if two equal chords of a circle intersect within the circle prove that the line joining the point of intersection to the centre makes equal angles with the chords.
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Step-by-step explanation:
In △OMX and △ONX,
∠OMX=∠ONX=90
∘
OX=OX(common)
OM=ON where AB and CD are equal chords and equal chords are equidistant from the centre.
△OMX≅△ONX by RHS congruence rule.
∴∠OXM=∠OXN
i.e.,∠OXA=∠OXD
Hence proved.
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See my last question
He explained me too
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