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

Step by Step Explanation:
1. Given:
$PA$ and $PB$ are the tangents to the circle with center $O$.
2. Here, we have to find the value of $\angle APB$.
Let us consider $\angle APB = x^\circ$.

We know that the tangents to a circle from an external point are equal.
So, $PA = PB$.
3. As $\triangle APB$ is an isosceles triangle$(PA = PB)$, the base angles of the triangle will be equal. $$\implies \angle PBA = \angle PAB$$
4. We know that the sum of the angles of a triangle is $180^\circ$. Thus, \begin{aligned} & \angle APB + \angle PAB + \angle PBA = 180^\circ \\ \implies & x^\circ + 2 \angle PAB = 180^\circ && \text{[As, } \angle PBA = \angle PAB] \\ \implies & \angle PAB = \dfrac { 1 }{ 2 } (180^\circ - x^\circ) = \bigg(90^\circ - \dfrac { 1 }{ 2 } x^\circ \bigg) \end{aligned}
5. $PA$ is a tangent and $OA$ is the radius of the circle with center $O$.
We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
\begin{aligned} \implies & \angle OAP = 90^\circ \\ \implies & \angle OAB + \angle PAB = 90^\circ \\ \implies & \angle OAB = 90^\circ - (90^\circ - \dfrac { 1 }{ 2 } x^\circ) \\ \implies & \angle OAB = \dfrac { 1 }{ 2 } x^\circ = \dfrac { 1 }{ 2 } \angle APB \\ \implies & \angle APB = 2 \angle OAB \end{aligned}
6. Hence, $\angle APB = 2 \angle OAB.$ 