Question

5. In a triangle ABC, the bisector of angle B and angle C intersect each other at Interior point
O of this triangle. Prove that angle BOC = 90°+1/2 LA​

Answer 1

Answer:

Given :

A △ ABC such that the bisectors of ∠ ABC and ∠ ACB meet at a point O.

To prove :

∠BOC=90

o

+

2

1

∠A

Proof :

In △ BOC, we have

∠1+∠2+∠BOC=180

o

....(1)

In △ ABC, we have,

∠A+∠B+∠C=180

o

∠A+2(∠1)+2(∠2)=180

o

2

∠A

+∠1+∠2=90

o

∠1+∠2=90

o

−

2

∠A

Therefore, in equation 1,

90

o

−

2

∠A

+∠BOC=180

o

∠BOC=90

o

+

2

∠A