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If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Given:
Two equal chords of a circle intersect within the circle
To do:
We have to prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Solution:
Let $AB$ and $CD$ be two equal cords which intersect at point $R$.
$PQ$ is the diameter of the circle.
From the centre of the circle, draw a perpendicular to $AB$ and another one perpendicular to $CD$
$OM \perp AB$
$ON \perp CD$.
Join $OR$.
From the diagram,
$OM$ bisects $AB$ and $OM \perp AB$
$ON$ bisects $CD$ and $ON \perp CD$
$AB = CD$
In triangles $OMR$ and $ONR$,
$\angle OMR = \angle ONR$
$OE = OE$ (Common side)
$OM = ON$ ($AB$ and $CD$ are equal and they are equidistant from the centre)
Therefore, by RHS congruency,
$\triangle OMR \cong \triangle ONR$
This implies,
$\angle ORM = ORN$..........(iii) (CPCT)
Therefore,
$\angle BRQ = CRQ$
Hence proved.
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