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Angle Equity

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 $\angle FDA = \angle EDA$.

Re: Angle Equity

Posted: Sat May 11, 2013 3:20 pm
by Nadim Ul Abrar
Note that ,
$A,Q,P,D$ are harmonic .
Now pencil $C(A,Q,P,D)$ imply , $F,Q,E,G(FQ \cap BC)$ are harmonic . $AD \perp BC$ lead to desired result .