Question

In the given figure O is the centre of the circle and OD BC. Then
prove that ∠BAC = ∠COD

Answer 1

Answer:

Join OB and OC

In △OBD and △OCD, we have

OB = OC [Each equal to radius of circumcircle]

∠ODB = ∠ODC [Each equal to 90˚]

and OD = OD [Common]

∴ △OBD ≅ △OCD

⇒ ∠BOD = ∠COD

⇒ ∠BOC = 2∠BOD = 2∠COD

Now, arc BC subtends ∠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