In figure, 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 \( ∠ \ AOB \ = \ 90° \).

Since tangents drawn from an external point to a circle are equal,

\(\because \) \( AP \ = \ AC \).
Thus in \( ∆ \ PAO \) and \( ∠ \ AOC \), we have

\( AP \ = \ AC \)
\( AO \ = \ AO \) [Common]
\( PO \ = \ OC \) [Radii of the same circle]
\( ∆ \ PAO \ \cong \ AOC \) [ By SSS-criterion of congruence]

\( ∠ \ PAO \ = \ ∠ \ CAO \ \)
\( \Rightarrow \ ∠ \ PAC \ = \ 2 ∠ CAO \)

Similarly, we can prove that

\( ∠ \ QBO \ = \ ∠ \ CBO \ \)
\( \Rightarrow \ ∠ \ CBQ \ = \ 2 ∠ \ CBO \)

Now, \( ∠ \ PAC \ + \ ∠ \ CBQ \ = \ 180°\)
[Sum of the interior angles on the same side of transversal is 180°]

\(\Rightarrow 2 ∠ \ CAO \ + \ 2 ∠ \ CBO \ = \ 180° \)

\( \Rightarrow ∠ \ CAO \ + \ ∠ \ CBO \ = \ 90° \)
\( \Rightarrow 180° \ - \ ∠ \ AOB \ = \ 90° \)
[\(\because \) \( ∠ \ CAO \ + \ ∠ \ CBO \ + \ ∠ \ AOB \ = \ 180° \)]

\( \Rightarrow \ ∠ \ AOB \ = \ 90° \)