Question

In figure, $PA$ and $PB$ are tangents from a point $P$ to a circle with centre $O$. Then the quadrilateral $OAPB$ must be a:

• A. square
• B. rhombus
• C. cyclic quadrilateral
• D. parallelogram

Answer 1

The correct option is C cyclic quadrilateral

A quadrilateral is cyclic if only opposites angles are supplementary.

Considering the angles of the quadrilateral $OAPB$.

The internal angles of a quadrilateral are $\angle OAP$,$\angle APB$,$\angle PBO$,and $\angle BOA$

Since in a quadrilateral, the sum of interior angles is ${360}^{\circ }$ we have,

$\angle OAP+\angle APB+\angle PBO+\angle BOA={360}^{\circ }$ ....(i)

we can rewrite the above equation as,

$\angle OAP+\angle OBP+\angle APB+\angle BOA={360}^{\circ }$ ....(ii)

Since $PA$ and $PB$ are the tangents to the circle, we have

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

Thus,

$\angle OAP+\angle OBP={180}^{\circ }$ .....(iii)

Substituting the value from equation (iii) in equation (ii), we have

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

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

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

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

Since ,$\angle APB+\angle BOA={180}^{\circ }$, the angles are supplementary.

Thus, the opposite angles in the quadrilateral $OAPB$, the angles $\angle APB$ and $\angle BOA$ and the angles $\angle OAP$ and $\angle OBP$ are supplementary.

Thus,the quadrilateral $OAPB$ is cyclic quadrilateral