Tangents $OA$ and $OB$ are drawn to a circle from an external point $O$. Through the point $A$, a chord $AC$ is drawn parallel to the tangent $OB$ and $OC$ passes through the circle at $E$.
I am required to show that A$F$ bisects $OB$.
My approach thus far has been to observe that triangles $AFO$ and $OFE$ are similar, and also $AEB$ is similar to $BEO$.
However I am not able to reach an appropriate proof... Please help!
Since $\angle EAO = \angle EOF$ we see that the line $FO$ is tangent to circumcircle $(AEO)$, so by PoP with respect to $F$ and $(AEO)$ we have $$FO^2 = FE\cdot FA$$
and by PoP on a given circle we have $$FB^2 = FE\cdot FA$$
so $FB = FO$.