M and N are the mid-points of two equal chords AB and CD respectively of a circle with center O. Prove that :

(i) ∠BMN = ∠DNM

(ii) ∠AMN = ∠CNM.

Join OM, ON, OB and OD.

Given,

M and N are the mid-points of two equal chords AB and CD.

∴ BM = $\dfrac{1}{2}AB$ and DN = $\dfrac{1}{2}CD$

Since, AB and CD are equal chords.

∴ BM = DN ……….(1)

We know that,

A straight line drawn from the center of a circle to bisect a chord, which is not a diameter, is at right angles to the chord.

∴ OM ⊥ AB and ON ⊥ CD.

∴ ∠OMB = ∠OND (Both equal to 90°) ……..(2)

(i) In triangle OMN,

⇒ ON = OM (Equal chords of a circle are equidistant from the center.)

⇒ ∠OMN = ∠ONM (Angles opposite to equal sides are equal) ……..(3)

Subtracting equation (2) from (3), we get :

⇒ ∠OMB - ∠OMN = ∠OND - ∠ONM

⇒ ∠BMN = ∠DNM.

Hence, proved that ∠BMN = ∠DNM.

(ii) From figure,

⇒ ∠OMA = ∠ONC (Both equal to 90°) ……………(4)

Adding equation (3) and (4), we get :

⇒ ∠OMA + ∠OMN = ∠ONC + ∠ONM

⇒ ∠AMN = ∠CNM.

Hence, proved that ∠AMN = ∠CNM.