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Question
If O is the circumcentre of a Δ ABC and OD ⊥ BC, prove that ∠ BOD = ∠A.
Solution
Join OB and OC.
In Δ OBD and Δ OCD, we have
OB = OC ....(Each equal to the radius of circumcircle)
∠ODB = ∠ODC ....(Each equal to 90°)
and OD = OD ....(Common)
∴ Δ OBD ≅ Δ OCD
⇒ ∠BOD = ∠COD
⇒ ∠BOC = 2∠BOD = 2∠COD
Now, arc BC substends ∠BOC at the centre and ∠BAC = ∠A at a point in the remaining part of the circle.
∴ ∠BOC = 2∠A
⇒ 2∠BOD = 2∠A ....( ∵∠BOC = 2∠BOD)
⇒ ∠BOD = ∠A
Hence proved.
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