Question

if the angle bisectors of angle abc and angle acb of a triangle abc meet at point o then angle boc=90°+½angle a​

Answer 1

Answer:

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,

∠1+∠2+∠BOC=180

o

In △ABC,

∠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,

90

o

−

2

∠A

+∠BOC=180

o

∠BOC=90

o

+

2

∠A

Step-by-step explanation:

thanks my answer please

Answer 2

Answer:

AAABC such that the bisectors of ZABC and ZACB meet at a point O.

To prove:

ZBOC = 90° + 1

Proof:

In ABOC,

21+22+ZBOC = 180°

In AABC,

ZA + 2B + C = 180⁰

ZA + 2(21) + 2(22) = 180⁰ ΖΑ 4+21+22=90⁰

2

21 +42 ZA 2 90⁰ -

Therefore, ΖΑ

90⁰ + ZBOC = 180⁰