In convex quadrilateral $ABCD$, where $AB=AD$, on $BD$ point $K$ is chosen. On $KC$ point $L$ is such that $\bigtriangleup BAD \sim \bigtriangleup BKL$. Line parallel to $DL$ and passes through $K$, intersect $CD$ at $P$. Prove that $\angle APK = \angle LBC$. @below edited