Question

In a ABC∆, the sides AB and AC are produced to P and Q respectively. The bisectors of ∠ PBC and ∠ QCB intersect at a point O. Prove that ∠ BOC =90 ° − 1/2 ∠A

Answer 1

Angle BOC  = 180 - (Angle CBO + Angle BCO) ----(1)

Since exterior angle of a triangle equal the sum of opposite interior angles, we have:-
Angle CBO = (1/2) angle CBP(Ext. angle)  = (1/2) (Angle A + Angle C)
Angle BCO = (1/2) angle BCQ (Ext. angle) = (1/2) (angle A + angle B)

==> angle CBO + angle BCO
       = (1/2)[(angle A+angleC) + (angle A + angle B)]
       = (1/2)[ (angle A + angle B + angle C) + angle A]
       = (1/2) ( 180 + angle A) = 90 + (1/2) angle A  ------(2)

From (1) and (2),
Angle BOC = 180 - [90 + (1/2) angle A] = 90 - (1/2) angle A

Answer 2

Given : bisectors of angle PBC and angle QCB intersect at O.

 To find : prove that ∠BOC = 90° -  (1/2) (∠A)  

Solution:

∠BAC = ∠A

Exterior angle = Sum of opposite interior angles

 ∠ PBC = ∠C + ∠A

∠ QCB = ∠B + ∠A

∠CBO = (1/2)∠PBC  = (1/2) (∠C + ∠A)

∠BCO = (1/2)∠QCB   = (1/2) (∠B + ∠A)

 ∠CBO  + ∠BCO  + ∠BOC = 180°   Sum of angles o a triangle )

=>  (1/2) (∠C + ∠A) +  (1/2) (∠B + ∠A)  +  ∠BOC = 180°

=> (1/2) (∠C + ∠A + ∠B) +  (1/2) (∠A)  +  ∠BOC = 180°

=> (1/2) (180°) +  (1/2) (∠A)  +  ∠BOC = 180°

=> 90°  +  (1/2) (∠A)  +  ∠BOC = 180°

=> ∠BOC = 90° -  (1/2) (∠A)  

QED

Hence Proved

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