Question

The sides AB and AC of triangle ABC are produced to P and Q respectively. if bisectors of angle PBC and angle QCB intersect at O. Prove that angle BOC = 90 - 1/2 angle A.
(with diagram)

Answer 1

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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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