In $Fig.$,$XY$ and $X'Y'$ are two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X'Y'$ at $B$. Prove that $\angle AOB = 90^{\circ}$.

(N/A) Join point $O$ to $C$.
In $\triangle OPA$ and $\triangle OCA$:
$OP = OC$ (Radii of the same circle)
$AP = AC$ (Tangents drawn from an external point $A$ are equal in length)
$AO = AO$ (Common side)
Therefore,$\triangle OPA \cong \triangle OCA$ by $SSS$ congruence criterion.
This implies $\angle POA = \angle COA$ ... $(i)$
Similarly,in $\triangle OQB$ and $\triangle OCB$:
$OQ = OC$ (Radii of the same circle)
$BQ = BC$ (Tangents drawn from an external point $B$ are equal in length)
$OB = OB$ (Common side)
Therefore,$\triangle OQB \cong \triangle OCB$ by $SSS$ congruence criterion.
This implies $\angle QOB = \angle COB$ ... $(ii)$
Since $POQ$ is a diameter of the circle,it is a straight line,so the sum of angles on one side is $180^{\circ}$:
$\angle POA + \angle COA + \angle COB + \angle QOB = 180^{\circ}$
Using equations $(i)$ and $(ii)$:
$2 \angle COA + 2 \angle COB = 180^{\circ}$
$2(\angle COA + \angle COB) = 180^{\circ}$
$\angle COA + \angle COB = 90^{\circ}$
Therefore,$\angle AOB = 90^{\circ}$.