Question

A circle having centre O circumscribe a ∆ABC and OD|BC. Prove that angle BOD=angle A​

Answer 1

in triangle BOC and triangle COD

BO = CO (both are radius)

∠BDO=∠CDO =90°

BD is common in both triangle

so both triangle are similar

so ∠BOD=∠COD

as we know angle at the center is double of the angle of corcumferece

∠BOC=2∠BAC

⇒2∠BOD=2∠BAC (as we proved above)

⇒∠BOD=∠A

hence proved.

Answer 2

Step-by-step explanation:

in triangle BOC and triangle COD

BO = CO (both are radius)

∠BDO=∠CDO =90°

BD is common in both triangle

so both triangle are similar

so ∠BOD=∠COD

as we know angle at the center is double of the angle of corcumferece

∠BOC=2∠BAC

⇒2∠BOD=2∠BAC (as we proved above)

⇒∠BOD=∠A

Hence Proved.