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