Prove with reason that two equal chords of any circle are equidistant from centre. (don't delete or report if you don't know the answer)
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Prove with reason that two equal chords of any circle are equidistant from centre. (don't delete or report if you don't know the answer)
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Step-by-step explanation:
The line OX is perpendicular to the chord AB and OY is perpendicular to the chord CD. We have to prove OX = OY. Also, the line OX is perpendicular to AB. Therefore, the theorem “equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres)”, is proved.
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Answer:
Step-by-step explanation:
Given : A circle have two equal chords AB and CD.
AB=CD and OM perpendicular to AB, ON perpendicular to CD
To Prove : OM=ON
Proof : AB=CD (Given)
∵ the perpendicular drawn from the centre of a circle to bisect the chord
1/2AB=1/2CD
⇒BM=DN
In △OMB and △OND
∠OMB=∠OND=90°
[Given]
OB=OD [Radii of same circle]
Side BM= Side DN [Proved above]
∴△OMB≅△OND [By R.H.S.]
∴OM=ON [By C.P.C.T]