BALKAN-SL 2009 G1

In the triangle $ABC, \angle BAC$ is acute, the angle bisector of $\angle BAC$ meets $BC$ at $D, K$ is the foot of the perpendicular from $B$ to $AC$ , and $\angle ADB = 45^o$ . Point $P$ lies between $K$ and $C$ such that $\angle KDP = 30^o$ . Point $Q$ lies on the ray $DP$ such that $DQ = DK$ . The perpendicular at $P$ to $AC$ meets $KD$ at $L$ . Prove that $PL^2 = DQ \cdot PQ$ .