Question

Chords equidistant from the center of a circle are equal in length. Explain this theorem with proof in detail please.

Answer 1

please refer the attachment

Answer 2

Statement : Chords equidistant from the centre of a circle are equal in length.

Given: OM and ON are perpendicular from the centre to the chords AB and CD and OM = ON.

To prove : chords AB = chord CD

Construction : Join OA and OC.

PROOF :

OM ⊥ AB $⇒ \frac{1}{2}$ AB = AM

ON ⊥ CD $⇒ \frac{1}{2}$ CD = CN

Consider, ΔAOM and ΔCON,

OA = OC . (radii of the circle)
OM = ON . (given)
∠OMA = ∠ONC = 90 ° . (given)

ΔAOM ≅ ∠CON . (RHS congruency)

AM = CN $⇒ \frac{1}{2}$ AB = $\frac{1}{2}$ CD $⇒$ AB = CD

The two chords are equal if they are equidistant from the centre.