OMM 2002 Problem 6

Let $ABCD$ be a quadrilateral with $\measuredangle DAB=\measuredangle ABC=90^{\circ}$ . Denote by $M$ the midpoint of the side $AB$ , and assume that $\measuredangle CMD=90^{\circ}$ . Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$ . The line $AK$ meets $BD$ at $P$ , and the line $BK$ meets $AC$ at $Q$ . Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$ .

[Moderator edit: The proposed solution can be found at http://erdos.fciencias.unam.mx/mexproblem3.pdf .]