Question

Prove that Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).

Answer 1

Given : A circle have two equal chords $AB$ and $CD$.

$AB=CD$ and $OM$ perpendicular to $AB$, $ON$ perpendicular to $CD$

To Prove : $OM=ON$

Proof : $AB=CD$ (Given)

$\because$ the perpendicular drawn from the centre of a circle to bisect the chord

$\therefore \frac{1}{2}AB=\frac{1}{2}CD$

$\Rightarrow BM=DN$

In $△OMB$ and $△OND$

$\angle OMB=\angle OND={90}^0$ [Given]

$OB=OD$ [Radii of same circle]

Side $BM=$ Side $DN$ [Proved above]

$\therefore △OMB\cong △OND$ [By R.H.S.]

$\therefore OM=ON$ [By C.P.C.T]