Question

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

Answer 1

OA is perpendicular to MN

OB is also perpendicular to QP

<OAM=90°

<OAN=90°

<OBQ=90°

<OAM=<OBP(alternate angles)  

<OAN=<OBQ( " )

Interior two angles are alternate, so MN//QP

Answer 2

$\huge \bf{Solution.}$

Given: PQ is a diameter of a circle with centre O. The lines AB and CD are the tangents at P and Q respectively.

To Prove: AB || CD

Proof: AB is a tangent to the circle at P and OP is the radius through the point of contact

∴ ∠OPA=90° ....(1)

(By tangent radius theorem)

∵ CD is a tangent to the circle at Q and OQ is the radius through the point of contact

∴ ∠ OQD=90° ....(2)

(By tangent radius theorem)

From (1) and (2),

∠OPA=∠OQD

But these angles from a pair of equal alternate interior angles.

∴ AB || CD.