O is the circumcentre of the triangle ANC and OD is perpendicular on BC. Prove that ∠BOD= ∠A
Answers
Given : O is the circumcentre of the ∆ ANC and OD is perpendicular on BC.
To prove : ∠BOD= ∠A
Proof :
In ΔOBD and ΔOCD ,
OB = OC [Radius]
∠ODB = ∠ODC [Each 90°]
OD = OD [Common]
Therefore, By RHS congruence criterion , we obtain,
ΔOBD ≅ ΔOCD
So, ∠BOD = ∠COD …………(1)
[By CPCT]
Since, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∴ ∠BOC = 2∠BAC
∠BOD + ∠COD = 2∠BAC
∠BOD + ∠BOD = 2∠BAC
[From eq (1)]
2∠BOD = 2∠BAC
∠BOD = ∠BAC
Hence proved that ∠BOD = ∠A
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Answer:-
Given :
O is the circumcentre of the ∆ ANC and OD is perpendicular on BC.
To prove :
∠BOD= ∠A
Proof :
In ΔOBD and ΔOCD ,
OB = OC [Radius]
∠ODB = ∠ODC [Each 90°]
OD = OD [Common]
Therefore, By RHS congruence criterion , we obtain,
ΔOBD ≅ ΔOCD
So, ∠BOD = ∠COD …………(1)
[By CPCT]
Since, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∴ ∠BOC = 2∠BAC
∠BOD + ∠COD = 2∠BAC
∠BOD + ∠BOD = 2∠BAC
[From eq (1)]
2∠BOD = 2∠BAC
∠BOD = ∠BAC
Hence proved that ∠BOD = ∠A
Hope it's help You❤️