Bisectors of the angles $B$ and $C$ of an isosceles triangle with $AB = AC$ intersect each other at $O$. $BO$ is produced to a point $M$. Prove that $\angle MOC = \angle ABC$.

(N/A) Given: In $\triangle ABC$,$AB = AC$. $BO$ and $CO$ are the angle bisectors of $\angle B$ and $\angle C$ respectively,intersecting at $O$. $BO$ is produced to $M$.
Proof:
In $\triangle ABC$,since $AB = AC$,we have $\angle ABC = \angle ACB$ (Angles opposite to equal sides are equal).
Since $BO$ and $CO$ are bisectors of $\angle B$ and $\angle C$,we have:
$\angle OBC = \frac{1}{2} \angle ABC$
$\angle OCB = \frac{1}{2} \angle ACB$
Since $\angle ABC = \angle ACB$,it follows that $\angle OBC = \angle OCB$.
In $\triangle OBC$,$\angle MOC$ is an exterior angle at vertex $O$ when side $BO$ is produced to $M$.
By the Exterior Angle Property of a triangle,the exterior angle is equal to the sum of the two interior opposite angles.
Therefore,$\angle MOC = \angle OBC + \angle OCB$.
Since $\angle OBC = \angle OCB$,we can write:
$\angle MOC = \angle OBC + \angle OBC = 2 \angle OBC$.
Substituting $\angle OBC = \frac{1}{2} \angle ABC$:
$\angle MOC = 2 \times (\frac{1}{2} \angle ABC) = \angle ABC$.
Hence,$\angle MOC = \angle ABC$.