# PROPERTIES OF CHORD OF A CIRCLE

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 Center 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)

## Perpendicular from the Center of a Circle to a Chord Bisects the Chord

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

(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.

## Equal Chords of a Circle are Equidistant from the Center

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

(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. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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