#### theorem

**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.