KöMaL 2005 B3816

The centre of the inscribed circle of the triangle ABC is $O$. The extensions of the line segments $AO$, $BO$, $CO$ beyond $O$ intersect the circumscribed circle at $A_1$, $B_1$, $C_1$, respectively. Prove that the area of the triangle $A_1 B_1 C_1$ is $\frac{R^2}{2}(\sin\alpha+\sin\beta+\sin\gamma),$ where $R$ is the radius of the circumscribed circle, and $\alpha$, $\beta$, $\gamma$ are the angles of the original triangle ABC .