Question

Two tangents PA and PB are drawn to the circle with centre O, such that $\angle$APB $={120}^{\circ }$. Prove that
OP = 2AP ?

Answer 1

In $△$OPB and $△$OPA,

$OA=OB$ [radius]

$\angle OBP=\angle OAP={90}^{\circ }$

$OP=OP$ [common]

$\Rightarrow △OPB\cong △OPA$
[RHS congruency]

Given, $\angle$APB = ${120}^{\circ }$

$\Rightarrow \angle$APO $=\angle$OPB $={60}^{\circ }$ [since, $△OPB\cong △OPA$]

In $△$OPA,

$\cos {60}^{\circ }=\frac{AP}{OP}$
$\Rightarrow \frac{1}{2}=\frac{AP}{OP}$

$\Rightarrow OP=2AP$

Hence Proved