Example 7 As shown in Figure 10, in $\triangle A B C$, $A B=A C, \angle A=20^{\circ}$, points $D$ and $E$ are taken on sides $A B$ and $A C$ respectively, such that $\angle E B C=60^{\circ}, \angle D C B=50^{\circ}$. Find the measure of $\angle B E D$.
Solution 1: Let $\angle B E D=x$, then $\angle C E D=40^{\circ}+x$.

Applying the trigonometric form of Ceva's theorem to $\triangle B C E$ and point $D$, we have
$$
\begin{array}{l}
1=\frac{\sin \angle B C D}{\sin \angle D C E} \cdot \frac{\sin \angle C E D}{\sin \angle D E B} \cdot \frac{\sin \angle E B D}{\sin \angle D B C} \\
=\frac{\sin 50^{\circ}}{\sin 30^{\circ}} \cdot \frac{\sin \left(x+40^{\circ}\right)}{\sin x} \cdot \frac{\sin 20^{\circ}}{\sin 80^{\circ}} . \\
\text { Then } \frac{\sin \left(x+40^{\circ}\right)}{\sin x}=\frac{\sin 80^{\circ}}{2 \cos 40^{\circ} \cdot \sin 20^{\circ}} \\
=2 \cos 20^{\circ}=\frac{\sin 70^{\circ}}{\sin 30^{\circ}}=\frac{\sin \left(30^{\circ}+40^{\circ}\right)}{\sin 30^{\circ}} .
\end{array}
$$

Since $\frac{\sin \left(x+40^{\circ}\right)}{\sin x}=\cos 40^{\circ}+\cot x \cdot \sin 40^{\circ}$
is a strictly decreasing function of $x$ in $(0, \pi)$, therefore,
$$
\angle B E D=x=30^{\circ} \text {. }
$$

Solution 2: Let $\angle B E D=x$, then,
$$
\angle E D B=160^{\circ}-x \text {. }
$$

Applying the trigonometric form of Ceva's theorem to $\triangle E D C$ and point $B$, we have
$$
\begin{array}{l} 
1=\frac{\sin \angle D E B}{\sin \angle B E C} \cdot \frac{\sin \angle E C B}{\sin \angle B C D} \cdot \frac{\sin \angle C D B}{\sin \angle B D E} \\
=\frac{\sin x}{\sin 40^{\circ}} \cdot \frac{\sin 80^{\circ}}{\sin 50^{\circ}} \cdot \frac{\sin 50^{\circ}}{\sin \left(160^{\circ}-x\right)} . \\
\text { Then } \frac{\sin \left(160^{\circ}-x\right)}{\sin x}=2 \cos 40^{\circ}=\frac{\sin 50^{\circ}}{\sin 30^{\circ}} \\
=\frac{\sin 130^{\circ}}{\sin 30^{\circ}}=\frac{\sin \left(160^{\circ}-30^{\circ}\right)}{\sin 30^{\circ}} .
\end{array}
$$

Thus, $\sin 160^{\circ} \cdot \cot x-\cos 160^{\circ}$
$$
=\sin 160^{\circ} \cdot \cot 30^{\circ}-\cos 160^{\circ} \text {. }
$$

Therefore, $\angle B E D=x=30^{\circ}$.
A careful reader might naturally ask, if the trigonometric form of Ceva's theorem is applied to $\triangle D B E$ and point $C$ or $\triangle D B C$ and point $E$, can the same result be obtained? The answer is yes, but the derivation becomes more complex. Specifically, a trigonometric identity must be used in the derivation, which is given here as a lemma.
Lemma $2 \sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 80^{\circ}=\frac{\sqrt{3}}{4}$.
Proof of the lemma: By the product-to-sum formulas, we have
$$
\begin{array}{l}
2 \sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 80^{\circ} \\
=2 \sin 20^{\circ} \cdot \cos 10^{\circ} \cdot \sin 40^{\circ} \\
=\left(\sin 30^{\circ}+\sin 10^{\circ}\right) \sin 40^{\circ} \\
=\frac{1}{2} \sin 40^{\circ}+\sin 10^{\circ} \cdot \sin 40^{\circ} \\
=\frac{1}{2} \sin 40^{\circ}+\frac{1}{2}\left(\cos 30^{\circ}-\cos 50^{\circ}\right) \\
=\frac{1}{2} \sin 60^{\circ}=\frac{\sqrt{3}}{4} .
\end{array}
$$

The detailed derivation of the above two angles is left to interested readers.