Question

If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that $\angle DBC={120}^0$ prove that BC+BD=BO, i.e, BO=2BC.

Answer 1

Sol. Given : A circle with centre O.

Tangents BC and BD are drawn from an external point B such that $\angle DBC={120}^{\circ }$

To prove : $BC+BD=BO,$ i.e., $BO=2BC$

Construction : join $OB,OC$ and $OD$.

Proof : In $\Delta OBC$ and $\Delta OBD$, we have

$OB=OB$ [common]

$OC=OD$ [Raddi of same circle]

$BC=BD$ [Tangents from an external point are equal in length ] __(i)

$\therefore \Delta OBC\cong \Delta OBD$ [By SSS criterion of congruence]

$\Rightarrow \angle OBC=\angle OBD\left(CPCT\right)$

$\therefore \angle OBC=\frac{1}{2}\angle DBC=\frac{1}{2}\times {120}^{\circ }$ [ $\because \angle CBD={120}^{\circ }$ given]

$\Rightarrow \angle OBC={60}^{\circ }$

OC and BC are radius and tangent respectively at contact point C.

So, $\angle OCB={90}^{\circ }$

Now, in right angle $\Delta OBC={60}^{\circ }$

$\therefore cos{60}^{\circ }=\frac{BC}{BO}$

$\Rightarrow \frac{1}{2}=\frac{BC}{BO}$

$\Rightarrow OB=2BC$

Hence, proved (ii) part,

$\Rightarrow OB=BC+BC$

$\Rightarrow OB=BC+BD$ [$\because BC=BD$ from (i)]

Hence, proved.