Question

M and N are the mid-points of two equal chords AB and CD respectively of a circle with centre O. Prove that $\angle AMN=\angle CNM$.

Answer 1

From the figure $OM$ bisects $AB$ and $ON$ bisects $CD$

(Perpendicular drawn from the centre of a circle to a chord bisects it)

$\Rightarrow BM=\frac{1}{2}AB=\frac{1}{2}CD=DN$ ........$\left(1\right)$

Applying Pythagoras theorem,

${OM}^2={OB}^2-{BM}^2={OD}^2-{DN}^2={ON}^2$ by $\left(1\right)$

$\therefore OM=ON$

$\Rightarrow \angle OMN=\angle ONM$ .......$\left(2\right)$

(Angles opposite to equal sides are equal)

$\left(i\right)\angle OMB=\angle OND$(both equal to ${90}^{\circ }$)

$\left(ii\right)\angle OMA=\angle ONC$(both equal to ${90}^{\circ }$)

Adding $\left(2\right)$ to above we get

$\angle OMB+\angle OMN=\angle OND+\angle ONM$

$\angle AMN=\angle CNM$