Question

In a triangle ABC the bisector of interior 2 A and the bisector of exterior 2 C meet at point O. Prove that angle AOC= 1/2 angle B
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Answer 1

Answer:

Exterior angle in a triangle=sum of two opposite interior angles.

⟹In△ABC,

∠BCT=∠A+∠B

⟹2α=2θ+∠B

=>α=θ+

2

∠B

−(1)

In△OAC

using the same angle property,

=>∠OCT=∠OAC+∠AOC

=>α=θ+∠AOC−(2)

comparing (1)& (2)

=>θ+∠AOC=θ+

2

∠B

∠AOC=

2

∠B

=

m

∠B

∴m=2