Question

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

Answer 1

Given:-

A A circle with center O and diameter AB.

Let PQ be the tangent at point A & RS be the tangent at point B.

To Prove: PQ∥RS

Proof:

Since PQ is tangent at point A

OA⟂PQ -------(Tangent at any point of a circle is perpendicular to the radius through point of contact)

∠OAP = 90° ... (i)

Similarly

RS is a tangent at point B

OB⟂RS --------(Tangent at any point of a circle is perpendicular to the radius through point of contact)

∠OBS = 90° ... (ii)

From (i) and (ii)

∠OAP=90° and ∠OBS = 90°

Therefore,

∠OAP=∠OBS

i.e, ∠BAP=∠ABS

For lines PQ and RS and transversal AB

∠BAP=∠ABS i.e., both alternate angles are equal

So, lines are parallel

∴ PQ∥RS Hence Proved!

Answer 2

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

Radius drawn to these tangents will be perpendicular to the tangents.

Thus, OA ⊥ RS and OB ⊥ PQ

∠OAR = 90º

∠OAS = 90º

∠OBP = 90º

∠OBQ = 90º

It can be observed that

∠OAR = ∠OBQ (Alternate interior angles)

∠OAS = ∠OBP (Alternate interior angles)

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