Question

In the given figure
the bisector of triangle ABC
and triangle BCA intersect each other at o. prove
that angle BOC = 90+ half of angle a
​

Answer 1

Answer:

hope it helps..........

Answer 2

Step-by-step explanation:

Solution

In triangle ABC,

∠A+∠B+∠C=180    (1)

OB and OC are bisectors of ∠B  and  ∠C

So, ∠B = 2∠OBC  

and  ∠C = 2∠OCB

Now equation (1) can be written as,

∠A+2(∠OBC+∠OCB)=180     (2)

In triangle OBC,

∠BOC+∠OBC+∠OCB=180

 

 ∠OBC+∠OCB=180  −∠BOC      (3)

From (2)  and (3),

∠A + 2(180   −∠BOC)=180

∠A + 360  − 2∠BOC=180  

∠A + 180  = 2∠BOC

1/2 ∠A+90  = ∠BOC