Question

in a triangle ABC the bisectors of angle B and angle C intersect each other at a point O. prove that angle BOC= 90      + half angle a

Answer 1

I will be using A, B, C not angle A or angle B.... Hope you will not mind it.

As we know A + B + C = 180
So, B + C = 180 - A
Or, (B/2) + (C/2) = 90 - (A/2) ....(1)

Now think in triangle BOC,
we have three angles, OBC, BOC and OCB
OBC = (B/2)
OCB = (C/2)
Let BOC = X

So, X + (B/2) + (C/2) = 180
From equation (1)

X + 90 - (A/2) = 180
X = 90 + A/2  (Proved !!!)

Answer 2

Answer:

As we know A + B + C = 180

So, B + C = 180 - A

Or, (B/2) + (C/2) = 90 - (A/2) ....(1)

Now think in triangle BOC,

we have three angles, OBC, BOC and OCB

OBC = (B/2)

OCB = (C/2)

Let BOC = X

So, X + (B/2) + (C/2) = 180

From equation (1)

X + 90 - (A/2) = 180

X = 90 + A/2  (Proved !!!)