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# Two Equal Chords Ab and Cd of a Circle with Centre O, Intersect Each Other at Point P Inside the Circle, Prove That: (I) Ap = Cp, (Ii) Bp = Dp - ICSE Class 10 - Mathematics

ConceptChord Properties - Equal Chords Are Equidistant from the Center

#### Question

Two equal chords AB and CD of a circle with centre O, intersect each other at point P inside the circle, prove that:
(i) AP = CP,
(ii) BP = DP

#### Solution

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 chord bisects it)

∴ MP = 1/2 AB = 1/2 CD = ND       ...........(i)

In rt Δ OMB, OM2= OB2 = MB2 ………………..(ii)

In rt Δ OND,  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)
MB + PM = ND + PN
⟹ BP = DP
Now, AB = CD
∴ AB – BP = CD – DP (∵ BP = DP)
⟹ AP = CP

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Solution Two Equal Chords Ab and Cd of a Circle with Centre O, Intersect Each Other at Point P Inside the Circle, Prove That: (I) Ap = Cp, (Ii) Bp = Dp Concept: Chord Properties - Equal Chords Are Equidistant from the Center.
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