Equal chords of a circle are equidistant from the centre of the circle | Maths Theorem | Class 9th

Описание: In this video, I proved this by making a working model of the theorem circle of class 9th.
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Theorem - “Equal chords of a circle are equidistant (equal distance) from the center of the circle.”
Construction: Join OB and OD

Proof: Given, In ∆OPB and ∆OQD

BP = 1/2 AB (Perpendicular to a chord bisects it) ……..(1)

DQ = 1/2 CD (Perpendicular to a chord bisects it) ……..(2)

AB = CD (Given)

BP = DQ (from eq 1 and 2)

OB = OD (Radii of the same circle)

∠OPB = ∠OQD = 90° (OP ⊥ AB and OQ ⊥ CD)

∆OPB ≅ ∆OQD ( By R.H.S Axiom of Congruency)

Hence,

OP = OQ ( By CPCT)

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