The diagonals of the inscribed quadrilateral $ABCD$ meet at the point $M$, $\angle AMB = 60^o$. Equilateral triangles $ADK$ and $BCL$ are built outward on sides $AD$ and $BC$. Line $KL$ meets the circle circumscribed ariound $ABCD$ at points $P$ and $Q$. Prove that $PK = LQ$.