In $\triangle ABC$, the internal bisector of $\angle ABC$ and the external bisector of $\angle ACB$ meet at $P$. If $\angle BAC = 40^\circ$ what is the measure of $\angle BPC$?
My try: i) Sum of angles of a triangle is $180^\circ$.
ii) Vertical opposite angles are equal.
We need to find $\angle BPC$. By i) we know $\angle BPC = 180^\circ - \angle PCA - \angle PKC$. So the line pass through points $P$ and $C$ is perpendicular to internal bisector of $\angle ACB$.