If from an external point $B$ of a circle with centre $O$,two tangents $BC$ and $BD$ are drawn such that $\angle DBC = 120^{\circ}$,prove that $BC + BD = BO$,i.e.,$BO = 2BC$.

(N/A) Two tangents $BD$ and $BC$ are drawn from an external point $B$.
To prove: $BO = 2BC$.
Join $OC$,$OD$,and $BO$.
Since $BC$ and $BD$ are tangents to the circle from an external point $B$,the line segment $BO$ bisects the angle $\angle DBC$.
Therefore,$\angle OBC = \angle DBO = \frac{1}{2} \angle DBC = \frac{1}{2} \times 120^{\circ} = 60^{\circ}$.
We know that the radius is perpendicular to the tangent at the point of contact. Thus,$OC \perp BC$ and $OD \perp BD$.
In the right-angled $\triangle OBC$,we have:
$\cos(\angle OBC) = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{BC}{BO}$
$\cos(60^{\circ}) = \frac{BC}{BO}$
$\frac{1}{2} = \frac{BC}{BO}$
$BO = 2BC$.
Since tangents drawn from an external point to a circle are equal in length,$BC = BD$.
Substituting $BD$ for $BC$,we get:
$BO = BC + BD$.