Let AB be the diameter of a circle with center O.

CD and EF are two tangents at ends A and B respectively.

To Prove : CD || EF

Proof :

OA ⏊ CD and OB ⏊ EF

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OAD = ∠OBE = 90°

∠OAD + ∠OBE = 90° + 90° = 180°

Considering AB as a transversal

⇒ CD || EF

[Two sides are parallel, if any pair of the interior angles on the same sides of transversal is supplementary]