Post Number:#2 by ahmedittihad » Wed Jan 11, 2017 4:55 pm
This was a nice problem. Let $C'$ be the reflection of $C$ w.r.t $BP$. Now, $\angle C'BC = 2*10$. So, $\angle QBC'=80-20=60$. As $QB=6, BC'=12$ and $\angle QBC' =60$ we see that $\triangle QBC'$ is a $30-60-90$ triangle. So, $\angle BQC'=90$. We get, $BQC'P$ is cyclic. So, $\angle CPQ= \angle BPC+\angle BPQ=\angle BPC+BC'Q=90+30=120$.
Q.E.D
Frankly, my dear, I don't give a damn.