Problem

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

Tags: geometry, circumcircle, MIT, college, angle bisector, geometry proposed, Iran



The incircle of triangle $ABC$ for which $AB\neq AC$, is tangent to sides $BC,CA$ and $AB$ in points $D,E$ and $F$ respectively. Perpendicular from $D$ to $EF$ intersects side $AB$ at $X$, and the second intersection point of circumcircles of triangles $AEF$ and $ABC$ is $T$. Prove that $TX\perp TF$. Proposed By Pedram Safaei