【Example 4.25】As shown in Figure 4.17, in $\triangle A B C$, $A B=A C, D$ is between $A, B$, $E$ is between $A, C$, $\angle A C D=A=20^{\circ}, \angle A B E=30^{\circ}$, find angle $\alpha$.
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\begin{array}{l}
\text { Solve } \frac{\sin \alpha}{\sin \beta}=\frac{E C}{A E}=\frac{B C \sin 50^{\circ}}{A B \sin 30^{\circ}}=\frac{\sin 20^{\circ}}{\sin 80^{\circ}} \cdot \frac{\sin 50^{\circ}}{\sin 30^{\circ}} \\
=\frac{2 \sin 20^{\circ} \cos 40^{\circ}}{\sin 80^{\circ}} \text {. } \\
\sin \alpha \cos 10^{\circ}=2 \sin 20^{\circ} \cos 40^{\circ} \sin \left(40^{\circ}+\alpha\right) \\
=\sin 20^{\circ}\left[\sin \left(80^{\circ}+\alpha\right)+\sin \alpha\right], \\
\sin \alpha=2 \sin 10^{\circ} \sin \alpha+2 \sin 10^{\circ} \cos \left(10^{\circ}-\alpha\right) \\
=2 \sin 10^{\circ} \sin \alpha+\sin \left(20^{\circ}-\alpha\right)+\sin \alpha, \\
\therefore 0=2 \sin 10^{\circ} \sin \alpha+\sin 20^{\circ} \cos \alpha-\cos 20^{\circ} \sin \alpha \text {, } \\
\therefore \tan \alpha=\frac{\sin 20^{\circ}}{\cos 20^{\circ}-2 \sin 10^{\circ}}=\frac{\sin 20^{\circ}}{\sin 70^{\circ}-\sin 10^{\circ}-\sin 10^{\circ}} \\
=\frac{\sin 20^{\circ}}{2 \sin 30^{\circ} \cos 40^{\circ}-\sin 10^{\circ}}=\frac{\sin 20^{\circ}}{\sin 50^{\circ}-\sin 10^{\circ}} \\
=\frac{\sin 20^{\circ}}{2 \sin 20^{\circ} \cos 30^{\circ}}=\frac{1}{\sqrt{3}} \text {, thus } \alpha=30^{\circ} \text {. } \\
\end{array}
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In the 1950s, American mathematician Thompson used a pure geometric method to solve this problem. Later, many people published various solutions to this problem, none of which were as straightforward as this one.
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