MathDB
Turkey TST 1998 Problem 5, AF=AC

Source: Turkey TST 1998 Problem 5

December 2, 2011
geometry proposedgeometry

Problem Statement

In a triangle ABCABC, the circle through CC touching ABAB at AA and the circle through BB touching ACAC at AA have different radii and meet again at DD. Let EE be the point on the ray ABAB such that AB=BEAB = BE. The circle through AA, DD, EE intersect the ray CACA again at FF . Prove that AF=ACAF = AC.
Back to ProblemsView on AoPS