Let $ABC$ be an acute triangle and $A_1, B_1$ and $C_1$, points on the sides $BC, CA$ and $AB$, respectively, such that $CB_1 = A_1B_1$ and $BC_1 = A_1C_1$. Let $D$ be the symmetric of $A_1$ with respect to $B_1C_1, O$ and $O_1$ are the circumcenters of triangles $ABC$ and $A_1B_1C_1$, respectively. If $A \ne D, O \ne O_1$ and $AD$ is perpendicular to $OO_1$, prove that $AB = AC$.