~~I.) ~~
~~to the rectangle con-~~
tained by AU and UI together with the ſquare on UI, the difference of

the ſquares on IF and UI, and conſequently the difference of the ſquares on

NF and NU is equal to the rectangle contained by AU and UI;

~~but the ~~
ſquare on NU has been proved equal to the rectangle contained by UI

and UE, therefore the ſquare on NF is equal to the rectangle contained

by EU and IU together with that contained by AU and IU, that is (

Eu. ~~II. ~~
~~1.) ~~
~~to the rectangle contained by AE and UI; ~~
~~wherefore AE is to NF ~~
as NF is to UI.

~~
~~

~~Laſtly, for the like reaſons which were urged above, the difference of ~~
the ſquares on NU and NG is equal to the difference of thoſe on GE and

UE:

~~now the ſquare on GE is equal to the rectangle contained by AE ~~
and UE, that is, to the rectangle contained by AU and UE together

with the ſquare on UE;

~~therefore the difference of the ſquares on GE ~~
and UE, or the difference of thoſe on NU and NG, is equal to the rect-

angle contained by AU and UE;

~~but the ſquare on NU is equal to the ~~
rectangle contained by UI and UE, therefore the ſquare on NG is

equal to the rectangle contained by AU and UE together with that

contained by UI and UE;

~~that is, to the rectangle contained by AI and ~~
UE, and ſo AI is to NG as NG is to UE.

~~Now FG is equal to the ſum ~~
of NF and NG;

~~therefore FG is equal to the ſum of two mean propor-~~
tionals between AE and UI, AI and UE.

~~
~~

~~Q. ~~
~~E. ~~
~~D.~~
~~
~~

PROBLEM VII. (Fig. 32, 33, 34, &c.)
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~

~~In any indeſinite ſtraight line let there be aſſigned the points A, E, I ~~
and U;

~~it is required to cut it in another point, O, ſo that the rectangle ~~
contained by the ſegments AO, UO may be to that contained by the ſeg-

ments EO, IO in the ratio of two given ſtraight lines, R and S.

~~
~~

Analysis. ~~Imagine the thing done, and O the point ſought: ~~
~~then will ~~
the rectangle AO, UO be to the rectangle EO, IO as R is to S.

~~Make ~~
UC to EC as R is to S;

~~and the rectangle AO, UO will be the rectangle ~~
EO, IO as UC is to EC.

~~Let now OB be taken a fourth proportional to ~~
UO, UC and IO:

~~then (~~
Eu. ~~V. ~~
~~15.) ~~
~~the rectangle AO, UO will be to ~~
the rectangle EO, IO as the rectangle UC, OB is to the rectangle EC, OB;

~~or (~~
Eu. ~~V. ~~
~~16.) ~~
~~the rectangle AO, UO is to the rectangle UC, OB as the ~~
rectangle EO, IO is to the rectangle EC, OB;

~~wherefore ſince UO is to UC ~~
as IO to OB, by conſtruction, AO will be to BO as EO to EC;

~~and ſo by ~~
compoſition or diviſion, CO is to CU as IB to BO, and AB is to BO as

CO to EC:

~~wherefore ex æquo perturb. ~~
~~& ~~
~~permut. ~~
~~AB is to IB as UC to ~~
EC, that is, in the given ratio;

~~and hence is given BC, the ſum or dif-~~
ference of CO and BO, as alſo the rectangle contained by them, equal to

the rectangle CU, IB, whence thoſe lines themſelves are given by the 85th

or 86th of the Data.

~~
~~

Synthesis. ~~Make AB to IB, and UC to EC in the given ratio, and de-~~
ſcribe on BC a circle;

~~erect, at B the indeſinite perpendicular BK, and take ~~
therein BD a mean proportional between AB and EC, or between IB and

and UC:

~~from D, draw DH, parallel to BC, if O be required any where ~~
between B and C;

~~but through F, the center of the circle on BC, if it be ~~
ſought any where without them, cutting the circle on BC in H.

~~Laſtly, ~~
draw HO perpendicular to DH, which will cut the indeſinite line in O,

the point required.

~~
~~

~~For it is plain from the conſtruction that HO and BD are equal, and ~~
(

Eu. ~~VI. ~~
~~17.) ~~
~~the rectangle AB, EC, or the rectangle IB, UC is equal to ~~
the ſquare on BD, and therefore equal to the ſquare on HO, which (

Eu. ~~III. ~~
~~35. ~~
~~36.) ~~
~~is equal to the rectangle BO, OC. ~~
~~Hence (~~
Eu. ~~VI. ~~
~~16.) ~~
~~AB ~~
is to BO as CO is to CE, and CO is to CU as IB is to BO;

~~whence, ~~
by compoſition or diviſion, AO is to BO as EO is to CE, and UO is to

~~
CU as IO is to BO; ~~
~~and ſo, by compound ratio, the rectangle AO, UO ~~
is to the rectangle BO, CU as the rectangle EO, IO is to the rectangle

BO, CE;

~~by permutation, the rectangle AO, UO is to the rectangle EO, ~~
IO, as the rectangle BO, CU is to the rectangle BO, CE;

~~or (~~
Eu. ~~V. ~~
~~15.) ~~
~~as CU is to CE; ~~
~~that is, by conſtruction as R to S.~~
~~
~~

~~Q. ~~
~~E. ~~
~~D.~~
~~
~~

Scholium. ~~In enumerating the ſeveral Caſes of this Problem I ſhall en-~~
deavour to follow the method which I conceive Apollonius did:

~~and there-~~
fore, notwithſtanding the preceding Analyſis and Conſtruction are general

for the whole, divide it into three Problems, each Problem into three Epi-

tagmas, or general Caſes, and theſe again into their ſeveral particular ones.

~~
~~

PROBLEM I. (Fig. 32 to 45.)
~~
~~~~
~~~~
~~
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~

~~Here O is ſought between the two mean points of the four given ones: ~~
~~and the three Epitagmas are, firſt, when A and U, the points which bound ~~
the ſegments containing the antecedent rectangle, are one an extreme, and

the other an alternate mean;

~~ſecondly, when thoſe points are one an ex-~~
treme and the other an adjacent mean;

~~thirdly, when they are both means, ~~
or both extremes.

~~
~~

Epitagma I, Conſiſts of eight Caſes, viz. ~~when the order of the given ~~
points is A, I, U, E;

~~U, E, A, I; ~~
~~A, E, U, I; ~~
~~or U, I, A, E, and the ~~
given ratio of a leſs to a greater, and four others wherein the order of the

points is the ſame as in thoſe, but the ratio of R to S, the ratio of a greater

to a leſs.

~~
~~

Case I. ~~Let the order of the given points be A, I, U, E, and the given ~~
ratio of a leſs to a greater;

~~and the Conſtruction will be as in Fig. ~~
~~32, where ~~
B is made to fall beyond A, with reſpect to I, and C beyond U with re-

ſpect to E, and DH is drawn through F, the center of the circle on BC.

~~That O, when this conſtruction is uſed, will fall between I and U is ~~
plain, becauſe CO is to CU as IB is to BO;

~~and therefore if CU be ~~
greater than CO, BO will be greater than IB, and if leſs, leſs;

~~but this, ~~
it is manifeſt, cannot be the Caſe if O falls either beyond I or U, and

therefore it falls between them.

~~
~~

Case II. ~~If the order oſ the given points be retained, but the ratio be of ~~
a greater to a leſs, B muſt ſall beyond I with reſpect to A, and C beyond

E, as in Fig.

~~33, and DH muſt be drawn as before: ~~
~~and it may be proved ~~
by reaſonings ſimilar to thoſe uſed in the ſirſt Caſe that O will fall between

I and U, as was required.

~~
~~

Cases III and IV. ~~Theſe caſes are conſtructed exactly in the ſame man-~~
ner as Caſes I and II reſpectively;

~~and the reaſonings to prove that O will ~~
fall as it ought arethe ſame with thoſe made uſe of in Caſe I, as will appear

by inſpecting Fig.

~~34. ~~
~~and 35.~~
~~
~~

Case V. ~~Let now the order of the given points be A, E, U, I, and the ~~
given ratio of a leſs to a greater.

~~Then muſt B be made to fall beyond A ~~
and C beyond U, as in Fig.

~~36, and DH is here to be drawn parallel to ~~
to BC:

~~and that O will fall as required may be made to appear thus. ~~
~~Draw ~~
EF and UG perpendicular to BC.

~~Now (~~
Eu. ~~VI. ~~
~~13. ~~
~~17.) ~~
~~the ſquare on ~~
EF is equal to the rectangle contained by EC and EB, the ſquare on UG

to the rectangle contained by UC and UB;

~~and, by conſtruction, the ~~
ſquare on HO to the rectangle contained by EC and AB, or to the rect-

angle contained by UC and IB;

~~and ſince, by ſuppoſition, EB is greater ~~
than AB, and UB leſs than IB, the rectangle EC, AB, or its equal, the

rectangle UC, IB will be leſs than the rectangle EC, EB, and greater than

the rectangle UC, UB;

~~and conſequently the ſquare on HO will be leſs ~~
than the ſquare on EF and greater than the ſquare on UG;

~~and therefore ~~
HO leſs than EF and greater than UG:

~~but this could not be the Caſe ~~
unleſs O fell between E and U, as was was required.

~~
~~

Case VI. ~~If the order of the points be retained; ~~
~~but the given ratio be ~~
of a greater to a leſs, B muſt then fall beyond I, and C beyond E;

~~and ~~
DH is drawn as in the preceding Caſe (See Fig.

~~37.) ~~
~~moreover, that O ~~
will fall between E and U may be made to appear, by reaſonings ſimilar to

thoſe there made uſe on the like occaſion.

~~
~~

Case VII. ~~Is conſtructed exactly in the ſame manner as Caſe V. ~~
~~and,~~

Case VIII. ~~As Caſe VI: ~~
~~the truth of which will appear by. ~~
~~barely in-~~
ſpecting Fig.

~~38 and 39.~~
~~
~~

Epitagma II. ~~In this Epitagma there are alſo eight Caſes, viz. ~~
~~when the ~~
order of the given points is A, U, E, I;

~~A, U, I, E; ~~
~~U, A, E, I; ~~
~~or ~~
U, A, I, E;

~~and the given ratio of a leſs to a greater: ~~
~~and there are ~~
four others wherein the order of the points are the ſame as in thoſe, but

~~
the ratio of a greater to a leſs; ~~
~~but theſe, I ſhall ſhew, may be reduced ~~
to four.

~~
~~

Case I. ~~The order of the given points being A, U, E, I; ~~
~~and the given ~~
ratio a leſs to a greater, the conſtruction will be eſſected by Fig.

~~40, ~~
wherein B is made to fall beyond A with reſpect to I and C beyond U, and

DH is drawn through the center of the circle on BC.

~~And O will fall ~~
between U and E for reaſons ſimilar to thoſe urged in the ſirſt Caſe of Epi-

tagma I.

~~It is moreover obvious that the conſtruction will not be eſſentially ~~
different ſhould the points E and I change places, and therefore need not

here be made a new Caſe.

~~
~~

Case II. ~~The order of the points being the ſame as in the laſt Caſe, let ~~
the given ratio be of a greater to a leſs;

~~then, as in Fig. ~~
~~41, B muſt fall ~~
beyond I, and C beyond E;

~~but DC muſt ſtill be drawn through the ~~
center of the circle on BC.

~~It is manifeſt that this conſtruction will ſerve ~~
for that Caſe wherein the points A and U change ſituations, if the ratio be,

as here, of a greater to a leſs.

~~
~~

Case III. ~~Here, let the order of the points be U, A, I, E, and the ~~
given ratio of a leſs to a greater, and the Conſtruction will be aſſected by

Fig.

~~42, in which B falls beyond A, and C beyond U with reſpect to I and ~~
E:

~~and the ſame conſtruction will ſerve if I and E change places, but the ~~
ratio remain the ſame.

~~
~~

Case IV. ~~If the poſition of the points be retained, but the ratio be ~~
made of a leſs to a greater;

~~then muſt B fall beyond I (Fig. ~~
~~43.) ~~
~~and C ~~
beyond E;

~~but DH drawn as before. ~~
~~That O muſt fall as was required, ~~
in theſe three laſt caſes, is obvious enough from what has been ſaid be-

fore on the like occaſion:

~~and it is alſo plain that the conſtruction will ~~
not be materially diſſerent though A and U change places.

~~
~~

Scholium. ~~That none of the Caſes of theſe two Epitagmas are ſubject ~~
to Limitations, might be proved with the utmoſt rigour of geometrical

reaſoning was it not ſuſſiciently manifeſt from conſidering that as the point

O approaches points A, or U, the ratio of the rectangle AO, OU to the

rectangle EO, OI will become very ſmall, and as it approaches the points

E, or I the ſaid ratio will become very great:

~~and nothing hinders that ~~
the ſaid point may ſall any where between thoſe.

~~
~~

Epitagma III. ~~There are here but four Caſes, viz. ~~
~~when the order of the ~~
given points is A, E, I, U;

~~A, I, E, U; ~~
~~E, A, U, I; ~~
~~or E, U, A, I; ~~
~~the two ſirſt of theſe are not poſſible unleſs the given ratio be the ratio of a ~~
greater to a leſs;

~~nor the two latter, unleſs it be of a leſs to a greater, and ~~
as theſe are reduced to the ſirſt two by reading every where E for A, I

for U, and the contrary, I ſhall omit ſpecifying them.

~~
~~

Case I. ~~If the order of the given points be A, E, I, U, the conſtruc-~~
tion will be effected by Fig.

~~44, wherein B is made to fall beyond I, and ~~
C beyond E, and DH is drawn parallel to BC.

~~That O, when this con-~~
ſtruction is uſed, will fall between E and I, is eaſily made appear by rea-

ſoning in a manner ſimilar to what was done in Caſe V.

~~of Epitagma I.~~
~~
~~

Case II. ~~The conſtruction of this Caſe, where the order of the points ~~
is A, I, E, U, is not materially different from that above exhibited as ap-

pears by Fig.

~~45, and that O will fall between I and E is manifeſt without ~~
farther illuſtration.

~~
~~

Limitation. ~~In theſe two Caſes the given ratio of R to S cannot be ~~
leſs than that which the ſquare on AU bears to the ſquare on a line which

is the difference of two mean proportionals between AI and EU, AE and

IU.

~~For by Lemma I. ~~
~~the leaſt ratio which the rectangle contained by AO ~~
and UO can have to the rectangle contained by EO and IO;

~~or, which ~~
is the ſame thing, that R can have to S, will be when the point O is

the interſection of the diameter AU, of a circle AYUV, with a ſtraight

line YV.

~~joining the tops of two perpendiculars EV, IY to the indeſi-~~
nite line, on contrary ſides thereof, and terminating in the periphery of

the circle.

~~Produce VE (Fig. ~~
~~27.) ~~
~~to meet the circle again in K, and ~~
draw the diameter KL;

~~join LY and KY, on which, produced, let fall ~~
the perpendicular UF.

~~Now, ſince by Lemma III. ~~
~~KF is a mean propor-~~
tional between AI and EU, and YF a mean proportional between AE

and IU:

~~it remains only to prove that the ratio of the rectangle con-~~
tained by AO and OU to the rectangle contained by EO and OI is the

ſame with the ratio which the ſquare on AU bears to the ſquare on KY,

which is the diſſerence between KF and YF.

~~Becauſe the angles E and ~~
KYL are both right, and the angles EVO and KYL equal (

Eu. ~~III. ~~
~~21.) ~~
~~the triangles EVO and YLK are ſimilar; ~~
~~and ſo VO is to EO as AU (LK) ~~
is to KY;

~~or the ſquare on VO is to the ſquare on EO as the ſquare on AU ~~
is to the ſquare on KY.

~~Now the triangles EVO, IYO being alſo ſimilar,
~~
~~
OY will be to OV as OI is to OE, and (Eu. ~~
~~V. ~~
~~15. ~~
~~16.) ~~
~~the rectangle ~~
contained by VO and YO or its equal, the rectangle contained by AO and

OU, is to the rectangle contained by EO and OI as the ſquare on OV is

to the ſquare on EO, as the ſquaree on AU is to the ſquare on KY.

~~
~~

~~Q. ~~
~~E. ~~
~~D.~~
~~
~~

Scholium. ~~It might be obſerved that in the two Caſes of this Epitagma ~~
where the points A and U are means, the limiting ratio will be a maxi-

mum inſtead of a minimum;

~~and that ratio will be the ſame with that which ~~
the ſquare on KY bears to the ſquare on EI, as is plain from what hath

been advanced above.

~~
~~

PROBLEM II. (Fig. 46 to 57.)
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~
~~
~~~~
~~~~
~~~~
~~

~~Where O is ſought between a mean and an extream point: ~~
~~and here, ~~
as in the firſt Problem, there are three Epitagmas.

~~Firſt when the points ~~
A and U, which bound the ſegments containing the antecedent rectangle,

are one an extreme, and the other an alternate mean;

~~ſecondly when they ~~
are both means, or both extremes;

~~thirdly when they are one an extreme, ~~
and the other an adjacent mean.

~~
~~

Epitagma I. ~~There are here eight Caſes, but they are conſtructed at ~~
four times, becauſe it is indifferent whether the given ratio be of a leſs to a

greater, or of a greater to a leſs.

~~
~~

Case I. ~~The order of the given points being A, E, U, I, as in Fig. ~~
~~46, ~~
make B to fall between A and I, C between U and E, and draw DH

through the center of the circle on BC;

~~and O will fall between A and ~~
E, becauſe AB is to BO as CO is to CE, and therefore, if AB be greater

than BO, CO muſt be greater than CE, and if leſs, leſs;

~~but this cannot ~~
be the caſe if O falls either beyond A or E:

~~and the like abſurdity fol. ~~
~~lows if o be ſuppoſed to fall otherwiſe than between I and U.~~
~~
~~

Case II. ~~Wherein the order of the points is U, I, A, E; ~~
~~and it is con-~~
ſtructed in the very ſame manner that Caſe I.

~~is, as appears by barely ~~
inſpecting Fig.

~~47.~~
~~
~~

Case III. ~~If the order of the given points be A, I, U, E (Fig. ~~
~~48.) ~~
~~the ~~
points B and C muſt be made to fall as in the two preceding Caſes;

~~but ~~
DH muſt be drawn parallel to BC, and O will fall as required.

~~For erect
~~
~~
at I, the perpendicular IG: ~~
~~by Eu. ~~
~~VI. ~~
~~13. ~~
~~17. ~~
~~the ſquare on IG is equal ~~
to the rectangle contained by IB and IC;

~~and the ſquare on HO is equal ~~
to the rectangle contained by IB and UC.

~~Now IC is by ſuppoſition ~~
greater than UC, and therefore the rectangle IB, IC is greater than the

rectangle IB, UC:

~~conſequently the ſquare on IG is greater than the ~~
ſquare on HO, and IG than HO;

~~whence O muſt fall between I and B, ~~
much more between I and A.

~~And in the ſame manner it may be proved ~~
that the point o falls between U and E.

~~
~~

Case IV. ~~In which the order of the given points is U, E, A, I; ~~
~~it is ~~
conſtructed exactly in the ſame manner as Caſe III, and is exhibited by

Fig.

~~49.~~
~~
~~

Epitagma II. ~~There are here only four Caſes, becauſe, as in Epi-~~
tagma I.

~~it is indifferent whether the given ratio be of a leſs to a greater, ~~
or of a greater to a leſs;

~~and the two laſt of thoſe, viz. ~~
~~where the order ~~
of the given points is E, A, U, I;

~~or E, U, A, I, being reducible to the ~~
two former by reading every where I for A, E for U, and the contrary,

I ſhall omit ſaying any thing of their conſtructions, except that they are

exhibited by Fig.

~~52 and 53.~~
~~
~~

Case I. ~~The order of the given points, being A,E,I,U, make B to ~~
fall between A and I, C between E and U, and draw DH through the

center of the circle on BC, as is done in Fig.

~~50; ~~
~~and O will fall as re-~~
quired for reaſons ſimilar to thoſe urged in Caſe I.

~~of the firſt Epitagma ~~
of this Problem.

~~
~~

Case II. ~~If the order of the given points be A, I, E, U, the conſtruc-~~
tion will be as in Fig.

~~51, where B and C are made to fall, and DH is ~~
drawn as in Caſe I.

~~
~~

Epitagma III. ~~Here there are eight Caſes, viz. ~~
~~four where in the ~~
order of the given points is A, U, E, I;

~~A, U, I, E; ~~
~~U, A, E, I; ~~
~~and ~~
U, A, I, E, and the given ratio of a greater to a leſs, when O will fall

between the two given points, which bound the conſequent rectangle;

~~and four others@ wherein the order of the given points is the ſame as ~~
here, but the given ratio of a leſs to a greater, and in which the point

O will fall between the points that bound the antecedent rectangle;

~~but ~~
as theſe laſt are reducible to the former by the ſame means which have been

uſed on former ſimilar occaſions, I ſhall not ſtop to ſpecify them.

~~
~~

~~The former four are all conſtructed by making B to fall between A and ~~
I, C between U and E, and drawing DH parallel to BC;

~~and it will ap-~~
pear by reaſonings ſimilar to thoſe uſed for the like purpoſe in Caſe III.

~~of ~~
Epitagma I.

~~that O muſt fall between E and I as was propoſed. ~~
~~See Fig. ~~
~~54, 55, 56 and 57.~~
~~
~~

Limitation. ~~In theſe four Caſes, the given ratio of R to S muſt not ~~
be leſs than that which the ſquare on the ſum of two mean proportionals

between AE and IU, AI and EU bears to the ſquare on EI.

~~For it has ~~
been proved (Lem.

~~II.) ~~
~~that when the ratio of the rectangle contained by ~~
AO and UO to that contained by EO and IO;

~~or, which is the ſame thing, ~~
the given ratio of R to S is the leaſt poſſible, the ſquare on EO will be to

the ſquare on IO as the rectangle contained by AE and UE is to that con-

tained by AI and UI;

~~and (~~
Lem. ~~V. ~~
~~Fig. ~~
~~31.) ~~
~~that FG will then be the ~~
fum of two mean proportionals between AE and UI, AI and UE:

~~it ~~
therefore only remains to prove that the rectangle contained by AO and

UO is to that contained by EO and IO as the ſquare on FG is to the

fquare on EI.

~~Now it has been proved in demonſtrating Lem. ~~
~~V. ~~
~~that the ~~
triangles EOG and IOF are ſimilar, and that the angle at V is right,

whence it follows that the triangles AOG and FOU are alſo ſimilar, and

conſequently that AO is to OG as OF is to UO;

~~therefore the rectangle ~~
contained by AO and UO is equal to that contained by GO and OF.

~~More-~~
over GO is to OF as EO is to IO, and ſo by compoſition and permutation,

FG is to EI as OG is to EO, and as OF is to IO:

~~hence by compound ~~
ratio the ſquare on FG is to the ſquare on EI as the rectangle contained

by (OG and OF) AO and UO is to that contained by EO and IO.

~~
~~

~~Q. ~~
~~E. ~~
~~D.~~
~~
~~

Scholium. ~~In the four Caſes, wherein the given ratio is of a leſs to a ~~
greater, and wherein the point O muſt fall between thoſe given ones which

bound the antecedent rectangle, the limiting ratio will be a maximum, and

the ſame with that which the ſquare on AU bears to the ſquare on FG.

~~
~~

PROBLEM III.
~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~

~~In this, the point O is ſought without all the given ones, and the three ~~
Epitagmas are as in Problem I.

~~
~~

Epitagma I. ~~There are here eight Caſes, viz. ~~
~~four when the order of ~~
the given points is the ſame as ſpecified in Epitagma I.

~~of Problem I, and ~~
O ſought beyond the given point which bounds the antecedent rectangle;

~~and four others when O is ſought beyond that which bounds the conſe-~~
quent one:

~~the Conſtructions of the four firſt are ſhewn by the ſmall ~~
letters b and o in Fig.

~~32, 34, 36 and 38; ~~
~~and the four latter ones by ~~
the ſame letters in Fig.

~~33, 35, 37 and 39; ~~
~~and the demonſtrations that ~~
o will fall as required by the Problem are exactly the ſame as thoſe made

uſe of in the laſt mentioned Epitagma.

~~It is farther obſervable, that the ~~
four firſt Caſes are not poſſible, unleſs the given ratio be of a leſs to a

greater;

~~nor the four latter, unleſs it be of a greater to a leſs, as is ma-~~
nifeſt without farther illuſtration.

~~
~~

Epitagma II. ~~Here, as in the ſecond Epitagma of Problem I, the points ~~
A and U are one an extreme, and the other an adjacent mean, and there

are eight Caſes;

~~but it will be ſufficient to exhibit the conſtructions of ~~
four of them, the others being not eſſentially different;

~~and theſe are ſhewn ~~
by the ſmall b and o in Fig.

~~40, 41, 42 and 43; ~~
~~the demonſtrations that o ~~
will fall as required need not be pointed out here;

~~but it may be neceſſary ~~
to remark that the firſt and third are not poſſible unleſs the given ratio be

of a leſs to a greater, nor the ſecond and fourth unleſs it be of a greater to

a leſs, as is obvious enough.

~~
~~

Epitagma III. ~~In which the points A and U are both means, or both ~~
extremes;

~~and there are here eight Caſes, viz. ~~
~~four wherem theſe points ~~
are extremes, and four others wherein they are means:

~~but theſe laſt being ~~
reducible to the former by the ſame method that was uſed in the third Epi-

tagmas of the two preceding Problems, I ſhall omit them.

~~
~~

~~All the Caſes of this Epitagma are conſtructed by making B fall beyond ~~
I, and C beyond E, with reſpect to A and U;

~~and drawing DH parallel to ~~
BC.

~~That O will fall beyond A in Fig. ~~
~~58 and 60, and beyond U in Fig. ~~
~~59 and 61 appears hence. ~~
~~Draw AG perpendicular to BC, meeting the ~~
circle on BC in G:

~~by Eu. ~~
~~VI. ~~
~~13. ~~
~~17, the ſquare on AG is equal to the
~~
~~
rectangle contained by AB and AC; ~~
~~but the ſquare on HO is equal to ~~
the rectangle contained by AB and EC:

~~now EC is, by ſuppoſition, ~~
greater than AC, therefore the rectangle AB, EC is greater than the

rectangle AB, AC, and the ſquare on HO greater than the ſquare on AG,

conſequently HO is itſelf greater than AG;

~~but this could not be the ~~
Caſe unleſs O fell beyond A.

~~In the ſame manner my it be proved that O ~~
will fall beyond U in Fig.

~~59 and 60.~~
~~
~~

Limitation. ~~In the above four Caſes the given ratio of R to S muſt ~~
not exceed that which the ſquare on AU bears to the ſquare on the ſum of

two mean proportionals between AI and UE, AE and UI.

~~For (Fig. ~~
~~30.) ~~
~~demit from A, on KO produced, the perpendicular AH. ~~
~~Now it has been ~~
proved (Lem.

~~III.) ~~
~~that the ratio of the rectangle continued by AO and UO ~~
to that contained by EO and IO, or which is the ſame thing, the given

ratio of R to S is the greateſt poſſible;

~~and (Lem. ~~
~~IV.) ~~
~~that KF is a mean ~~
proportional between AI and UE, alſo that YF is a mean proportional

between AE and UI:

~~but HK is equal to YF, therefore HF is equal to ~~
the ſum of two mean proportionals between AI and UE, AE and UI;

it only then remains to prove, that the rectangle contained by AO and UO

is to that contained by EO and IO as the ſquare on AU is to the ſquare

on HF.

~~The triangles OEK, OHA, OIY and OUF are all ſimilar; ~~
~~con-~~
ſequently OK is to OE as OA is to OH, as OY is to OI, and therefore

by compound ratio, the rectangle contained by AO and UO (OK and

OY) is to that contained by EO and IO as the ſquare on AO is to the

ſquare on OH;

~~but alſo AO is to UO as HO is to EO, and by compoſi-~~
tion and permutation, AU is to HF as AO is to HO, or (Eu.

~~VI. ~~
~~22.) ~~
the ſquare on AU is to the ſquare on HF as the ſquare on AO is to the

ſquare on HO, and ſo by equality of ratios, the rectangle contained by AO

and UO is to that contained by EO and IO as the ſquare on AU is to the

ſquare on HF.

~~
~~

~~Q.~~
~~E.~~
~~D.~~
~~
~~

Scholium. ~~In the four Caſes wherein the points A and U are means, ~~
the limiting ratio will be a minimum, and the ſame with that which the

ſquare on HF bears to the ſquare on EI.

~~
~~

THE END.

A
SYNOPSIS OF ALL THE DATA FOR THE Conſtruction of Triangles, FROM WHICH GEOMETRICAL SOLUTIONS
Have hitherto been in Print.
~~
~~

~~With References to the Authors, where thoſe ~~
Solutions are to be found:~~
~~

By JOHN LAWSON, B. D. Rector of Swanscombe, in KENT.
ROCHESTER:

~~Printed by T. ~~
Fisher; ~~and Sold by J. ~~
Nourse, B. White, T. Payne, and J. Wilkie, in London@

MDCCLXXIII. [Price One Shilling.]

ADVERTISEMENT.
~~
~~

~~IT is but few years ago ſince the Compiler of this Synopſis ~~
conceived his firſt idea of the uſeſulneſs of ſuch an undertaking,

and he exhibited a ſmall ſpecimen thereof in a periodical Work

then publiſhing under the Title of

The British Oracle. ~~Here-~~
upon he received ſeveral letters from Mathematical friends, ex-

preſſing their ſenſe of the great propriety of ſuch a collection, and

ſtrongly encouraging him to purſue the undertaking.

~~Since that ~~
time it has been a growing work, and would continue ſo, were

the publication delayed ever ſo long, as freſh Problems are con-

tinually propoſed to the public.

~~He has therefore now determined ~~
to ſend it abroad, as complete as he can make it to the preſent

period, and leave additions to be made by future collectors.

~~
~~

AN EXPLANATION OF THE SYMBOLS made uſe of in this SYNOPSIS.
H. # repreſents the Hypothenuſe of a right-angled triangle. V. # Vertical angle. B. # Baſe or ſide oppoſite V. P. # Perpendicular from V on B. S&s. # Sides about V; S the greater, s the leſs. A & a. # Angles at B; A the greater, a the leſs. m & n. # Segments of B by P; m the greater, n the leſs. Ar. # Area. Per. # Perimeter. L. # Line from V to B, biſecting V. λ. # Line from V to B, cutting B in a given ratio. l. # Any other line ſpecified how drawn. R. # Radius of inſcribed circle. ☉(sun). # Circle. ◻. # Square. : # Ratio: thus, S: s ſignifies the ratio of the ſides.

~~
~~~~
~~~~
~~

~~N. ~~
~~B. ~~
~~Between each of the Data a full ſtop is placed. ~~
~~Moreover, m and n are ~~
ſometimes uſed for different ſegments than thoſe of B by P, but then it is ſignified in

words:

~~- the ſame alſo is to be obſeved of P.~~
~~
~~

~~Obſerve likewiſe, for the more ready finding any propoſed Problem in the Synop-~~
ſis, that the data are ranged in the ſame order as the Symbols here recited;

~~viz. ~~
~~all data whereof V is one are placed firſt; ~~
~~next thoſe where B is given, either ſim-~~
ply by itſelf, or combined with any other datum;

~~next P, &~~
~~c.~~
~~
~~

~~Moreover, when in the references you find this mark*, it ſignifies that ſuch Authors ~~
have only conſtructed the Problem partially, and not generally as propoſed in the Synop-

ſis, e.

~~g. ~~
~~for a right-angled triangle, when it is propoſed for a triangle in general; ~~
~~and again, for a line biſecting another, when it is propoſed to cut it ſo that the ſeg-~~
ments may be in any given ratio.

~~
~~

INDEX OF THE Authors refered to in the SYNOPSIS.
ANDERSONI Var. Prob. Practice, cum Supplemento Apollonii # Redivivi, 4to. # Pariſiis 1612 Anderſoni Exercitationum Math. Deas ima. 4to. # ibid. 1619 Aſhby’s Algebra, 2d Edit. 12mo. # Lond. 1741 Britiſh Oracle (Vol. I. being all that was publiſhed) 12mo. # ibid. 1769 Caſtillioneus inNewtoni Arith. Univerſalem, 4to. # Amſtelod. 1761 Clavius in Euclidem, var. Ed. Court Magazine. D’Omerique (Hugonis) Analyſis Geometrica, 4to. # Gadibus 1699 Diarian Repoſitory, Periodical Work, printed for Robinſon, 4to. # Lond. 1770, &c. Foſter’s Miſcellanies, or Math. Lugubrations, fo. # ibid. 1659 General Magazine. Gentleman’s Magazine. Gentleman’s Diary. Ghetaldi (Marini) Var. Prob. Collectio, 4to. # Venetiis 1607 Ghetaldus de Reſolutione & Compoſitione Math. fo. # Romæ 1640 Gregorius a Sancto Vincentio, fo. # Antverp. 1647 Herigoni Curſus Math. Lat. & Gallicè, 8vo. 5 Tom. # Paris 1644 Hutton’s Ladies Diaries. # Mathematical Miſcellany. Imperial Magazine. Ladies Diaries. Martin’s Math. Correſpondence, in his Magazine. Mathematical Magazine. Mathematician, Periodical Work. 8vo. # Lond. 1751
Miſcellanea Curioſæ, Periodical Work. 8vo. 6 Nos. # rork 1734-5 Miſcellanea Cur. Math. Period Work, by Holliday, 4to. 2 vols. # Lond. 1745 Miſcellanea Scientiſica Curioſa, Period. Work, 4to. # ibid. 1766 Oughtred’s Clavis, var. Editions Lat. & Eng. Palladium, Periodical Work. Pappus Alexandrinus Commandini, fo. # Bononiæ 1660 Regiomontanus de Triangulis, fo. # Baſiliæ 1561 Renaldinus (Carolus) de Res. & Comp. Math. fo. # Patavii 1668 Ronayne’s Algebra, 2d Edit. 8vo. # Lond. 1727 Rudd’s Practical Geometry in 2 Parts, 4to. # ibid. 1650 Saunderſon’s Algebra, 2 vols. 4to. # Camb. 1740 Schooten’s (Franciſcus à) Exercitationes Math. 4to. # Lug. Bat. 1657 Simpſon’s Algebra, 8vo. 3d Edit. \\ 1ſt Edit. # Lond. 1767 \\ ibid. 1745 # Select Exerciſes, 8vo. # ibid. 1752 # Geometry. 8vo. 2d Edit. # ibid. 1760 Supplement to Gentleman’s Diary, 3 Nos. 12mo. # ibid. 1743, &c. Town and Country Magazine. Turner’s Mathematical Exerciſes, Periodical Work, 8vo. # ibid. 1750 Univerſal Magazine. Vietæ Opera, fo. # Lug. Bat. 1646 Weſt’s Mathematics, 2d Edit. 8vo. # Lond. 1763 Wolfius’s Algebra, tranſlated by Hanna, 8vo. # ibid. 1739

Lately was publiſhed by the ſame Author;
[Price Six Shillings in Boards.]
~~
~~

~~APOLLONIUS concerning ~~
Tangencies, as reſtored by Vieta & ~~Ghetaldus, ~~
with a Supplement;

~~the 2d Edit. ~~
~~To which is now added a Second Supplement, ~~
being Fermat’s Treatiſe on

Spherical Tangencies. ~~Likewiſe Apollonius con-~~
cerning

Determinate Section, as reſtored by Willebrordus Snellius; ~~to which ~~
is added an entire new Work, being the ſame reſtored by Mr.

~~W. ~~
~~Wales.~~
~~
~~

SYNOPSIS.
1. V. B. P. # SIMPSON’s Alg. pr. 5. - Mis. Cur. Math. Vol. I. \\ page 31. - Vieta Iſt. Ap. to Apollonius Gallus, pr. 5. \\ - *Aſhby’s Alg. pag. 111. - *Rudd’s Pract Geo. part 2d, qu. \\ 4.-L. Diary, qu. 160. 2. V. B. P: m. # Town and Country Mag. Nov. and Dec. 1772. 3. V. B. P±m. # Mathematician, pr. 77. - Univerſal Mag. Mar. 1749. 4. V. B. S: s. # Simpſon’s Alg. pr. 3. - Simp. Geom. pr. 13. - Pappus Lib. \\ VII. pr. 155. - Herigon App. Geometriæ planæ, pr. 13. - \\ D’Omerique Lib. III. pr. 35. - Court Mag. July, 1762. 5. V. B. S + s. # Simpſon’s Alg. pr. 1. - Ghetaldus var. prob. 13 & *7. - Ghetaldus \\ deRes. & Comp. Math. Lib. V. c. 4, pr. 4, pag. 337, & *Lib. II. \\ pr. 9, pag. 93. - Renaldinus pag. 318, 326, 524, 79. - \\ Saunderſon’s Alg. art. 332. - *Rudd’s Prac. Geom. part \\ 2d, qu. 38. - *Aſhby’s Alg. pag. 102. 6. V. B. S - s. # Simpſon’s Alg. pr. 2. - Ghetaldus var. prob. 12 & *6. - \\ Ghetaldus de Res. & Comp. Lib. V. c. 4, pr. 3, and *Lib. II. pr. 8. \\ - Renaldinus pag. 317, 326, 527, 79. - *Simpſon’s Sel. Ex. pr. \\ 2. - *Wolfius’s Alg. pr. 12 8. - *Aſhby’s Alg. pag. 106. 7. V. B. S + s x S. # *Town and Country Mag. Jan. and Feb. 1769. 8. V. B. S + s + P. # Arith. Univ. Caſtillionei, pr. 5. 9. V. B. Ar. # *Simpſon’s Alg. pr. 33. - Simp. Geom. pr. 5. - *Saunder - \\ ſon’s Alg. art. 329. - *D’Omerique L. III. pr. 36. - *Oughtred’s

# Clavis, ch. 19, pr. 24. - *Vieta Geo. Eff. pr. 20. - *Herigon \\ App. Geo. planæ, pr. 10. - *Rudd’s Pract. Geom. part 2d, \\ qu. 47. - *Palladium, 1754, pa. 22. 10. V. B. Per. # Reducible to V. B. S + s. 11. V. B. L. # Simpſon’s Alg. pr. 72. - Simp. Geom. pr. 21. 12. V. B. λ. # *Simpſon’s Alg. pr. 58. - *Ghetaldus var. prob. 3. - Regio - \\ montanus de triangulis, Lib. II. pr. 29. 13. V. B. Direction \\ of l thro’ V. # }Simpſon’s Sel. Ex. pr. 48. 14. V. B. R. # *Simpſon’s Sel. Ex. pr. 29. - Br. Oracle, qu. 67, Cor. - \\ *Weſt’s Mathematics, 2d. Ed. pag. 45. 15. V. B. Side of \\ ins. ◻. # }Mathematician, pr. 25. - Br. Oracle, qu. 19. 16. V. B: P. S±s. # Brit. Oracle, qu. 31. 17. V. B±P. S±s. # Arith. Un. Caſtillionei, pr. 7. - *Simpſon’s Sel. Ex. pr. 31. - \\ *Simpſon’s Alg. Iſt. Ed. pr. 81. 18. V. B±P. Ar. # *Turner’s Math. Ex. pr. 18. 19. V. B±S. s. # Ghetaldus var. prob. 16, 17. - *Idem, 10, II. - Idem de \\ Res. &. Comp. Lib. V. cap. 4, pr. 7, 8. - *Anderſon var. prob. 2; \\ - *Simp. Sel. Ex. pr. 1. - Twyſden in Foſter’s Miſcell. pr. I. \\ - Renaldinus pag. 526, 529, *437. - *Mathematician, pr. \\ 42. - *Aſhby’s Alg. pr. 31, 32. - and in other places. 20. V. B±S. S±s. # Br. Oracle, qu. 102. - *Court Mag. Nov. 1761. - L. Diary, \\ qu. 661. - *Hutton’s L. Diary, qu. 147. 21. V. B±S. B±s. # *Clavius’s Euclid at end of B. II. - *Ghetaldus var. prob. 23. \\ - *Schooten pr. 37. - *Oughtred ch. 19, pr. 17. - *Saun - \\ derſon, art. 327. - *Anderſon var. prob. 15, 16. - *L. Diary, \\ 1770, p. 35. 22. V. B2: m2. P. # *Mis. Scient. Cur. qu. 54.

23. V. Point in B. \\ : \\ {S±s.\x} # }Simpſon’s Sel. Ex. pr. 43. - Simp. Geom. pr. 19, 20. 24. V. S + s - B. P. # Hutton’s Miscellany, qu. 5. 25. V. S + s - B. Ar. # Math. Mag. No. III. pr. 4. 26. V. B + S - s. \\ S + s + m - n. # }Gent. Mag. 1768, pag. 428, 519. 27. V. B±S. Per. # *Anderſon var. prob. 1, 3. - See V. B±S. s. 28. V. B ‖ to a given \\ line. Ar. # }Simpſon’s Sel. Ex. pr. 41. - Simp. Geom. pr. 4. 29. V. Pointin B. Ar. # Simpſon’s Sel. Ex. pr. 42. 30. V. ∠ of B with λ. λ. # *Mathematician, pr. II. - *York Mis. Cur. qu. 15. 31. V. P. S: s. # Simp. Alg. pr. 3. - Court Mag. Octo. 1762. 32. V. P. S±s. # Simp. Alg. pr. 80, 78. - *Oughtred ch. 19, pr. 9, 10. - \\ *Ghetaldus var. prob. 20, 21. - *De Res. & Comp. Lib. III. \\ pr. 3, 4. - Diarian Repoſitory, pag. 24. - *Hutton’s L. Diaries, \\ qu. 22. - *Wolfius’s Alg. pr. 127. - *Ronayne’s Alg. B. II. c. 2, \\ pr. 2. 33. V. P. S x s. # D’Omerique Lib. I. pr. 31. - Gent. D. qu. 149. 34. V. P. m: n. # Simp. Alg. pr. 11. 35. V. P. m - n. # Simp. Alg. pr. 10. - Foſter’s Lug. pr. 2. - *Rudd’s Pr. \\ Geom. part 2d, qu. 5. 36. V. P. Per. # Simp. Alg. pr. 60. - Ronayne’s Alg. B. II. ch. 1, pr. 2. - \\ Arith. Univ. Caſtillionei, pr. 4. - Mis. Cur. Math. qu. 61. 37. V. P. λ. # *Town and C. Mag. 1769, pag. 296, 381. 38. V. P. Ifrom A or a \\ to bis. S or s. # }*Br. Oracle, qu. 74. 39. V. P. R. # *Br. Oracle, qu. 51. 40. V. P. Ra. of cir- \\ cum. ☉(sun). # }Gent. Diary, 1767, qu. 300. 41. V. P + s. S - s: \\ m - n. # }Gent. Diary, 1751, qu. 108.

42. V. S or s. S x s: \\ m x n, ſegts. by L. # }Brit. Oracle, qu. 87. - Renaldinus, pa. 337. 43. V. S x s. m x n. # Brit. Oracle, qu. 60. 44. V. S: s. m - n. # Simpſon’s Alg. pr. 3. # ☞ When V and S: s are given, the triangle is given in \\ ſpecies, and therefore may be conſtructed with any other \\ datum which does not affect the angles. 45. V. S: s. R. # Br. Oracle, qu. 21. 46. V. S±s. m: n. # Simpſon’s Alg. pr. 7, 9. - *York Mis. Cur. qu. 46. 47. V. S±s. m - n. # Simpſon’s Alg. pr. 6, 8. - *Oughtred ch. 19, pr. 12, 13. - \\ *Ghetaldus var. pr. 8, 9. - Idem, pr. 14, 15. - Renaldinus \\ pag. 319, 529. 48. V. S±s. Ar. # Anderſon var. prob. 22. - *Simp. Alg. pr. 34. 49. V. S±s. L. # Deducible from Simpſon’s Geo. pr. 19, 20. 50. V. S - s. R. # *Br. Oracle, qu. 20. 51. V. S2 + s2. λ. # Hutton’s Mis. qu. 24. 52. V. m. n. # Simpſon’s Alg. pr. 4. - York Mis. Cur. qu. 17, Caſe 3d. - \\ *Rudd’s Pract. Geo. part 2d, qu. 2. - Court Mag. Feb. 1763, \\ m and n being ſegts. by L. 53. V. Ar. Per. # *D’Omerique L. III. pr. 34. - *Simpſon’s Alg. pr. 35. - \\ *Simpſon’s Sel. Ex. pr. 30. - Arith. Uni. Caſtillionei, pr. 8, \\ *3. - *Rudd’s Prac. Geo. part 2d, qu. 7. - L. Diary, 1761, \\ qu. 480. - Mathematician, qu. 49. - Gent. Diary, 1744, qu. \\ 40. - *Wolfius’s Alg. pr. 113. 54. V. Ar. Side of \\ ins. ◻. # }L. Diary, 1763, qu. 507. - Court Mag. Dec. 1762. 55. V. Per. L. # Reducible to V. P. Per. 56. V. Per. R. # *Rudd’s Prac. Geo. part Iſt, qu. 19. - Reducible to \\ V. B. S + s. 57. V. L. m: n, \\ ſegts. by L. # }Mis. Cur. Math. V. I. qu. 26. 58. V. L. m - n, \\ ſegts. by L. # }L. Diary, 1773, qu. 662.

59. V. L. Side of \\ ins. ◻. # }T. and Country Mag. Nov. and Dec. 1772. 60. V. λ. Ra. of \\ circ. ☉(sun). # }*Gent. Diary, 1766, qu. 282. - Reducible to V. B. λ. 61. B. P. S: s. # D’Omerique L. 1. 49. - Vieta 1ſt. App. Apoll. Galli, pr. 2. - \\ Ghetaldus de Res. & Comp. L. II. pag. 48, 49, &c - Simp. Alg. \\ pr. 23. - Simp. Sel. Ex. pr. 19. - Simp. Geom. pr. 13. - \\ Schooten, pr. 22. - Turner’s Math. Exer. prob. 57. - Hutton’s \\ Miſcel. qu. 58. 62. B. P. S + s. # D’Omerique L. III. 25. - Vieta ib. pr. 3. - Gregorius a S. \\ Vinc. pr. 82, pag. 48. - Anderſon var. prob. 20, Cor. - \\ Simp. Alg. pr. 77. - Simp. Geom. pr. 15. - Arith. Un. \\ Caſtillionei, pr. 9. 63. B. P. S - s. # D’Omerique L. III. 26. - Vieta ib. pr. 4. - Simp. Alg. pr. \\ 76. - Simp. Sel. Ex. pr. 20. - Simp. Geom. pr. 15. 64. B. P. S x s. # D’Omerique L. I. 31. - Vieta ib. pr. 1. - Ghetaldus de Res. \\ & Comp. pag. 52. - Simpſ. Sel. Ex. pr. 21. - Simp. Geom. \\ pr. 16. 65. B. P. A - a. # Simpſ. Alg. pr. 15. 66. B. P. L. # Anderſon var. prob. 21. 67. B. P. Supp. of \\ A=Comp. of a. # }Math. Mag. No. I. pr. 1. 68. B. P: S. S + s. # D’Omerique L. III. 30. 69. B. S - P. s - P. # Ghetaldus de Res. & Comp. pag. 264. 70. B. ∠ of P with S. s. # Imperial Mag. Sep. 1760. 71. B. S±s. A or a. # Reducible to V. B±S. s. 72. B. S: s. A. # Gent. Diary, 1749, qu. 81. 73. B. S: s. A=2a. # Hutton’s Miſcel. qu. 16. 74. B. S: s. A - a. # D’Omerique L. I. 50. - Simp. Alg. pr. 14. 75. B. S±s. A - a # Simp. Alg. pr. 12, 13. 76. B. S + s. m: n. # D’Omerique L. III. 32. - Renaldinus, pag. 331. 77. B. S - s. Ar. # York Mis. Cur. qu. 32. - See B. P. S-s.

78. B. S±s. Side \\ of ins. ◻. # }Reducible to B. P. S±s. 79. B. A. Ar. # Gent. Diary, 1741, qu. 5. 80. B. A - a. λ. # *Simpſ. Algebra, pr. 59. 81. B. A or a. R. # Imperial Mag. Nov. 1760. 82. B. R. Ra. of \\ circums. ☉(sun). # }Mathematician, pr. 66. 83. B±P. A - a. m - n. # L. Diary, qu. 646. 84. B±S. All the \\ angles. # }*Britiſh Oracle, qu. 50. 85. B + s. S. n. # D’Omerique L. III. 29. 86. B - S. S±. m - n. # Oughtred ch. 19, pr. 14. - Ghetaldus var. prob. 5. - Ghe- \\ taldus de Res. & Comp. pag. 66. 87. B + S - s. Ar. Per. # Gent. Diary, 1749, qu. 89. 88. ∠ of B with L. \\ S: s. # }Town and C. Mag. 1769, pag. 606, 662. 89. ∠ of B with L. \\ S: s. λ. # }*Palladium, 1752, qu. 47. 90. ∠ of B with L. \\ m. n, \\ ſegts. by L. # }Ghetaldus var. prob. 4. - Regiomontanus de triang. L. II. \\ 33. - Miſ. Scient. Cur. pr. 27. 91. B x La max. S. s. # L. Diary, 1773, qu. 656. 92. B x λ a max. S. s. # *L. Diary, 1762, qu. 495. 93. P. P. P. # Mis. Cur. Math. Vol. I. pag. 30. - Rudd’s Pract. Geom. part \\ 2d, qu. 43. 94. P. S: s. A - a. # Simp. Alg. pr. 14. 95. P. S±s. A - a. # Simp. Alg. pr. 79. 96. P. S: s. m: n. # D’Omerique L. III. 33. - Simp. Alg. pr. 25. 97. P. S: s. m - n. # Simpſ. Alg. pr. 24. 98. P. S + s. m: n. # D’Omerique L. III. 31.

99. P. S + s. m - n. # Oughtred ch. 19, pr. 15. - Ghetaldus var. prob. 2. - Ghe- \\ taldus de Res. & Comp. pag. 56. - D’Omerique L. III. 27. \\ - Renaldinus, pag. 455, 456. 100. P. S - s. m - n. # Oughtred ch. 19, pr. 16. - Ghetaldus var. prob. 1. - Ghetal- \\ dus de Res. & Comp. pag. 36. - D’Omerique L. III. 28. - \\ Renaldinus, pag. 460. 101. P. A or a. Per. # Reducible to B. S + s. A or a. 102. P. A - a. m: n. # Simp. Alg. pr. 17. 103. P. A - a. m - n. # Simp. Alg. pr. 19. 104. P. m - n. Ra. \\ of circum. ☉(sun). # } Martin’s Mag. qu. 395. 105. P. L. λ. # *Mathematician, pr. 10. - *L. Diary, qu. 270. 106. P: S. P - n. m - n. # Mathematician, pr. 64. 107. P: L. S - s. \\ Ra. of cir@um. ☉(sun). # }Gent. Diary, qu. 363. 108. S. s. Ar. # Mis. Cur. Math. qu. 121. - Saunderſon, art. 333. 109. S. s. λ. # *Mathematician, pr. 9. - *Gent. Diary, 1759, qu. 186. - \\ Simp. Sel. Ex. pr. 33. - Schooten, pr. 23. - *Rudd’s Prac. \\ Geometry, part 2d, qu. 14, 16. 110. S. L. n. \\ or s. L. m, \\ ſegts. by L. # }Br. Oracle, qu. 81. 111. S: s. A. L. # Town and C. Mag. Aug. and Sep. 1770. 112. S: s. A - a. m - n. # Simp. Alg. pr. 14. 113. S±s. A - a. m: n. # Simp. Alg. pr. 17. 114. S±s. A - a. m - n. # Simp. Alg. pr. 16, 18. 115. S: s. m. n. # Simp. Alg. pr. 22. 116. S±s. m. n. # Simp. Alg. pr. 20, 21.

117. S - s. m - n. L. \\ Segts. by L. # }Gent. Mag. 1768, pag. 471, 570. 118. Sxm. s x n. L. \\ Segts. by L. # }L. Diary, qu. 622. 119. S - s. m - n. R. # Br. Oracle, qu. 61. 120. A or a. R. Side \\ of ins. ◻. # } *Simpſ. Sel. Ex. pr. 27. 121. A - a. Per. L. # Simp. Alg. pr. 61. 122. m. n. L. # Simp. Alg. pr. 57. 123. Per. All the \\ angles. # }Mathematician, pr. 44. # ☞ When all the angles are given, the triangle is given in \\ ſpecies, and therefore may be conſtructed, by ſimilar triangles, \\ with any other datum.

Continuation of the Synopsis, Containing ſuch Data as cannot readily be expreſſed by the Symbols before uſed without more words at length.
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~~

~~124. ~~
~~IT is required to conſtruct an iſoceles triangle ſuch, as to have its equal legs ~~
to a given line, and moreover ſuppoſing a circle inſcribed therein, and a di-

ameter thereof drawn parallel to the baſe, and continued to meet the equal legs;

~~ſuch line ſhall divide the area in a given ratio.~~
~~
~~

~~Rudd’s Prac. ~~
~~Geom. ~~
~~part 2d, qu. ~~
~~4~~
6. ~~- Supp. ~~
~~to Gent. ~~
~~Diary, 1741, 1742, qu. ~~
~~II.~~
~~
~~

~~125. ~~
~~In a plane triangle, ABC, there is given the angle ~~

~~to conſtruct the triangle ſo, that if BD be drawn, the ~~
angle ABD may be a maximum, and BC to EC in

a given ratio.

~~
~~

~~L. ~~
~~Diary, 1773, qu. ~~
~~659.~~
~~
~~

~~126. ~~
~~V. ~~
~~B. ~~
~~Line from A or a to divide S or s in a given ratio.~~
~~
~~

~~Town and Coun. ~~
~~Mag. ~~
~~Dec. ~~
~~1772.~~
~~
~~

~~127. ~~
~~V. ~~
~~B. ~~
~~Difference of two lines from the angles at the baſe to the centre of the ~~
inſcribed ☉(sun).

~~
~~

~~*Simpſon’s Sel. ~~
~~Ex. ~~
~~pr. ~~
~~18.~~
~~
~~

~~128. ~~
~~V. ~~
~~B. ~~
~~Two lines from A and a meeting in a point O given in poſition, as alſo ~~
the angle of a line from V to O with either of the ſides about the vertical angle.

~~
~~

~~Mis. ~~
~~Cur. ~~
~~Math. ~~
~~qu. ~~
~~107.~~
~~
~~

~~129. ~~
~~V. ~~
~~A point in B. ~~
~~Rect. ~~
~~of the ſegments of B made by that point.~~
~~
~~

~~Simp. ~~
~~Sel. ~~
~~Ex. ~~
~~pr. ~~
~~44.~~
~~
~~

~~130. ~~
~~Poſition of a line through V. ~~
~~B. ~~
~~S - s.~~
~~
~~

~~Simp. ~~
~~Sel. ~~
~~Ex. ~~
~~pr. ~~
~~49.~~
~~
~~

~~131. ~~
~~Poſition of a line through V. ~~
~~B. ~~
~~S, line biſect. ~~
~~B, & ~~
~~s in geometrical progreſſion.~~
~~
~~

~~Simp. ~~
~~Sel. ~~
~~Ex. ~~
~~pr. ~~
~~51.~~
~~
~~

~~132. ~~
~~V. ~~
~~S. ~~
~~The angle of a line from extreme of S with the baſe, and ſegment of ~~
s cut off thereby adjacent to the baſe.

~~
~~

~~*Mis. ~~
~~Cur. ~~
~~Math. ~~
~~Vol. ~~
~~II. ~~
~~qu. ~~
~~30.~~
~~
~~

~~133. ~~
~~V. ~~
~~The ſquare of the baſe equal to the rectangle of one ſide and a given line.~~
~~
~~

~~General Mag. ~~
~~qu. ~~
~~61.~~
~~
~~

~~134. ~~
~~V. ~~
~~P. ~~
~~The angle of two lines from extremes of B to middle of P.~~
~~
~~

~~Mathematician, pr. ~~
~~65.~~
~~
~~

~~135. ~~
~~V. ~~
~~B. ~~
~~The angle of two lines from extremes of B to the middle of P.~~
~~
~~

~~Brit. ~~
~~Oracle, qu. ~~
~~91.~~
~~
~~

~~136. ~~
~~V. ~~
~~S. ~~
~~The ratio of the ſquare of a line, drawn in a given direction from V to B, ~~
to the rectangle of the ſegments of B made thereby.

~~
~~

~~Anderſoni Exercitationes Math. ~~
~~No. ~~
~~8.~~
~~
~~

~~137. ~~
~~V. ~~
~~S or s. ~~
~~m: ~~
~~n, theſe being ſegts. ~~
~~by a line from V to B dividing V into ~~
given angles.

~~
~~

~~Hutton’s La. ~~
~~Diary, qu. ~~
~~139.~~
~~
~~

~~138. ~~
~~V. ~~
~~A line from V to B dividing V in a given ratio. ~~
~~Area a minimum.~~
~~
~~

~~L. ~~
~~Diary, 1761, qu. ~~
~~479.~~
~~
~~

~~139. ~~
~~V. ~~
~~Line from A biſecting S. ~~
~~Line from a biſecting s.~~
~~
~~

~~*Simpſon’s Sel. ~~
~~Ex. ~~
~~pr. ~~
~~15.~~
~~
~~

~~140. ~~
~~V. ~~
~~Per. ~~
~~Line parallel to B biſecting the area.~~
~~
~~

~~*Gent. ~~
~~Di. ~~
~~1750, qu. ~~
~~98. ~~
~~- Reducible to V. ~~
~~B. ~~
~~S + s.~~
~~
~~

~~141. ~~
~~S. ~~
~~s. ~~
~~Line from V to centre of ins. ~~
~~☉(sun).~~
~~
~~

~~L. ~~
~~Diary, 1771, qu. ~~
~~635.~~
~~
~~

~~142. ~~
~~Baſe. ~~
~~One ſide. ~~
~~Ratio of a line from V, making a given angle with ſaid ſide, ~~
to alternate part of B.

~~
~~

~~Hutton’s Mis. ~~
~~pag. ~~
~~63, Cor.~~
~~
~~

~~143. ~~
~~B. ~~
~~Point of contact therein of ins. ~~
~~☉(sun). ~~
~~mxn.~~
~~
~~

~~Hutton’s L. ~~
~~Diaries, 1722, qu. ~~
~~94.~~
~~
~~

~~144. ~~
~~L. ~~
~~A perpendicular thereto from one of the angles at the baſe. ~~
~~The other ~~
angle at the baſe.

~~
~~

~~L. ~~
~~Diary, 1769, 1770, qu. ~~
~~604.~~
~~
~~

~~145. ~~
~~L. ~~
~~A perpendicular thereto from A. ~~
~~Another from a.~~
~~
~~

~~L. ~~
~~Diary 1768, 1769, qu. ~~
~~588.~~
~~
~~

~~146. ~~
~~One of the angles at the baſe. ~~
~~Perpendicular therefrom to oppoſite ſide a max-~~
imum.

~~A line from the other angle at the baſe biſecting its oppoſite ſide.~~
~~
~~

~~*L. ~~
~~Diary, 1769, 1770, qu. ~~
~~607.~~
~~
~~

~~147. ~~
~~S. ~~
~~s. ~~
~~m: ~~
~~n, theſe being ſegts. ~~
~~by a line from V to B, and the ratio of this line to ~~
m or n.

~~
~~

~~Rudd’s Prac. ~~
~~Geom. ~~
~~part 2d, qu. ~~
~~40.~~
~~
~~

~~148. ~~
~~S. ~~
~~s. ~~
~~Line from V making an angle with S=A.~~
~~
~~

~~Diarian Repoſitory, qu. ~~
~~197.~~
~~
~~

~~149. ~~
~~S. ~~
~~m. ~~
~~n, theſe being ſegments of B by l from V making a given angle with S.~~
~~
~~

~~T. ~~
~~and Country Mag. ~~
~~1769, pag 662.~~
~~
~~

~~150. ~~
~~One angle at the baſe. ~~
~~The ſi le adjacent. ~~
~~The ratio of the other ſide to a line ~~
drawn from V to an unknown point in B, and the length of a line drawn from

the ſaid point parallel to the given ſide to terminate in the unknown ſide.

~~
~~

~~Hutton’s Miſc. ~~
~~qu. ~~
~~23.~~
~~
~~

~~151. ~~
~~A - a. ~~
~~R. ~~
~~Line from centre of ins. ~~
~~circle to middle of B.~~
~~
~~

~~Gent. ~~
~~Diary, 1771 - 2, qu. ~~
~~349.~~
~~
~~

~~152. ~~
~~Three lines from the angles biſecting the oppoſite ſides.~~
~~
~~

~~Mathematician, pr. ~~
~~48. ~~
~~- Simp. ~~
~~Sel. ~~
~~Ex. ~~
~~pr. ~~
~~22. ~~
~~\\ - Palladium, 1752. ~~
~~qu. ~~
~~43.~~
~~
~~

~~153. ~~
~~Sxs: ~~
~~mxn, ſegts by L. ~~
~~m - n. ~~
~~Angle made by L & ~~
~~I biſecting the baſe.~~
~~
~~

~~Brit. ~~
~~Oracle, qu. ~~
~~93.~~
~~
~~

~~154. ~~
~~The baſe of an iſoceles triangle, and the diſtance of the vertical angle from ~~
the foot of a perpendicular from one of the equal angles upon the oppoſite ſide.

~~
~~

~~Brit. ~~
~~Oracle, qu. ~~
~~8.~~
~~
~~

~~155. ~~
~~P - n. ~~
~~m of an iſoceles triangle, m and n being ſegts. ~~
~~by a perpendicular from ~~
one of the angles at the baſe on one of the equal ſides.

~~
~~

~~Court Mag. ~~
~~Sep. ~~
~~1761.~~
~~
~~

~~156. ~~
~~V. ~~
~~Line biſecting A or a. ~~
~~Neareſt diſtance from V to periphery of ins. ~~
~~☉(sun).~~
~~
~~

~~*Gent. ~~
~~Diary, qu. ~~
~~129.~~
~~
~~

~~157. ~~
~~V. ~~
~~The ſegments of S made by a line drawn from A to make a given angle ~~
with B.

~~
~~

~~*Mis. ~~
~~Cur. ~~
~~Math. ~~
~~Vol. ~~
~~I. ~~
~~qu. ~~
~~79.~~
~~
~~

~~158. ~~
~~V. ~~
~~B. ~~
~~Line from A or a to the centre of inſcribed circle.~~
~~
~~

~~Mis. ~~
~~Sci. ~~
~~Cur. ~~
~~qu. ~~
~~53.~~
~~
~~

~~159. ~~
~~B. ~~
~~L. ~~
~~Line from extremity of L parallel to S or s.~~
~~
~~

~~Math. ~~
~~Mag. ~~
~~No. ~~
~~III. ~~
~~prob. ~~
~~7.~~
~~
~~

~~160. ~~
~~V. ~~
~~The ſegments of B made by a line dividing V into given angles.~~
~~
~~

~~Mis. ~~
~~Cur. ~~
~~Math. ~~
~~Vol. ~~
~~II. ~~
~~qu. ~~
~~48.~~
~~
~~

~~161. ~~
~~V. ~~
~~One ſide. ~~
~~Ratio of the ſegments of the baſe made by a line dividing V ~~
into given angles.

~~
~~

~~Palladium, 1756, pa: ~~
~~43.~~
~~
~~

SYNOPSIS
~~
~~~~
~~ 1. H. H x P + P2. 2. H. P2 - n2. 3. H. m2 - P2. 4. H. S±m, or H. s±n. 5. H. S2 + m2, or H. s2 + n2. 6. H. S2 + n2. 7. H. m2 - n2. 8. H. l biſecting A or a. # Univer. Muſeum, July, 1767. - Ladies \\ Diary, 1772, qu. 633, Cor. - Mis. \\ Scient. Curioſa, pag. 196, Cor. II.

9. H2: S x s. Any other datum. 10. H2: S2 + m2, or H2: s2 + n2. any other. 11. H + P. H x P. 12. H + P. H x P + P2. 13. H + P. S2 + n2. 14. H + P. m2 + n2. 15. H x P + P2. P. 16. H x P: S2 + n2. any other. 17. H x P + P2: S2 + n2. any other. 18. {1/2}H - P. S - s. # Turner’s Math. Ex. pr. 37. 19. H±S. m, or H±s. n. 20. H + S. n, or H + s. m. 21. H2 + S2: P2, or H2 + s2: P2. any other. 22. H2 + S2: S2 + m2, or H2 + s2: s2 + n2. any other. 23. H2 + S2. m, or H2 + s2. n. 24. H2 + S2. n, or H2 + s2. m. 25. H2 + S2: m2, or H2 + s2: n2. any other. 26. H x S. s. or Hxs. S. # D’Omerique L. III. pr. 37. - Vieta Geo. \\ Eff. pr. 18. - Oughtred, ch. 19, pr. \\ 25. - Herigon Geo. planæ, pr. 9. - \\ Schooten, pr. 38. 27. H x S: s2, or H x s: S2. any other 28. H + S + m. H2 + S2 + m2. 29. H + S + m. H x S + S x m + S2. 30. H + S + m. H x S + S x m. 31. H + S + m. H2 + m2. 32. H2 + S2 + m2: H x S + S x m. any other. 33. H2 + S2 + m2: H x S + S x m + S2. any other.

34. H2 + S2 + m2. H + m. 35. H2 + S2 + m2: H2 - m2. any other. 36. H2 + S2 + m2: H2 + 2m2. any other. 37. H2 + S2 + m2: P2. any other. 38. H2 + S2 + m2. S. 39. H2 + S2 + m2: s2. any other. 40. H2 + S2 + m2: 2S2 + m2. any other. 41. H2 + S2 + m2.: S2 + 2m2. any other. 42. H2 + S2 + m2: Sxn. any other. 43. H2 + S2 + m2. n. 44. H x S + S x m + S2. H + m. 45. H x S + S x m + S2. S. 46. H + m. H2 - m2. 47. H + m. P. 48. H + m. s. 49. H2 + 2m2: P2. any other. 50. H2±m2: S2. any other. 51. H2 - m2: S2 + 2m2. any other. 52. H2 - m2. n. 53. P. m2 + n2. 54. P2: m2±n2. any other. 55. P + m. n, or P + n. m. 56. P - n. m. 57. m - P. n. 58. P + m. m - n. 59. P + n. m - n. 60. P2 + s2: m2 - P2. any other. 61. P2 + s2: S2 + n2. any other.

62. P2 - n2. m2 - P2. 63. P2 - n2: S2. any other. 64. m2 - P2: s2. any other. 65. m2 - P2: S2 + n2. any other. 66. P2 - n2: S2 + n2. any other. 67. P2 - n2. m. 68. m2 - P2. n. 69. P2 - n2: m2. any other. 70. m2 - P2: n2. any other. 71. P2 - n2. m - n. 72. m2 - P2. m - n. 73. P2 - n2: m2 + 2n2. any other. 74. Pxm - P x n: S2 + n2. any other. 75. S. n, or s. m # Oughtred ch. 19, pr. II. - Ronayne’s Alg. \\ B. II. ch. 2, pr. I - Ghetaldus var. prob. 19. \\ - Idem de Res. & Comp. L. III. pr. 2. - \\ Renaldinus. pag. 518. 76. S. m - n, or s. m - n. # Oughtred ch. 19, pr. 8. - D’Omerique, L. \\ III. pr. 24. - Renaldinus, pag. 412. - Ghe- \\ taldus var. prob. 18. - Idem de Res. & Comp. \\ L. III. pr. I. - Foſter’s Math. Lug. pr. 18. 77. S2: n2, or s2: m2. any other. 78. S2: s2 + n2, or s2: S2 + m2. any other. 79. S2: 2s2 + n2, or s2: 2S2 + m2. any other. 80. S + m. n, or s + n. m. 81. S2 + m2. n, or s2 + n2. m. 82. S2 + n2. m, or S2 + n2. n. 83. S2 + n2: s2 + n2. any other.

84. S2 + n2. m - n. 85. S2 + n2: m2 - n2. any other. 86. S2 + n2: m2 + 2n2. any other. 87. s2 + n2: m2 - n2. any other. 88. m - n. m2 - n2.

~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~~~
~~

~~Of Data for Right-angled Triangles which have not yet ~~
been conſtructed in general, the vertical angle being

ſuppoſed acute or obtuſe.

~~
~~

~~☞ The much greater part of theſe problems are purpoſely leſt without any reference. ~~
~~The Compiler has ſeen an Author from whom Conſtructions to them all may ~~
be derived, but he forbears to name him, in order to leave them as Exerciſes

for young

Geometricians.~~
~~

~~89. ~~
~~Ar. ~~
~~Sides in arithmetical progreſſion.~~
~~
~~

~~Simp. ~~
~~Alg. ~~
~~pr. ~~
~~36. ~~
~~- Simp. ~~
~~Sel. ~~
~~Ex. ~~
~~pr. ~~
~~45.~~
~~
~~

~~90. ~~
~~Ar. ~~
~~Sides in geom. ~~
~~progreffion.~~
~~
~~

~~Wolfiu’s Alg. ~~
~~pr. ~~
~~114. ~~
~~- Simps. ~~
~~Sel. ~~
~~Ex. ~~
~~pr. ~~
~~46. ~~
~~- Vieta Iſt. ~~
~~App. ~~
~~to Apoll. ~~
~~Gallus, pr. ~~
~~7.~~
~~
~~

~~91. ~~
~~Per. ~~
~~Sides in geom. ~~
~~progreſſion.~~
~~
~~

~~Simps. ~~
~~Alg. ~~
~~pr. ~~
~~39.~~
~~
~~

~~92. ~~
~~I biſecting an acute angle. ~~
~~I from right angle biſecting the foregoing given line.~~
~~
~~

~~Gent. ~~
~~Diary, qu. ~~
~~266.~~
~~
~~

~~93. ~~
~~H. ~~
~~Part of S adjacent to the right angle intercepted by a perpendicular to H ~~
from middle of H.

~~
~~

~~L. ~~
~~Diary, qu. ~~
~~633.~~
~~
~~

~~94. ~~
~~One leg and a line parallel thereto intercepted by the hypothenuſe and ~~
the other leg being given;

~~to determine the triangle ſuch, that the rectangle un-~~
der the hypothenuſe and a line from the acute angle, adjacent to the given leg,

to the point of interſection of the parallel and the other leg may be of a given

magnitude.

~~
~~

~~L. ~~
~~Diary. ~~
~~qu. ~~
~~648.~~
~~
~~

~~N. ~~
~~B. ~~
~~In all the Nos. ~~
~~from 28 to 52 incluſive, S may be changed for s, and m ~~
for n, though this be not expreſſed as in others.

~~
~~

FINIS.