Question

In a Δ ABC, it is given that AB = AC and the bisectors of ∠B and ∠C intersect at O. If M is a point on BO produced, prove that ∠MOC = ∠ABC.

Answer 1

Given that in Δ ABC,

AB = AC and the bisectors of ∠B and ∠C intersect at O and M is a point on BO produced.

We have to prove ∠MOC = ∠ABC

Since,

AB = AC

Δ ABC is isosceles

∠B = ∠C

Or,

∠ABC = ∠ACB

Now,

BO and CO are bisectors of ∠ABC and ∠ACB respectively.

∠ABO = ∠OBC = ∠ACO = ∠OCB = \(\frac{1}{2}\)∠ABC = \(\frac{1}{2}\)∠ACB (i)

We have, in ΔOBC

∠OCB + ∠OBC + ∠BOC = 180°(ii)

And also,

∠BOC + ∠COM = 180°(iii) [Straight angle]

Equating (ii) and (iii), we get

∠OCB + ∠OBC + ∠BOC = ∠BOC + ∠COM

∠OBC + ∠OBC = ∠MOC

2∠OBC = ∠MOC

2( \(\frac{1}{2}\)∠ABC) = ∠MOC [From (i)]

∠ABC = ∠MOC

Therefore,

∠MOC = ∠ABC