Points $C_1, B_1$ on sides $AB, AC$ respectively of triangle $ABC$ are such that $BB_1 \perp CC_1$. Point $X$ lying inside the triangle is such that $\angle XBC = \angle B_1BA, \angle XCB = \angle C_1CA$. Prove that $\angle B_1XC_1 =90^o- \angle A$. (A. Antropov, A. Yakubov)