65. In $\triangle A B C$, $A B=A C$, extend the angle bisector $B D$ of $\angle A B C$ to point $E$, such that $D E=A D$. If $A B+C E=B C$, then $\angle A B C=$ $\qquad$ is solved as
$$
\alpha=40^{\circ} \text { . }
$$
Answer: $40^{\circ}$
Solution: As shown in the figure, intercept $\mathrm{CF}=\mathrm{CE}$ on $\mathrm{BC}$.
Since
Therefore
It is easy to prove
We can get
So
Thus

Let $\angle A B C=\alpha$, then we have
$$
\begin{array}{c}
A B+C E=B C, \quad B F+C F=B C, \\
\mathrm{AB}=\mathrm{BF}, \\
\triangle A B D \cong \triangle F B D, \\
\mathrm{AD}=\mathrm{DF}, \\
\triangle E D C \cong \triangle F D C, \\
\angle A D B=\angle F D B=\angle F D C=\angle E D C=60^{\circ} \text {. } \\
\angle A=180^{\circ}-2 \alpha, \\
\angle B D C=\left(180^{\circ}-2 \alpha\right)+\frac{\alpha}{2}=120^{\circ},
\end{array}
$$