Given : $PA$ and $PB$ are tangents from external point $P$ to a circle with centre $O$.
To Prove : $\angle APB = 2\angle OAB$.

Step 1: Let $\angle APB = \theta$.
Step 2: Tangents from an external point are equal, so $PA = PB$.
Therefore $\triangle PAB$ is isosceles and
$$\angle PAB = \angle PBA$$
Step 3: By angle sum property in $\triangle PAB$ :
$$\angle PAB + \angle PBA + \theta = 180^\circ$$
$$2\angle PAB = 180^\circ - \theta \Rightarrow \angle PAB = 90^\circ - \frac{\theta}{2}$$
Step 4: $OA \perp PA$ (radius is perpendicular to tangent at the point of contact), so
$$\angle OAP = 90^\circ$$
Step 5: $$\angle OAB = \angle OAP - \angle PAB = 90^\circ - \left(90^\circ - \frac{\theta}{2}\right) = \frac{\theta}{2}$$
Step 6: Therefore
$$\theta = 2\angle OAB \Rightarrow \angle APB = 2\angle OAB$$
Hence proved.