Equal Chords are Equidistant from the Center of Circle
Equal Chords are Equidistant from the Center of Circle
Theorem 1 Equal chords of a circle are equidistant from the centre
Given Two chords AB and CD of a circle C (O,r) such that AB = CD and OL To prove Chords AB and CD are equidistant from the centre O i.e, OL = OM. CONSTRUCTION Join OA and OC. PROOF Since the perpendicular from the centre to a chord bisects the chord Therefore, and, But,
Now, in right triangles OAL and OCM, we have and, So, by RHS criterion of convergence, we have Hence, equal chords of a circle are equidistant from the centre. |  |
Theorem 2 Chords of a circle which are equidistant from the centre are equal.
Given Two chords AB and CD of a circle C(O,r) which are equidistant from its centre i.e., OL = OM, where OL AB and OM
CD.
TO PROVE Chords are equal i.e AB = CD CONSTRUCTION Join OA and OC PROOF Since the perpendicular from the centre of a circle to a chord bisects the chord. Therefore,
and,
In, triangles OAL and OCM , we have
and, So, by RHS< criterion of convergence, we have
Hence, the chords of a circle which are equidistant from the centre are equal. | |
Illustration:
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| Right Option : A | |||||
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If two triangles on the same side of the common base have same base and equal areas , then which of the following statements is true? | |||
| Right Option : B | |||
| View Explanation | |||
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| Right Option : C | |||||
| View Explanation | |||||
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