IOM 2018 Problem 6

The incircle of a triangle $ABC$ touches the sides $BC$ and $AC$ at points $D$ and $E$ , respectively. Suppose $P$ is the point on the shorter arc $DE$ of the incircle such that $\angle APE = \angle DPB$ . The segments $AP$ and $BP$ meet the segment $DE$ at points $K$ and $L$ , respectively. Prove that $2KL = DE$ .

Dušan Djukić