Question

$M$ and $N$ are the mid-points of two equal chords $AB$ and $CD$ respectively of a circle with centre $O$ then
(i) $\angle BMN=\angle DNM$
(ii) $\angle AMN=\angle CNM$.

• A. True
• B. False

Answer 1

Answer: (A)

Drop $OM\perp AB$ and $ON\perp CD$

$\therefore OM$ 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,

${OM}^2={OB}^2-{MB}^2$

$={OD}^2-{DN}^2$

$={ON}^2$

$\therefore OM=ON$

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

(Angles opposite to equal sides are equal)

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

Subtracting $\left(2\right)$ from above

$\angle BMN=\angle DNM$

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

adding $\left(2\right)$ from above

$\angle AMN=\angle CNM$

Hence the given statement is true.