Czech-Slovak-Polish Match 2004 P-3

[Phlembac Adib Hasan]

In the interior of a cyclic quad $ABCD$, a point $P$ is given such that $\angle BPC=\angle BAP+ \angle PDC$. Denote by $E, F, G$ the feet of perpendiculars from $P$ to $AB,AD,CD$, respectively. Prove that $\triangle PBC\sim \triangle FEG$

Hint:

Draw circumcircles of $PAB$ and $PCD$

[asif e elahi]

Let $S$ be a point on $BC$ such that $\angle BPS=\angle BAP$.Then $\angle CPS=\angle CDP$.As $\angle BPS=\angle BAP$,by alternate segment theorem $PS$ is tangebt to the circumcircle of $\bigtriangleup APB$.Similarly $PS$ is tanget to the circumcircle of $\bigtriangleup CPD$.So $\bigodot APB$ and $\bigodot CPD$ touches each other at $P$.Let $AB$ intersects $CD$ at $Q$.As $ABCD$ cyclic,$QA\times QB=QD\times QC$.So $Q$ lies on the radical axis of $\bigtriangleup APB$ and $\bigtriangleup CPD$.So $S,P,Q$ collinear. Let $PQ$ intersects $\bigodot BCQ$ at $Q$ and $R$.Again $\angle QEP=\angle QGP=90^{\circ}$.So $EPGQ$ cyclic.
$\angle PEG=\angle PQG=\angle RQC$
As $\angle AEP+\angle AGP=90^{\circ}+90^{\circ}=180^{\circ}$,$AEPF$ cyclic.So $\angle BPR=\angle BAP=\angle EAP=\angle EFP$
$\angle BRQ=\angle BCQ=\angle QAD= \angle QAF=\angle EPF$
So in $\bigtriangleup BRP$ and $\bigtriangleup EFP$, $\angle BRP=\angle EPF$ and $BPR=EFP$.So $\angle RBP=\angle PEF$.Again $\angle RBC=\angle PEG$.So $\angle PBC=\angle EFG$.Similarly we can prove $\angle PCB=\angle EGF$.
So $\bigtriangleup BPC\sim \bigtriangleup EFG$.