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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