, 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. $\square$