37th Iranian Mathematical Olympiad

1. Extend $AP$ and $DP$ until they meet $DC$ and $AB$ at $X$ and $Y$, respectively. Since $\angle PAB=\angle PDC=90^{\circ }$, $DAYX$ is a cyclic quadrilateral. Further, $DAYL$ and $DXLA$ are both cyclic since $$\angle XAL=\angle XDL=\angle YAL=\angle YDL=45^{\circ }.$$ Hence, $DAYLX$ is cyclic and $L$ is the midpoint of the small arc $XY$. Moreover, $XY$ is antiparallel to $DA$ which is parallel to $BC$. Thus, $XY$ is parallel to $BC$. It follows that $K$ is the midpoint of $XY$. However, $KL\perp XY\Rightarrow KL\perp BC$.