Full text: Pergaeus, Apollonius: The two books of Apollonius Pergaeus, concerning tangencies, as they have been restored by Franciscus Vieta and Marinus Ghetaldus

17. 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.

18. 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.

19. 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.

20. 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.

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