Question 3
Let $ABC$ be an acute triangle with $\angle BAC=30^\circ$. The internal and external bisectors of $\angle ABC$ meet line $AC$ at $B_1,B_2$, respectively; those of $\angle ACB$ meet line $AB$ at $C_1,C_2$, respectively. Suppose that the circles with diameters $B_1B_2$ and $C_1C_2$ meet at a point $P$ inside triangle $ABC$. Prove $\angle BPC=90^\circ$.
New configuration:
New configuration: