Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Let we have a circle as shown and connect two points, P and Q, such that PQ becomes the diameter of the circle. Now, draw two tangents, AB and CD, at points P and Q, respectively.

Now, both PO and OQ are radii to circle and are perpendicular to the tangents.

So, OP is perpendicular to AB, and OQ is perpendicular to CD.

So, ∠OPA = ∠OPB = ∠OQC = ∠OQD = 90°

From the above figure, ∠OPB and ∠OQC are alternate interior angles.

Therefore, ∠OPB = ∠OQC = 90°

Since they are also alternate interior angles and are equal, so it can be said that line AB and line CD will be parallel to each other.

(Hence Proved).