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.