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

Step by Step Explanation:
1. Given:
$PA$ and $PB$ are the tangents to the circle with center $O$ from an external point $P$.

Here, we have to check if quadrilateral $AOBP$ is cyclic or not.

We know that in a cyclic quadrilateral the sum of opposite angles is $180^\circ$.
2. Also, the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Thus, \begin{aligned} & PA \perp OA \implies \angle OAP = 90^\circ \\ & \text{ and } \\ & PB \perp OB \implies \angle OBP = 90^\circ \end{aligned} So, $$\angle OAP + \angle OBP = 90^\circ + 90^\circ = 180^\circ \ldots \text{(i)}$$
3. Now, the sum of all the angles of a quadrilateral is $360^\circ$. Thus, \begin{aligned} & \angle AOB + \angle OAP + \angle APB + \angle OBP = 360^\circ \\ & \angle AOB + \angle APB + ( \angle OAP + \angle OBP )= 360^\circ \\ \implies & \angle AOB + \angle APB = 180^\circ && \text{ [Using } eq \text{(i)] } \end{aligned}
4. Both pairs of opposite angles have the sum $180^\circ$. Thus, we can say that quadrilateral $AOBP$ is $cyclic.$