$\angle ACD$ is an exterior angle of $\Delta ABC$. The bisector of $\angle ABC$ and exterior $\angle ACD$ intersect each other at point $E$. Prove that,$\angle BEC = \frac{1}{2} \angle BAC$.

(N/A) In $\Delta ABC$,$\angle ACD$ is an exterior angle. By the exterior angle theorem,$\angle ACD = \angle BAC + \angle ABC$.
In $\Delta EBC$,$\angle ECD$ is an exterior angle. Therefore,$\angle ECD = \angle EBC + \angle BEC$.
Since $BE$ is the bisector of $\angle ABC$,$\angle EBC = \frac{1}{2} \angle ABC$.
Since $CE$ is the bisector of $\angle ACD$,$\angle ECD = \frac{1}{2} \angle ACD = \frac{1}{2} (\angle BAC + \angle ABC)$.
Substituting these into the equation for $\Delta EBC$: $\frac{1}{2} (\angle BAC + \angle ABC) = \frac{1}{2} \angle ABC + \angle BEC$.
Simplifying,$\frac{1}{2} \angle BAC + \frac{1}{2} \angle ABC = \frac{1}{2} \angle ABC + \angle BEC$.
Thus,$\angle BEC = \frac{1}{2} \angle BAC$.