Question

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.

Answer 1

We draw $OM\perp AB\&ON\perp CD$

In $△OMX$ and $△ONX$

$\angle OMX=\angle ONX$ [Both ${90}^o$]

$OX=OX$ [Common]

$OM=ON$ [AB and CD are equal chords and equal chords are equidistant from the centre]

$\therefore △OMX\cong △ONX$ [RHS congruency]

$△OPX\cong △OQX$

$\therefore \angle OXM=\angle OXN$ [CPCT]

i.e, $\angle OXA=\angle OXD$

Hence proved.