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 : $\angle AMN=\angle CNM and \angle BMN=\angle DNM$
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)
$$\begin{gathered} In \Delta OMN, \\ OM=ON \\ \angle OMN=\angle ONM \left(angles opposite to equal sides\right) .....\left(1\right) \\ \angle OMA=\angle ONC \left(each {90}^0\right) .....\left(2\right) \\ And, \angle OMB=\angle OND \left(each {90}^0\right) .....\left(3\right) \\ Subtracting \left(1\right) and \left(2\right), we have, \\ \angle OMA-\angle OMN=\angle ONC-\angle ONM \\ \Rightarrow \angle AMN=\angle CNM \\ And, \\ Adding \left(1\right) and \left(3\right), we have, \\ \angle OMB+\angle OMN=\angle OND+\angle ONM \\ \Rightarrow \angle BMN=\angle DNM \end{gathered}$$
Hence Proved.