Let the triangle $ABC$ be acute. Let us take in the segment $BC$ two points $F$ and $G$ such that $BG > BF = GC$ and an interior point$ P$ to the triangle on the bisector of $\angle BAC$. Then are drawn through $P$, $PD\parallel AB$ and $PE \parallel AC$, $D \in AC$ and $E \in AB$, $\angle FEP = \angle PDG$. prove that $\vartriangle ABC$ is isosceles.