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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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