A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

ON CONTACTS OR TANGENCIES 183
solved, i.e. suppose DA, FA drawn to the circle cutting it in
points B, C such that BC produced passes through F.
Draw BG parallel to DF; join GC
and produce it to meet DF in H.
Then
IBAC= IBGC
= ICHF
= supplement of Z CHD
;
therefore A, D, H, C lie on a circle, and
DF.FH=AF.FC.
Now AE .EC is given, being equal to the square on the
tangent from E to the circle ; and DF is given ; therefore HE
is given, and therefore the point H.
But F is also given ; therefore the problem is reduced to
drawing HC, FC to meet the circle in such a way that, if
HC, FC produced meet the circle again in G, B, the straight
line BG is parallel to HF: a problem which Pappus has
previously solved. 1
Suppose this done, and draw BK the tangent at B meeting
HF in K. Then
Z KBC — ABGC, in the alternate segment,
= ICHF.
Also the angle CFK is common to the two triangles KBF,
CHF\ therefore the triangles are similar, and
CF:FH = KF:FB,
or
HF.FK = BF.FC.
Now BF .FC is given, and so is HF;
therefore FK is given, and therefore K is given.
The synthesis is as follows. Take a point H on DE such
that DE . EH is equal to the square on the tangent from E to
the circle.
Next take K on HF such that HF . FK = the square on the
tangent from F to the circle.
Draw the tangent to the circle from K, and let B be the
point of contact. Join BF meeting the circle in C, and join
1
Pappus, vii, pp. 830-2.
184 APOLLONIUS OF PERGA
HG meeting the circle again in G. It is then easy to prove
that BG is parallel to DF.
Now join EG, and produce it to meet the circle again at A
;
join AB.
We have only to prove that AB, BD are in one straight line.
Since DE .
EH = AE .EG, the points A, D, H, G are concyclic.
Now the angle GHF, which is the supplement of the angle
GHD, is equal to the angle BGG, and therefore to the
angle BAG.
Therefore the angle BAG is equal to the supplement of
angle DUG, so that the angle BAG is equal to the angle DAG,
and AB, BD are in a straight line.
The problem of Apollonius is now easy. We will take the
case in which the required circle touches all the three given
circles externally as shown in the figure.
Let the radii of the