Question

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

Answer 1

REF.image

To Prove: OP = 2AP

In $\Delta OAPand\Delta OBP$

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

(radius is perpendicular to tangent)

OA = OB ....(radius)

OP = OP ....(common side)

$\therefore \Delta OAP\cong \Delta OBP...$ (SAS test of congruence)

$\angle OPA=\angle OBP=\frac{{120}^{\circ }}{2}={60}^{\circ }$ ...(cpct)

In $\Delta OAP,$

$cos\left(\angle OPA\right)=cos{60}^{\circ }=\frac{AP}{OP}$

$\frac{1}{2}=\frac{AP}{OP}$

$\therefore OP=2AP$