The bisectors of ∠B and ∠C of an isosceles triangle with AB = AC intersect each other at a point O.

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

To prove: ∠MOC = ∠ABC

Proof:

In ∆ABC,

∵">∵∵AB = AC             (Given)
∴">∴∴∠ABC = ∠ACB   (Angles opposite to equal sides are equal)

$\Rightarrow \frac{1}{2} \angle A B C=\frac{1}{2} \angle A C B$

⇒">⇒⇒∠OBC = ∠OCB         (Given, OB and OC are the bisectors of ∠B and ∠C, respectively)    .....(i)

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