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Incircle and perpendicular

Source: Iran 3rd round 2012-Geometry exam-P4

September 20, 2012
geometrycircumcircleMITcollegeangle bisectorgeometry proposedIran

Problem Statement

The incircle of triangle ABCABC for which AB≠ACAB\neq AC, is tangent to sides BC,CABC,CA and ABAB in points D,ED,E and FF respectively. Perpendicular from DD to EFEF intersects side ABAB at XX, and the second intersection point of circumcircles of triangles AEFAEF and ABCABC is TT. Prove that TX⊥TFTX\perp TF.
Proposed By Pedram Safaei
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