As the figure shows, rotate $\triangle ABC$ by $50^o$ counterclockwise to get $\triangle PQC$. Find $\angle BQC$.
By $\angle ABC = \angle PQC$, we know that points $P, B, Q, C$ form a circle, finding $\angle BQC$ is equivalent as finding arc $BPC$.
That's all I've got. How can I continue?
Since $\triangle CAP$ is an isosceles and $P$ is on $AB$, $\angle APC=65^\circ$.
So $\angle BQC = \angle APC = 65^\circ$.
Proof) $P,B,Q,C \text{ concyclic} \implies \angle BQC = \angle APC$:
The sum of the central angles of arcs $BPC$ and $BQC$ is $360^\circ$, so the sum of their inscribed angles $\angle BQC + \angle BPC$ would be one half of $360^\circ$, which is $180^\circ$. (By the inscribed angle theorem)
Also, looking at the angles around $P$, we see $\angle BPC + \angle APC = 180^\circ$.
Combining those two yields $\angle BQC = \angle APC$.