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Geometry Difficulty 7.6 National olympiad, round 2 Prove it Saudi Arabia

Let $ABC$ be a triangle with incenter $I$ , and let $D, E, F$ be the midpoints of sides $BC, CA, AB$ , respectively. Lines $BI$ and $DE$ meet at $P$ , and lines $CI$ and $DF$ meet at $Q$ . Line $PQ$ meets sides $AB$ and $AC$ at $T$ and $S$ , respectively. Prove that $AS = AT$ .

Solution

Because sides of triangles $ABC$ and $DEF$ are parallel (homothetic triangles), we have
$\begin{aligned} \angle BPD & = 180^\circ - \angle EDF - \angle FDB - \frac{1}{2} \angle CBA \\ & = 180^\circ - \angle BAC - \angle ACB - \frac{1}{2} \angle CBA \\ & = \frac{1}{2} \angle CBA = \angle DBP. \end{aligned}$
We deduce that triangle $DPB$ is isosceles and therefore $DP = DB$ .

Similarly, we prove that $DQ = DC$ . But $DB = DC$ . We conclude that triangle $DPQ$ is isosceles, and therefore, its bisector at $D$ is perpendicular to $PQ$ .

But the bisector of triangle $ABC$ at $A$ is parallel to the bisector of triangle $DEF$ at $E$ since the two triangles are homothetic. We deduce that the bisector of triangle $ATS$ at $A$ is perpendicular to $ST$ and therefore $AS = AT$ .

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