Properties of chord of a circle :
Here we are going to see the proof of three important properties of chord.
Equal chords of a circle subtend equal angles at the centre.
chord AB = chord CD (<AOB = <COD)
Theorem 1 :
Perpendicular from the centre 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
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
Hence, AC = BC
Theorem 2 :
Equal chords of a circle are equidistant from the centre.
Given: A cirle with centre O and radius r such that
chord AB = chord CD.
To prove: OL = OM
Draw OL = AB and OM = CD. Join OA and OC
(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 .
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