#### Question

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

(i) ∠BMN = ∠DNM.

(ii) ∠AMN = ∠CNM.

#### Solution

Drop OM⊥AB and ON⊥CD

∴ OM bisects AB and ON bisects CD

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

⇒ BM =`1/2` AB = `1/2` CD =DN .............(1)

Applying Pythagoras theorem,

OM^{2}= OB^{2} -BM^{2}

= OD^{2} - DN^{2} (by (1))

= ON^{2}

∴ OM=ON

⇒ ∠OMN = ⇒ ∠ONM ……………(2)

(Angles opp to equal sides are equal)

(i) ∠OMB=∠OND (both 90°)

Subtracting (2) from above,

∠BMN= ∠DNM

(ii) ∠OMA = ONC (both 90°)

Adding (2) to above,

∠AMN = ∠CNM

Is there an error in this question or solution?

Solution M and N Are the Mid-points of Two Equal Chords Ab and Cd Respectively of a Circle with Centre O. Prove That: (I) ∠Bmn = ∠Dnm. (Ii) ∠Amn = ∠Cnm. Concept: Chord Properties - Equal Chords Are Equidistant from the Center.