A history of Greek mathematics - Wilbourhall.org

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