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: $2(\angle ABE +\angle CBF)=\angle AOC$.

My Attempt:

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.

  • $\begingroup$ So $A$ and $B$ lie on the circle, while $C$ could be inside the circle (for instance, $OC$ could be half the radius). Have I understood correctly? Can $C$ be anywhere, or are there restrictions? $\endgroup$ – Arthur Oct 16 '16 at 6:14
  • $\begingroup$ @Arthur, Sorry for the typo. All these points $A,B and C$ lie on the circumference of circle. $\endgroup$ – pi-π Oct 16 '16 at 6:19
  • $\begingroup$ @Arthur, Please check the updated version $\endgroup$ – pi-π Oct 16 '16 at 6:20
  • $\begingroup$ Is this your homework? $\endgroup$ – user261263 Oct 16 '16 at 6:31
  • $\begingroup$ ##Eugen Covaci, No. Its not. $\endgroup$ – pi-π Oct 16 '16 at 6:33

Let $D$ be a point on the circle on the arc $AC$ not containing $B$

We know that $\angle ADC = \frac{1}{2} \cdot \angle AOC$.

Connect $BC$. By Alternate Segment Theorem, $\angle EBA = \angle ADB$ and $\angle FBC = \angle CDB$. Thus $\angle ADC = \angle EBA + \angle FBC$ which implies that $\angle AOC = 2(\angle EBA + \angle FBC)$


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