Question

In fig., O is the center of the circle, PA and PB are tangent segments. Show that the quadrilateral AOBP is cyclic.

Answer 1

Since tangent at a point to a circle is perpendicular to the radius through the point.

Therefore,OA $\perp$ AP and OB $\perp$ to BP

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

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

In quadrilateral OAPB, we have

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

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

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

$\angle$APB + $\angle$AOB = ${180}^{\circ }$ .... (ii)

From equations (i) and (ii), we can say that the quadrilateral AOBP is cyclic.