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प्रश्न
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
उत्तर
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.
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