Question

chords equidistant from the centre of the circle are equal in length

Answer 1

To Prove: AB = CD

Construction: Join OA and OC.

Proof:

OL ⊥ AB

⇒AL = BL [ Since, the ⊥cular from the centre of a circle to a chord bisects the chord ]

and, OM ⊥ CD

⇒ CM = DM [ Since, the ⊥cular from the centre of a circle to a chord bisects the chord 

=> OA=OC (Common hypotenuse)

Therefore..♢ALO ~= ♢CMO( SSS rule)

NOW....AL=CM (C.P.C.T)

AND 2AL=2 CM

Therefore..AB=CD (Hence, Proved)
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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.