Question

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

Answer 1

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,
OM² = OB² - BM²
         = OD² - DN²                         ....( By 1 )
         = ON²
∴ 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.