Question

Prove that the chords equidistant from the center of a circle are equal in length.​

Answer 1

Answer:-

★ Given:-

• A circle with center O.

• AB and CD are two chords

• OM ⊥ AB , ON ⊥ CD

• OM = ON

★ To Prove :-

• AB = CD

★ Theorem Used :-

▪︎Perpendicular line from the chord to the center of circle bisects the chord.

★ Construction:-

• Join OA

• Join OC

★ Proof:-

In ∆ OMA and ∆ ONC

• ∠ OMA = ∠ ONC [ Each 90° ]

• OA = OC [ Each equal to radii ]

• OM = ON [ Given ]

Hence,

By R.H.S. Congruence Rule

∆ OMA ≅ ∆ ONC

AM = CN ( C.P.C.T. )

⇨ 2AM = 2CN ( Multiplying both sides)

According to theoream, Perpendicular line from the chord to the center of circle bisects the chord;

So,

2AM = 2CN

⇨ AB = CD

Hence Proved.