Question

$PA$ and $PB$ are tangents to the circle with centre $O$ from an external point $P$, touching the circle at $A$ and $B$ respectively. Show that the quadrilateral $AOBP$ is cyclic.

Answer 1

Since tangents to a circle is perpendicular to the radius .

$\therefore OA$ is perpendicular $AP$ and $OB$ is perpendicular $BP$

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

$\Rightarrow \angle OAP+\angle OBP={90}^{\circ }+{90}^{\circ }={180}^{\circ }$ ......$\left(1\right)$

In quadrilateral $OAPB$,

$\angle OAP+\angle APB+\angle AOB+\angle OBP={360}^{\circ }$

$\Rightarrow (\angle APB+\angle AOB)+(\angle OAP+\angle OBP)={360}^{\circ }$

$\Rightarrow \angle APB+\angle AOB+{180}^{\circ }={360}^{\circ }$

$\Rightarrow \angle APB+\angle AOB={360}^{\circ }-{180}^{\circ }={180}^{\circ }$ ......$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, the quadrilateral $AOBP$ is cyclic.