Question

Angle BOC=90degree+1/2 angleA. Prove​

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

Step-by-step explanation:

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