Example 1 As shown in Figure 1, in Rt $\triangle A B C$, $\angle C=90^{\circ}, \angle A=30^{\circ}$, the angle bisector of $\angle C$ intersects the external angle bisector of $\angle B$ at point $E$, and $A E$ is connected. Then $\angle A E B$ is equal to ()$^{[2]}$.
Figure 1
(A) $50^{\circ}$
(B) $45^{\circ}$
(C) $40^{\circ}$
(D) $35^{\circ}$
(2003, Shandong Province "KLT Kailingtong Cup" Junior High School Mathematics Competition)

Reference [2] directly provides the answer (B) without giving the solution process. However, it is not difficult to provide a solution that reflects the problem's intent and is easy for junior high school students to understand by comprehending the conditions of the problem.
Solution 1: As shown in Figure 2, draw $E D \perp C B$ at $D$, and draw $E H \perp A B$ at $H$. Since $B E$ is the bisector of $\angle A B D$, we have
$\angle H B E=\angle D B E$.
Also, $B E=B E$, hence

Rt $\triangle H B E \cong \mathrm{Rt} \triangle D B E$.
Thus, $E H=E D$.
Draw $E F \perp C A$ at $F$, similarly we get
Rt $\triangle F C E \cong$ Rt $\triangle D C E$.
Thus, $E F=E D$.
From (1) and (2), we have $E H=E F$.
Since $A E=A E$, we have
$\mathrm{Rt} \triangle E H A \cong \mathrm{Rt} \triangle E F A$.
Therefore, $\angle E A H=\angle E A F=\frac{1}{2} \angle H A F$
$=\frac{1}{2}\left(180^{\circ}-30^{\circ}\right)=75^{\circ}$.
In $\triangle A B E$,
$\angle A E B=180^{\circ}-\angle B A E-\angle A B E$
$=180^{\circ}-75^{\circ}-60^{\circ}=45^{\circ}$.
Evaluation: This solution involves drawing perpendiculars (auxiliary lines) from the angle bisector, proving congruence of triangles three times, and finally using the triangle angle sum theorem to reach the conclusion. The process is correct, the answer is correct, and all conditions are used. However, the use of the angle bisector is limited to its definition, and the structural characteristics of the figure are revealed but not deeply or consciously. Some readers may have noticed that the application of congruent triangles repeats the proof of the "angle bisector theorem and its converse." If we take one step further and use the angle bisector theorem, we can save the application of congruent triangles.
Solution 2: As shown in Figure 2, the auxiliary lines are the same as in Solution 1.
Since $E B$ is the bisector of $\angle A B D$ and $E C$ is the bisector of $\angle A C B$, we have
$E D=E H, E D=E F$.
Thus, $E H=E F$.
Therefore, point $E$ lies on the bisector of $\angle F A H$.
The rest is the same as Solution 1.
Evaluation: This problem can also be proven by drawing other auxiliary lines (including the excircle, etc.). However, as long as the condition $\angle A=30^{\circ}$ is used to reach the conclusion, the problem has not been fully understood. Figure 2 already shows that the essence of the problem is: folding the square $C D E F$ along $E B$ and $E A$ (symmetry transformation) makes $E D$ and $E F$ coincide with $E H$, then
$$
\begin{array}{l}
\angle A E B=\angle A E H+\angle H E B \\
=\frac{1}{2} \angle F E H+\frac{1}{2} \angle H E D \\
=\frac{1}{2} \angle F E D=45^{\circ} .
\end{array}
$$
Clearly, this result is independent of whether $\angle A$ is $30^{\circ}$. In this sense, $\angle A=30^{\circ}$ is an unnecessary condition. Of course, providing this condition can reduce the difficulty of solving the problem, but not using this condition can lead to a more concise proof, which is a "clever and elegant solution" for middle school students.
1.2 Deepening Understanding
Solution 3: As shown in Figure 2, the auxiliary lines are the same as in Solution 1.
Since $B E$ is the bisector of $\angle A B D$, we have
$E D=E H$,
$\angle H E B=\angle D E B=90^{\circ}-\frac{1}{2} \angle H B D$.
Also, since $C E$ is the bisector of $\angle F C D$, we have
$E D=E F, \angle C E D=\angle C E F=45^{\circ}$.
Therefore, $E H=E D=E F$,
$\angle D E F=90^{\circ}, \angle H E B=\frac{1}{2} \angle H E D$.
By Rt $\triangle A H E \cong$ Rt $\triangle A F E(\mathrm{HL})$, we have
$\angle A E H=\angle A E F=\frac{1}{2} \angle F E H$.
From (1) and (2), we get
$\angle A E B=\angle A E H+\angle H E B$
$=\frac{1}{2} \angle F E H+\frac{1}{2} \angle H E D$
$=\frac{1}{2} \angle F E D=45^{\circ}$.
Solution 4: As shown in Figure 2, the auxiliary lines are the same as in Solution 1.
Since $B E$ is the bisector of $\angle A B D$ and $C E$ is the bisector of $\angle A C B$, we have
$E D=E H, E D=E F$,
$\angle C E F=\angle C E D=90^{\circ}-\frac{1}{2} \angle F C D=45^{\circ}$.
(1) Therefore,
$E H=E F$, and points $\dot{A} 、 H 、 E 、 F$ are concyclic, so
$\angle A E H=90^{\circ}-\frac{1}{2} \overparen{H E^{\circ}}$
$=90^{\circ}-\frac{1}{2} \widehat{E F}^{\circ}=\angle A E F$.
Similarly, by $E D=E H$ and points $H 、 B 、 D 、 E$ being concyclic, we have
$\angle H E B=\angle B E D$.
From (1), (2), and (3), we get
$\angle A E B=\angle A E H+\angle H E B$
$=\frac{1}{2} \angle F E H+\frac{1}{2} \angle H E D$
$=\frac{1}{2} \angle F E D=45^{\circ}$.
Evaluation: This problem can also be solved in more ways without using $\angle A=30^{\circ}$ (not detailed here), which delves deeper into the underlying structure of the figure and provides a more essential understanding of the figure. Not only that, but we can also generalize and "discover" more conclusions.