In a circle, $\ O$ is the centre of the circle. $\ O$, $\ A$, $\ B$ and $\ C$ are joined consecutively such that $\ OABC$ quad is formed inside the circle with $OA=OC$, being the radii of the circle. A tanget to the circle $EF$, passes through the point of contact $B$. Then prove that: $2(\angle ABE +\angle CBF)=\angle AOC$.
I tried by extending $A$ and $C$ to a point $P$ on the circumference of the circle, but it did not work.
Please help me to prove this.
Help much appreciated.