Question

the Bisectors of angle b & angle c intersect each other at a point O. prove that angle BOC =90°+ half angle A​.

Answer 1

TO PROVE

Answer: PROVED THAT ∠BOC = 90° + 1/2 ∠A

Explanation:

Please Refer to Attached figure.

As given Bisector of ∠ ABC that is BO and Bisector CO of ∠ACB meets at O

now consider Δ OBC

using ANGLE SUM PROPERTY OF TRIANGLE we can say that

∠OBC + ∠OCB + ∠BOC = 180°

(1/2) ∠ABC + (1/2) ∠ACB + ∠BOC = 180° [ using OB is bisector , ∠OBC = (1/2)   ∠ABC and OC is bisector so ∠OCB  = (1/2) ∠ACB ]

=> ∠BOC = 180° - (1/2) ∠ABC - (1/2) ∠ACB  

=> ∠BOC = 180° -  ( (1/2) ∠ABC + (1/2) ∠ACB )                          ----------eq(1)

Now applying Angle sum property of triangle on ΔABC we get

∠ABC +∠ACB + ∠BAC = 180°

On dividing both sides by 2 we get

(∠ABC +∠ACB + ∠BAC )/2 = 180°/2

⇒(1/2)∠ABC +  (1/2)∠ACB + (1/2)∠BAC = 90°

⇒ (1/2)∠ABC +  (1/2)∠ACB = 90° - (1/2)∠BAC

On substituting (1/2)∠ABC +  (1/2)∠ACB as 90° - (1/2)∠BAC in equation(1)

we get

∠BOC = 180° -  ( 90° - (1/2)∠BAC )    

⇒∠BOC = 180° -  90° + (1/2)∠BAC

⇒∠BOC = 90° + 1/2 ∠A     [ since ∠BAC is same as ∠A ]

Here comes the most important and satisfied word of this problem "HENCE PROVED".

#answerwithquality

#BAL