Bulgarian Winter Tournament

From the fact that $C$ lies on the circle circumscribed about $△FEQ$ and $CE=CF$ it follows that $CQ$ is an exterior bisector of $\angle FQE$ and it remains to prove that $QP$ is a bisector of $\angle FQE$, i.e. $QF:QE=FP:PE$. From $\angle QFC=\angle QEC$ and $\angle QAC=\angle QBC$ it follows that $△QFA\sim △QEB$, i.e. $$QF:QE=AF:BE=AD:BD.$$ Let $I$ be the center of the circle $k$ inscribed in $△ABC$ and the line $DP$ intersects $k$ for the second time at point $R$. Then $$\angle REF=\angle RDF=90^{\circ }-\angle DFE=90^{\circ }-\angle DIB=\angle IBA$$ and analogously $\angle RFE=\angle IAB$, i.e. $△FER\sim △ABI$. But $RP\perp EF$, $ID\perp AB$, i.e. $P$ and $D$ are corresponding elements in similar triangles and $FP:PE=AD:BD$, which completes the proof. $\square$