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Equal Chords and Their Distances from the Centre

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Theorem: Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).
Consider a circle of centre O . It has PQ and RS two chords of equal length.

Now , draw a perpendicular from centre to PQ and RS intersecting them at L and M .
To Prove: OL = OM
Join OQ and OS.
Proof:  We know that PQ =RS  (given)
QL = `(PQ )/2` (perpendicular drawn from a centre to a chord bisect the chord)
Similarly, SM =`(RS)/2` 
Therefore , QL = SM
Consider ∆OQL and ∆OSM
OQ = OS  (radius of the same circle)
QL = SM (proved)
∠OLQ = ∠OMS = 90°  (given )
∆OQL ≅ ∆OSM    (SSA rule)
OL = OM (By CPCT).

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

Shaalaa.com | Theorem: Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).

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Theorem: Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres). [00:05:28]
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Theorem: Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).
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Theorem: Chords equidistant from the centre of a circle are equal in length.
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