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Question
Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle.
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Solution
Given: PQ is a diameter of circle which Bisects
Chord AB at C
To prove: PQ bisects ∠AOB
Proof:
In ΔAOC and ΔBOC
OA=OB
OC=OC
AC=BC
Then Δ AOC ≅ BOC
∴∠AOC = ∠BOC
Hence PQ bisects ∠AOB
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