Selection tests for the Balkan Mathematical Olympiad 2013

Because sides of triangles $ABC$ and $DEF$ are parallel (homothetic triangles), we have $$\begin{array}{rl}\angle BPD& =180^{\circ }-\angle EDF-\angle FDB-\frac{1}{2}\angle CBA\\ & =180^{\circ }-\angle BAC-\angle ACB-\frac{1}{2}\angle CBA\\ & =\frac{1}{2}\angle CBA=\angle DBP.\end{array}$$ 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 $AS=AT$.