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प्रश्न
Two equal chords AB and CD of a circle with center O, intersect each other at point P inside the circle.
Prove that: (i) AP = CP ; (ii) BP = DP
उत्तर
Drop OM and ON perpendicular on AB and CD.
Join OP, OB, and OD.
∴ OM and ON bisect AB and CD respectively. ....( Perpendicular drawn from the centre of a circle to a chord bisects it. )
∴ MB = `1/2"AB" = 1/2"CD" = "ND"`....(i)
In right ΔOMB,
OM2 = OB2 - MB2 ....(ii)
In right ΔOND,
ON2 = OD2 - ND2 ....(iii)
From (i), (ii), and (iii),
OM = ON
In ΔOPM and ΔOPN,
∠OMP = ∠ONP ....( both 90° )
OP = OP ....( common )
OM = ON ....( proved above )
By Right Angle-Hypotenuse-Side criterion of congruence,
∴ ΔOPM ≅ ΔOPN ....( by RHS )
The corresponding parts of the congruent triangles are congruent.
∴ PM = PN ....( c.p.c.t. )
Adding (i) to both sides,
MB + PM = ND + PN
⇒ BP = DP
Now, AB = CD
∴ AB - BP = CD - DP ...( ∵ BP = DP )
⇒ AP = CP.
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