2012 BMOSL 2012 G6 wrote:
Let $P$ and $Q$ be points inside a triangle $ABC$ such that $\angle PAC = \angle QAB$ and $\angle PBC = \angle QBA$. Let $D$ and $E$ be the feet of the perpendiculars from $P$ to the lines $BC$ and $AC$, and $F$ be the foot of perpendicular from $Q$ to the line $AB$. Let $M$ be intersection of the lines $DE$ and $AB$. Prove that $MP \perp CF$
It is equivalent to proving this Property.
Property:- Let $P$ be a point in the plane of $\triangle ABC$ and $\triangle DEF$ is the Pedal triangle of $P$ WRT $\triangle ABC$, if $EF\cap BC=M$ and let the $P-$ Pedal Circle $(\omega)$ Intersect $BC$ at $K$. Then $AK\perp MP$
Let $T\in AK$ such that $PT\perp AK$. Then Applying Radical Axis Theorem on $\odot(AP),\odot(PK)$ and $\omega$ we get that $EF,PT,BC$ concurs at $M$. Hence, $MP\perp AK$. $\blacksquare$