**Property 1 :**

Equal chords of a circle subtend equal angles at the centre.

<AOB = <DOC

**Property 2 :**

Perpendicular from the centre of a circle to a chord bisects the chord.

AC = BC

**Property 3 :**

Equal chords of a circle are equidistant from the centre.

OM = OL

**Property 4 :**

If two chords intersect inside a circle, then the measure of each angle formed is one half the sum of the measures of the arcs intercepted by the angle and its vertical angle.

m<1 = (1/2) (Measure arcCD+measure of arcAB)

m<2 = (1/2) (Measure arcBC+measure of arcCD)

**Property 5 :**

If two chords intersect inside a circle, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.

EA ⋅ EB = EC ⋅ ED

Equal chords of a circle subtend equal angles at the centre.

chord AB = chord CD (<AOB = <COD)

**Converse of the Result :**

If the angles subtended by two chords at the centre of a circle are equal, then the chords are equal.

<AOB = <COD (chord AB = chord CD)

**Theorem 1 :**

Perpendicular from the center of a circle to a chord bisects the chord.

Given : A circle with centre O and AB is a chord of the circle other than the diameter and OC ⊥ AB

To prove : AC = BC

Construction : Join OA and OB

**Proof :**

In triangles OAC and OBC

(i) OA = OB

(Radii of the same circle)

(ii) OC is common

(iii) <OCA = <OCB (Each 90°, since OC ⊥ AB)

(iv) Triangle OAC ≡ Triangle OBC

(RHS congruency.)

Hence, AC = BC

**Converse of Theorem 1 :**

The line joining the center of the circle and the midpoint of a chord is perpendicular to the chord.

**Theorem 2 :**

Equal chords of a circle are equidistant from the center.

Given : A circle with center O and radius r such that

chord AB = chord CD.

To prove : OL = OM

Construction :

Draw OL = AB and OM = CD. Join OA and OC

**Proof :**

(i) AL = (1/2) AB and CM = (1/2) CD

(Perpendicular from the centre of a circle to the chord bisects the chord.)

AB = CD (1/2) AB = (1/2) CD ==> AL = CM

(ii) OA = OC (radii)

(iii) <OMC= <OLA (Each 90°)

(iii) Triangle OLA ≡ Triangle OMC (RHS congruence.)

OL = OM

Hence AB and CD are equidistant from O .

**Converse of Theorem 2 :**

The chords of a circle which are equidistant from the center are equal.

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