A circle having centre o circumscibes a triangle abc and od perpendicular bc. Prove that angle bod=angle a
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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.
IF IT'S HELPFUL TO YOU PLEASE MARK IT AS BRAINLIEST ❤
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.
IF IT'S HELPFUL TO YOU PLEASE MARK IT AS BRAINLIEST ❤
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