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ICSE Class 7 Answered

The sides AB and AC of triangle ABC are produced to P and Q respectively. The bisectors of exterior angles at B and C of triangle ABC meet at O. Prove that angle BOC = 90 degree -1/2 angle A
Asked by naitik892006 | 04 Nov, 2018, 12:54: PM
answered-by-expert Expert Answer
In triangle ABC
(angle ABC + angle ACB)=180-angle A
now 
angle CBP = angle A + angle ACB
so angle CBO = 1/2 angle A + 1/2 angle ACB
similarly
angle BCO = 1/2 angle A + 1/2 angle ABC
in triangle BOC
angle BOC = 180 - angle CBO - angle BCO
angle BOC = 180 - (angle CBO + angle BCO)
angle BOC = 180 - (1/2 angle A + 1/2 angle ABC+1/2 angle A + 1/2 angle ACB)
angle BOC = 180 - (angle A + 1/2 angle ABC + 1/2 angle ACB)
angle BOC = 180 - (angle A + 90 - 1/2 angle A)
angle BOC = 90 - 1/2 angle A
Answered by Arun | 05 Nov, 2018, 11:19: AM
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