prove that line segment joining midpoints of two equal chords of a circle makes equal angles with the chord

Given that : AB and CD are two equal chords. And, M, N are mid point of chord AB and CD respectively.

To prove : $\angle AMN=\angle CNMand\angle BMN=\angle DNM$

Construction : Join OM and ON

Proof :

Since the line segment joining the centre of a circle to the midpoint of a chord is perpendicular to the chord. Therefore,

Since AB and CD are equal chords. So, they are equidistant from the other. (i.e., OM =ON)

$In\Delta OMN,\phantom{\rule{0ex}{0ex}}OM=ON\phantom{\rule{0ex}{0ex}}\angle OMN=\angle ONM\left(anglesoppositetoequalsides\right).....\left(1\right)\phantom{\rule{0ex}{0ex}}\angle OMA=\angle ONC\left(each{90}^{0}\right).....\left(2\right)\phantom{\rule{0ex}{0ex}}And,\angle OMB=\angle OND\left(each{90}^{0}\right).....\left(3\right)\phantom{\rule{0ex}{0ex}}Subtracting\left(1\right)and\left(2\right),wehave,\phantom{\rule{0ex}{0ex}}\angle OMA-\angle OMN=\angle ONC-\angle ONM\phantom{\rule{0ex}{0ex}}\Rightarrow \angle AMN=\angle CNM\phantom{\rule{0ex}{0ex}}And,\phantom{\rule{0ex}{0ex}}Adding\left(1\right)and\left(3\right),wehave,\phantom{\rule{0ex}{0ex}}\angle OMB+\angle OMN=\angle OND+\angle ONM\phantom{\rule{0ex}{0ex}}\Rightarrow \angle BMN=\angle DNM$

Hence Proved.

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