Question

in a ∆abc the angle bisectors bo and co. prove that 2angle box= 180+ angle a​

Answer 1

Answer:

$As BO and CO are the angle bisectors of external angles of△ABC, Then</p><p></p><p>∠1=∠2∠4=∠3</p><p></p><p>We know, ∠A+∠ABC+∠ACB=180∘…eqn(1)</p><p></p><p>And ∠ABC=180−2∠1∠ACB=180−2∠4</p><p></p><p>Putting it in the eqn (1), we get</p><p></p><p>∠A+180−2∠1+180−2∠4=180⇒∠1+∠4=90+21∠A…eqn(2)</p><p></p><p>Also we know from the figure, ∠BOC+∠1+∠4=180∘</p><p></p><p>∠BOC=180−∠1−∠4</p><p></p><p>From eqn (2)</p><p></p><p>∠BOC=180−90−21∠A⇒∠BOC=90∘−21</p><p></p><p></p><p>$

As BO and CO are the angle bisectors of external angles of△ABC, Then

∠1=∠2

∠4=∠3

We know, ∠A+∠ABC+∠ACB=180

∘

…eqn(1)

And ∠ABC=180−2∠1

∠ACB=180−2∠4

Putting it in the eqn (1), we get

∠A+180−2∠1+180−2∠4=180

⇒∠1+∠4=90+

2

1

∠A…eqn(2)

Also we know from the figure, ∠BOC+∠1+∠4=180

∘

∠BOC=180−∠1−∠4

From eqn (2)

∠BOC=180−90−

2

1

∠A

⇒∠BOC=90

∘

−

2

1

Explanation:

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