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

Question

Philoid · Accepted Answer

The figure is given below:

Let O be the center of the circle and AB and CD be the tangents at points P and Q respectively.

OP and OQ are the radii to the circle.

To prove: AB ∥ CD.

As AB is a tangent at point P, therefore AB is perpendicular to OP.

i.e. AB ⊥ OP. (∵ tangent at any point of circle is ⊥ to the radius through point of contact.)

∴ ∠OPA = 90°

Also, CD is tangent at point Q, therefore CD is perpendicular to OQ.

i.e. CD ⊥ OQ. (∵ Tangent at any point of circle is ⊥ to the radius through point of contact.)

∴ ∠OQD = 90°

∴ ∠OPA = ∠OQD = 90°

i.e. ∠QPA = ∠PQD = 90°

Now, for the lines AB and CD, PQ is the transversal; and

∠QPA = ∠PQD i.e. alternate angles are equal.

Therefore, the lines are parallel.

Thus, AB ∥ CD.