Question 121
Let $ABC$ be a triangle and $H$ its orthocenter. The circumscribed circle of $ABH$, center $O$, intersects the line $(BC)$ in $D$, the line $(DH)$ Intersects the line $(AC)$ in $P$. Show that the center of the circumscribed circle of $APD$ is on the circumscribed circle of $ODB$. 