In the figure, the sides AB and  AC of   are produced to P and  Q respectively. The bisectors of   and   intersect at O. Prove that .

The correct answer is: Hence proved

In triangle ABC

(∠ABC + ∠ ACB )= 180  -  ∠ A(sum of angles in triangle)

now , ∠CBP  = ∠  A  +  ∠ ACB (exterior angle property)

so ,∠ CBO = ×∠ A + × ∠ ACB (OB is angular bisector)
similarly,

∠BCO  =  ×   ∠ A + × ∠ ABC
In triangle BOC,

∠BOC = 180  -  ∠CBO  -  ∠BCO (sum of angles in a triangle)

∠ BOC = 180 - (∠CBO  +  ∠ BCO)

∠ BOC = 180 - (×  ∠ A +  × ∠ ABC + × ∠ A  +  × ∠ ACB)

∠ BOC = 180 - (∠A +  ×  ∠ ABC +  × ∠ ACB)

∠BOC = 180 - (∠A + 90 -  × ∠A)

∠BOC = 90 - × ∠ A
Hence prove