A history of Greek mathematics - Wilbourhall.org

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.