Let $K, L$ be the midpoints of the diagonals $AC, BD$, respectively. Then $KN, LM$ are the midlines of triangles $ABC, ABD$, respectively. Therefore, $MKNL$ is a rhombus. and so the points $K$ and $L$ are equidistant from $M$ and $N$, so, they lie on the line $PQ$. Moreover, $\angle BPQ= \angle CQP$. Let the line through the point $D$ parallel to $PQ$ intersects $AB$ at the point $X$. Then $BP = PX$. In addition, $XPQD$ is a isosceles trapezoid with equal angles at the vertices $P, Q$. Therefore, $QD = PX = BP$ so $AP = CQ$.

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