Posted: Fri May 10, 2013 11:12 pm
$AD$ is the altitude of an acute $\triangle ABC$. Let $P$ be an arbitrary point on $AD$. $BP,CP$ meet $AC,AB$ at $M,N$ respectively. $MN$ intersects $AD$ at $Q$. $F$ is an arbitrary point on side $AC$. $FQ$ intersects line $CN$ at $E$. Prove that $\angle FDA = \angle EDA$.