Question

in a ∆ABC internal angle bisectors of angle B and angle C meet at O . prove that angle BOC= 90° + 1/2 of angle BAC

Answer 1

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Answer 2

Answer

In Triangle BOC:

/_BOC = 180 - 1/2(/_B + /_C) [Angle Sum Property of a triangle]

=> /_B + /_C = 360 - 2/_BOC

In Triangle ABC:

/_B + /_C = 180 - /_A [Angle Sum Property of a triangle]

.°. 360 - 2/_BOC = 180 - /_A

=> 180 = 2/_BOC - /_A

=> /_BOC = 90 + /_A/2

HENCE PROVED