**Solution:**

Let AB be a diameter of the circle. Two tangents PQ and RS are drawn at points A and B respectively.

Radii of the circle to the tangents will be perpendicular to it.

∴ OB ⊥ RS and,

∴ OA ⊥ PQ

∠OBR = ∠OBS = ∠OAP = ∠OAQ = 90º

From the figure,

∠OBR = ∠OAQ (Alternate interior angles)

∠OBS = ∠OAP (Alternate interior angles)

Since alternate interior angles are equal, lines PQ and RS will be parallel.

**Hence Proved that the tangents drawn at the ends of a diameter of a circle are parallel.**