Prove that the line draw through the centre of circle to bisects a chord is in pedicular to the chord
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Given a circle with centre O and AC = BC
To Prove: OC ⊥ AB
Construction: Join points O, A and O, B.
Proof:
In Δ’s OCA and OCB
OA = OB (Radii of the circle)
OC = OC (Common side)
AC = BC (Given)
Hence ΔOCA ≅ ΔOCB (SSS Congruence rule)
∴ ∠OCA = ∠OCB (CPCT)
From the figure, ∠OCA + ∠OCB = 180° (Linear pair)
∠OCA + ∠OCA = 180°
2∠OCA = 180°
∴ ∠OCA = 90°
Thus OC ⊥ AB
Given a circle with centre O and AC = BC
To Prove: OC ⊥ AB
Construction: Join points O, A and O, B.
Proof:
In Δ’s OCA and OCB
OA = OB (Radii of the circle)
OC = OC (Common side)
AC = BC (Given)
Hence ΔOCA ≅ ΔOCB (SSS Congruence rule)
∴ ∠OCA = ∠OCB (CPCT)
From the figure, ∠OCA + ∠OCB = 180° (Linear pair)
∠OCA + ∠OCA = 180°
2∠OCA = 180°
∴ ∠OCA = 90°
Thus OC ⊥ AB
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