CMO 2017

Let $\angle ABC=m$ and let $O$ be the circumcenter of triangle $DPQ$. Since $P$ and $Q$ are in the interior of $ABCD$, it follows that $m=\angle ABC>60^{\circ }$ and $\angle DAB=180^{\circ }-m>60^{\circ }$ which together imply that $60^{\circ }<m<120^{\circ }$. Now note that $\angle DAP=\angle DAB-60^{\circ }=120^{\circ }-m$, $\angle DCQ=\angle DCB-60^{\circ }=120^{\circ }-m$ and that $\angle PBQ=60^{\circ }-\angle ABQ=60^{\circ }-(\angle ABC-60^{\circ })=120^{\circ }-m$. This combined with the facts that $AD=BQ=CQ$ and $AP=BP=CD$ implies that triangles $DAP$, $QBP$ and $QCD$ are congruent. Therefore $DP=PQ=DQ$ and triangle $DPQ$ is equilateral. This implies that $\angle ODA=\angle PDA+30^{\circ }=\angle DQC+30^{\circ }=\angle OQC$. Combining this fact with $OQ=OD$ and $CQ=AD$ implies that triangles $ODA$ and $OQC$ are congruent. Therefore $OA=OC$ and, if $M$ is the midpoint of segment $AC$, it follows that $OM$ is perpendicular to $AC$. Since $ABCD$ is a parallelogram, $M$ is also the midpoint of $DB$. If $K$ denotes the intersection of the line through $P$ perpendicular to $DP$ and the line through $Q$ perpendicular to $DQ$, then $K$ is diametrically opposite $D$ on the circumcircle of $DPQ$ and $O$ is the midpoint of segment $DK$. This implies that $OM$ is a midline of triangle $DBK$ and hence that $BK$ is parallel to $OM$ which is perpendicular to $AC$. Therefore $K$ lies on the altitude from $B$ in triangle $ABC$, as desired. $\square$