In the trapezoid $ABCD$ , the base $AB$ is smaller than the $CD$ base. The point $K$ is chosen such that $AK$ is parallel to BC and $BK$ is parallel to $AD$. The points $P$ and $Q$ are chosen on the $AK$ and $BK$ rays respectively, such that $\angle ADP = \angle BCK$ and $\angle BCQ = \angle ADK$. (a) Show that the lines $AD, BC$ and $PQ$ go through the same point. (b) Assuming that the circumscribed circumferences of the $APD$ and $BCQ$ triangles intersect at two points, show that one of those points belongs to the line $PQ$.