Question

In ∆ ABC , the bisector of ∆B and ∆C meets at O . prove that ∆ BOC =90° + ∆ A /2.

Answer 1

Hope this answer helps you.

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