Question 109
Let $ABC$ be a right-angled triangle with a hypotenuse $AB$ and longer leg $BC$. Let $D$ be a foot of an altitude from the vertex $C$. Circle $k$ with the center $D$ and the radius $CD$ intersects the leg $BC$ in a point $Q$ and line $AB$ in points $E$ and $F$ ($E \neq F$), where $F$ is a point on the hypotenuse $AB$. Segment $QE$ intersects the leg $AC$ in a point $P$. Prove that $PE = QF$. (Jaroslav Švrček)