Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A.

To do:

We have to prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A.

Solution:

$AB$ is the diameter of the circle.

A tangent is drawn at point A.

Draw a chord $PQ$ parallel to the tangent $XAY$.

This implies,

$PQ$ is a chord of the circle and $OA$ is a radius of the circle.

$\angle XAO = 90^o$     (Tangent at a point on a circle is perpendicular to the radius through the point)

$\angle PCO = \angle XAO$         (Corresponding angles are equal)

This implies,

$\angle PCO = 90^o$

$CO$ bisects $PQ$       (Perpendicular from the centre of a circle to chord bisects the chord)

Similarly,

The diameter $AB$ bisects all the chords which are parallel to the tangent at the point A.

Hence proved.

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Updated on: 10-Oct-2022

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