# 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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