Question 49
In an acute triangle $ABC$, the extension of the altitude $AD$ over $D$ intersects the circumcircle at $E$. The midpoint of $CE$ is $F$. The circumcircles of $ABC$ and $DEF$ intersect at $G \neq E$. The foot of the altitude drawn from $A$ to $FG$ is $P$. Prove that $DA = DP$.