$ABCD$ is a rectangle and $P$ is a point in the same plane. If the perpendicular through $C$ to $AP$ and the perpendicular through $B$ to $DP$ intersect at $Q$, prove that $PQ \parallel AD$.
One can easily find a solution using coordinate-bashing but a synthetic solution would be appreciated. Thanks!
Let there be a translation alone $\overline{AD}$ that maps $P$ to $P'$. Then, $Q$ is the orthocenter of $ \triangle BCP' $. Hence, $ \overline{P'Q} \parallel \overline{AD} $ and $ \overline{PP'} \parallel \overline {AD} $, which gives $ \overline {PQ} \parallel \overline {AD} $. $\Box$