Question

in triangle ABC the bisectors of angle B and angle C meet at O.Prove that angle BOC=90 degree+ angle A/2

Answer 1

Here BO, CO are the angle bisectors of ∠PBC & ∠QCB intersect each othe at O.

∴ ∠1 = ∠2  and ∠3 = ∠4

Side AB and AC of ΔABC are produced to P and Q respectively.

∴ Exterior of ∠PBC = ∠A + ∠C --------------(1)

And Exterior of ∠QCB = ∠A + ∠B --------------(2)

Addiing (1) and (2) we get

∠PBC + ∠QCB = 2 ∠A + ∠B + ∠C.

2∠2 + 2∠3 = ∠A + 180°

∠2 + ∠3 = (1 /2)∠A + 90°  ----------(3)

But in a ΔBOC = ∠2 + ∠3 + ∠BOC = 180° --------( 4)

From equ (3) and (4) we get

(1 /2)∠A + 90° + ∠BOC = 180°

∠BOC = 90° - (1 /2)∠A

Answer 2

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Ur answer is here ↪️↪️↪️↪️↪️↪️↪️↪️↪️↪️
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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 !!!)

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. HOPE ITS HELP UH!!!