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प्रश्न
Given two equal chords AB and CD of a circle with center O, intersecting each other at point P.
Prove that:
(i) AP = CP
(ii) BP = DP
उत्तर
In ΔOMP and ΔONP,
OP = OP ...( common sides )
∠OMP = ∠ONP ...( both are right angles )
OM = OM ...( side both the chords are equal, so the distance of the chords from the centre are also equal )
ΔOMP ≅ ΔONP ...( RHS congruence criterion )
⇒ MP = PN ...(c.p.c.t ) ....( a )
(i) Since AB = CD ...( given )
⇒ AM = CN ...( drawn from the centre to the chord bisects the chord )
⇒ AM + MP = CN + NP .....( from a )
⇒ AP = CP ....( b )
(ii) Since AB = CD
⇒ AP + BP = CP + DP
⇒ BP = DP ....( from b )
Hence proved.
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