Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

(N/A) Let us consider a circle centered at point $O$. Let $P$ be an external point from which two tangents $PA$ and $PB$ are drawn to the circle,touching the circle at points $A$ and $B$ respectively. $AB$ is the line segment joining the points of contact $A$ and $B$,which subtends $\angle AOB$ at the center $O$ of the circle.
It can be observed that:
$OA$ (radius) $\perp PA$ (tangent)
Therefore,$\angle OAP = 90^{\circ}$
Similarly,$OB$ (radius) $\perp PB$ (tangent)
Therefore,$\angle OBP = 90^{\circ}$
In the quadrilateral $OAPB$,the sum of all interior angles is $360^{\circ}$.
$\angle OAP + \angle APB + \angle PBO + \angle BOA = 360^{\circ}$
$90^{\circ} + \angle APB + 90^{\circ} + \angle BOA = 360^{\circ}$
$\angle APB + \angle BOA = 180^{\circ}$
Hence,it is proved that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the center.