For, let the circle thus described touch the given straight line at G; join EG meeting the given circle at H,
and join DH. Then the triangles EHD and EGG are similar; and therefore the rectangle EC, ED is equal to the rectangle EG, EH (III. 31, VI. 4, VI. 16). Thus the rectangle EA, EF is equal to the rectangle EH, EG; and therefore H is on the circumference of the described circle (III. 36, Corollary). Take K the centre of the described circle; join KG, KH, and BH. Then it may be shewn that the angles KHG and EHB are equal (I. 29, I. 5). Therefore KHB is a straight line; and therefore the described circle touches the given circle.
Two solutions will be obtained, because there are two solutions of the problem in 6; the circles thus described touch the given circle externally.
By joining DA instead of EA we can obtain two solutions in which the circles described touch the given circle internally.
