Question

The bisectors of the exterior angles angle CBX and angle BCY meet at O. Prove that angle BOC = 90-1/2 angle A

Answer 1

Given :  The bisectors of the exterior angles angle CBX and angle BCY meet at O

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

Solution:

Exterior angle = Sum of opposite interior angles

∠CBX  = ∠C + ∠A

∠BCY = ∠B + ∠A

∠CBO = (1/2)∠CBX  = (1/2) (∠C + ∠A)

∠BCO = (1/2)∠BCY  = (1/2) (∠B + ∠A)

=> ∠CBO  + ∠BCO  + ∠BOC = 180°

=>  (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)  

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