Question

Question 4
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.

Answer 1

Two tangents BD and BC are drawn from an external point B.

To Prove: BO = 2BC
Given: $\angle$ DBC = ${120}^{\circ }$
Join OC, OD and BO.
Since, BC and BD are tangents.
OC $\perp$ BC and OD $\perp$ BD
We know, OB is an angle bisector of $\angle$ DBC.
$\therefore \angle OBC=\angle DBO={60}^{\circ }$
In right angle $\angle OBC,cos{60}^{\circ }=\frac{BC}{OB}$
$\Rightarrow \frac{1}{2}=\frac{BC}{OB}$
$\Rightarrow$ OB = 2BC
Also, BC = BD.
[Tangent drawn from internal point to circle are equal]
$\therefore$ OB =BC+BC
$\Rightarrow$ OB = BC + BD