Question

O is the circumcentre of the triangle ANC and OD is perpendicular on BC. Prove that ∠BOD= ∠A

Answer 1

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 2

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

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