$\angle ACD$ is an exterior angle of $\Delta ABC$ and the bisector of $\angle A$ intersects $BC$ at $E$. Prove that $\angle ABC + \angle ACD = 2 \angle AEC$.

(N/A) Let $\angle ABC = \angle B$,$\angle ACB = \angle C$,and $\angle BAC = \angle A$.
Since $AE$ is the bisector of $\angle BAC$,we have $\angle BAE = \angle CAE = \frac{1}{2} \angle A$.
In $\Delta ABC$,the exterior angle $\angle ACD = \angle ABC + \angle BAC = \angle B + \angle A$.
In $\Delta AEC$,the exterior angle $\angle AEC$ is not directly applicable,so we use the angle sum property: $\angle AEC = 180^\circ - (\angle C + \angle CAE) = 180^\circ - (\angle C + \frac{1}{2} \angle A)$.
Alternatively,in $\Delta AEC$,$\angle AEC = \angle B + \angle BAE = \angle B + \frac{1}{2} \angle A$.
Multiplying by $2$,we get $2 \angle AEC = 2 \angle B + \angle A$.
We know $\angle ABC + \angle ACD = \angle B + (\angle B + \angle A) = 2 \angle B + \angle A$.
Thus,$\angle ABC + \angle ACD = 2 \angle AEC$.