Question

In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB. Also, CO is joined and produced to meet the circle D. If $\angle$ACD = $y^{\circ }$ and $\angle$AOD = $x^{\circ }$, prove that x = 3y.

Answer 1

We have: $OB=BC$, $\angle BOC=\angle BCO=y$
External $\angle OBA=\angle BOC+\angle BCO=\left(2y\right)$
Again, OA = OB, $\angle OAB=\angle OBA=\left(2y\right)$
External $\angle AOD=\angle OAC+\angle ACO$
Or x = $\angle OAB+\angle BCO$
Or x = (2 y) + y = 3y
Hence, x = 3y