Point $M$ is the midpoint of the base $BC$ of trapezoid $ABCD$. On base $AD$, point $P$ is selected. Line $PM$ intersects line $DC$ at point $Q$, and the perpendicular from $P$ on the bases intersects line $BQ$ at point $K$. Prove that $\angle QBC = \angle KDA$.
My solution:
Let $BQ$ cuts $AD$ at $H$. So by Thales's theorem,
$\frac{PD}{MC}=\frac{QP}{QM}=\frac{HP}{BM}$. But $BM=MC$, so $HP=PD$. We also have that $KP\perp HD$, hence $\Delta KHD$ is isoceles at $K$. So $\angle KDA=\angle KHD=\angle QBC$
Sincerely,
XH
Let $QB$ intersects $AD$ at $X$ .Enough to show: $\angle KXD=\angle KDX$ or $P$ is the midpoint of $XD$ .notice that $\triangle XQD$ and $\triangle QBC$ are homothetic and $Q,P,M$ are collinear so $P$ is the midpoint of $XD$..