Question

The bisectors of ∠B and ∠C of an isosceles triangle with AB = AC intersect each other at a point O. BO is produced to meet AC at a point M. Prove that ∠MOC = ∠ABC.

Answer 1

Given: In isosceles $∆$ABC, AB = AC; OB and OC are bisectors of ∠B and ∠C, respectively.

To prove: ∠MOC = ∠ABC

Proof:

In ∆ABC,

$\because$AB = AC (Given)
$\therefore$∠ABC = ∠ACB (Angles opposite to equal sides are equal)
$\Rightarrow \frac{1}{2}$∠ABC = $\frac{1}{2}$∠ACB
$\Rightarrow$∠OBC = ∠OCB (Given, OB and OC are the bisectors of ∠B and ∠C, respectively) .....(i)

Now, in ∆OBC, ∠MOC is an exterior angle
$\Rightarrow$∠MOC = ∠OBC + ∠OCB (An exterior angle is equal to the sum of two opposite interior angles)
$\Rightarrow$∠MOC = ∠OBC + ∠OBC [From (i)]
$\Rightarrow$∠MOC = 2∠OBC
Hence, ∠MOC = ∠ABC (Given, OB is the bisector of ∠B)