Question

in triangle abc,if bisector of angel b and angel c meet at point o,prove that angel boc=90+1/2La​

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

Answer 2

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

Step-by-step explanation:

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