Circle with center $I$, inscribed in a triangle $ABC$ , touches the sides $BC$ and $AC$ at points $A_1$ and $B_1$ respectively. On rays $A_1I$ and $B_1I$, respectively, let be the points $A_2$ and $B_2$ such that $IA_2=IB_2=R$, where $R$is the radius of the circumscribed circle of the triangle $ABC$. Prove that: a) $AA_2 = BB_2 = OI$ where $O$ is the center of the circumscribed circle of the triangle $ABC$, b) lines $AA_2$ and $BB_2$ intersect on the circumcircle of the triangle $ABC$.