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A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ.
Given:
A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle.
To do:
We have to prove that R bisects the arc PRQ.
Solution:
Let chord $PQ$ be parallel to the drawn at point R.
Proof:
$PQ\ \parallel\ XY$ and $PR$ is the transversal.
This implies,
$ \angle XRP= \angle RPQ$ (Alternate angles are equal)
$\angle XRP = \angle PQR$ (Angle between tangent and chord is equal to angle made by the chord in the alternate segment)
$\angle RPQ = \angle PQR$
This implies,
$PR = QR$ (Sides opposite to equal angles are equal)
$PR = QR$
Therefore,
$R$ bisects arc $PRQ$.
Hence proved.
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