Question 107
Given a Triangle $ABC$ with $\angle BAC = 60^\circ$. Let $BB_1$ and $CC_1$ Be the Angle bisectors in this triangle. Prove that the point symmetric to $A$ with respect to $B_1C_1$ lies on side $BC$. in a triangle $ABC$ angle $A$ equal to $60^\circ$. Let him go $BB_1$ and $CC_1$ - the bisectors of this triangle. Prove that the point symmetrical to the vertex $A$ relatively straight $B_1C_1$, lies on the side $BC$. (D. Prokopenko)