Question

in the picture two equal chords meet at a point inside the circle. Prove that the diameter through this point bisects the angle between the chords. ​

Answer 1

Answer:

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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