$ M$ is the midpoint of base $ BC$ in a trapezoid $ ABCD$. A point $ P$ is chosen on the base $ AD$. The line $ PM$ meets the line $ CD$ at a point $ Q$ such that $ C$ lies between $ Q$ and $ D$. The perpendicular to the bases drawn through $ P$ meets the line $ BQ$ at $ K$. Prove that $ \angle QBC = \angle KDA$.
Proposed by S. Berlov
Such an easy problem
Let $ N \equiv QB \cap AD.$ Then $ QM$ is the Q-median of $ \triangle QBC.$ Since $ BC \parallel ND,$ then $ P$ is midpoint of $ ND.$ The triangle $ \triangle NDK$ is isosceles. So if $ R \equiv KD \cap BC,$ then $ \triangle KBR$ is also isosceles $ \Longrightarrow$ $ \angle QBC = \angle BRK = \angle KDA.$
Let $B' = QB%Error. "capAD" is a bad command.
$ and $R = KD%Error. "capBC" is a bad command.
$.
Notice that we only need to prove that $\triangle KB'D$ is isosceles, because if this happens, $\triangle KBM$ is also isosceles(because $B'D // BC$), and so $\angle KBR=\angle QBC = \angle KRB = \angle CRD=\angle RDA=\angle KDA$.
But $BC // B'D$, and so there're an homothety that goes from $\triangle QBC$ to $\triangle QND$, and if $M$ is the midpoint of $BC$, $P$ is the midpoint of $B'D$, what implies that $\triangle KND$ is isosceles.
Q.E.D.
From Luis's solution we get $KP$ external bisector of angle $\angle QKD$, hence the parallel to the bases of trapezoid through $K$ is the internal angle bisector of $\angle QKD$, done.
Best regards,
sunken rock