Question

Two tangents PA and PB are drawn to a circle with center O from an external point P. Prove that $\angle APB=2\angle OAB$ [3 MARKS]

Answer 1

Concept: 1 Mark
Application: 2 Mark

$OA\perp PA$ (tangent and the radius at the point of contact)

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

Let $\angle OAB=x$

$\angle PAB={90}^{\circ }-x\dots \left(1\right)$

PA = PB (tangent from same external point)

$△APB$ is isosceles

$\angle PAB=\angle PBA={90}^{\circ }-x$

In $△APB$, By angle sum property

$\angle APB$ + $\angle PAB$ + $\angle PBA={180}^{\circ }$

$\angle P+{90}^{\circ }-x+{90}^{\circ }-x={180}^{\circ }$

$\angle P+{180}^{\circ }-2x={180}^{\circ }$

$\angle P=2x$

$\angle APB=2x$

i.e., $\angle APB=2\angle OAB$