Two tangents $BC$ and $BD$ are drawn. Prove that $OB=2BC$

Question

Two tangent segments $BC$ & $BD$ are drawn to a circle with centre $O$ such that $\angle{CBD}=120^{\circ}$. Prove that $OB=2BC$.

What I've tried,

$BC=BD$[two tangents drawn from a single point to the same circle are equal]

$\therefore$ We can prove that $\triangle$ OBC$\cong$$\triangle$OBD

But, this gets me nowhere.

Answer 1

You already found $\Delta OBC \sim \Delta OBD$. Since $\angle CBD = 120^\circ$, you know $\angle CBO = \angle OBD = 60^\circ$. Furthermore, $\angle BCO = \angle BDO = 90^\circ$ since $BC$ and $BD$ are tangent lines. Thus $\Delta OBC$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle, so $OB = 2 \cdot BC$.