Question 92
Let $H$ be the orthocenter of an acute triangle $ABC$, and a circle $\omega$ that passes through all of the points $H$, $A$, $B$ meet the line segment $BC$ at the point $D (\neq B)$. Let $P$ be the intersecting points of the line $DH$ and the line segment $AC$, and $Q$ be the circumcenter of the triangle $ADP$. Show that the center of the circle $\omega$ lies on the circumscribed circle of the triangle $BDQ$.