Just angle-chasing. Let $\angle{PAB}=x,\angle{QAB}=y\implies \angle{BPQ}=\angle{CPQ}=x,\angle{BQP}=\angle{CQP}=y$. So $A,P,Q,C$ concyclic and thus $\angle{BAQ}=y=\angle{CQP}=\angle{CAP}$.

The A-Humpty point lemma.

Dear Mathlinkers, a link with http://www.artofproblemsolving.com/Forum/viewtopic.php?p=9076. can be helpfull. Sincerely Jean-Louis

ThE-dArK-lOrD wrote: Just angle-chasing, we let $\angle{PAB}=x,\angle{QAB}=y\Rightarrow \angle{BPQ}=\angle{CPQ}=x,\angle{BQP}=\angle{CQP}=y$ So $A,P,Q,C$ concyclic, we get $\angle{BAQ}=y=\angle{CQP}=\angle{CAP}$, we are done. ThE-dArK-lOrD wrote: Just angle-chasing, we let $\angle{PAB}=x,\angle{QAB}=y\Rightarrow \angle{BPQ}=\angle{CPQ}=x,\angle{BQP}=\angle{CQP}=y$ So $A,P,Q,C$ concyclic, we get $\angle{BAQ}=y=\angle{CQP}=\angle{CAP}$, we are done. can you explain why A,P,Q and C are concyclic?

WizardMath wrote: The A-Humpty point lemma. can you show me what is that lemma,i couldn't find the source!!!

Actually, the name is not that well known, but anyways, here's the link: https://drive.google.com/file/d/0B3gLVLnxtyRvejlpenRmZGh6SDQ/view See the one in the section on Symmedian related properties.

please add figure here.

http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYS8yLzQ1ZTcyZDUyOTE1YTE0MDM2ZGRjZmZiZDE4NzM2N2RmNjM2OGY1LmpwZWc=&rn=aW1hZ2UuanBlZw==

anantmudgal09 wrote: Two circles $C_1$ and $C_2$ intersect each other at points $A$ and $B$. Their external common tangent (closer to $B$) touches $C_1$ at $P$ and $C_2$ at $Q$. Let $C$ be the reflection of $B$ in line $PQ$. Prove that $\angle CAP=\angle BAQ$. What is meant by the reflection of a point in a line ??

Very easy to proceed, after noticing that $B$ is the $A-\text{HM}$ point in $\Delta APQ$, Anyways, here's a solution with just angle chasing, $\angle PCQ=\angle PBQ=180^{\circ}-\angle BPQ - \angle BQP=180^{\circ}-\angle BAP-\angle BAQ=180^{\circ}-A$ $\implies$ $APCQ$ is cyclic, $\angle BAQ$ $=$ $\angle BQP$ $=$ $\angle PQC$ $=$ $\angle PAC$

Here's a solution that uses only angle chasing. Let $\angle BPQ=\theta$, and $\angle BQP=\phi$. Construct the lines $CP$ and $CQ$. We have by congruency, $\angle PQB=\angle PQC=\phi$. Similarly, we have $\angle QPB-\angle QPC=\theta$. Now because $PQ$ is tangent to $\mathbf{C_1}$ and $\mathbf{C_2}$ at $P$ and $Q$ respectively, we have $\angle PAB=\theta$ and $\angle BAQ=\phi$. But $\angle PCQ=180^{\circ}-\theta-\phi$. So, we have that points $A,P,C,Q$ are concyclic. Now, we have $\angle PQC=\angle PAC=\angle CAP=\phi$, and $\angle BAQ=\phi$ too. Our proof is thus complete.