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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Updated on: 10-Oct-2022

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