10.1.21 ** Given $\triangle A B C$, point $E$ is an interior point, $B E$ extended intersects $A C$ at point $D$, it is known that $\angle D C E=10^{\circ}, \angle D B C=30^{\circ}, \angle E C B=20^{\circ}, \angle A B D=40^{\circ}$, find $\angle B A E$.
Parse as shown in the figure, the specific steps are as follows: Since $\angle B D A=60^{\circ}$, our goal is to prove $A B^{2}=B E \cdot B D$. Let's assume $B C=1$, then $B D=\frac{1}{\sqrt{3}}$, and by the Law of Sines, $\frac{B E}{B C}=\frac{\sin 20^{\circ}}{\sin 130^{\circ}}$, which means $B E=\frac{\sin 20^{\circ}}{\sin 50^{\circ}}$. Also, $\frac{A B}{B C}=\frac{\sin 30^{\circ}}{\sin 80^{\circ}}$. The problem then reduces to proving $\frac{\sin ^{2} 30^{\circ}}{\sin ^{2} 80^{\circ}}=\frac{1}{\sqrt{3}} \frac{\sin 20^{\circ}}{\sin 50^{\circ}}$ or proving $\sin 20^{\circ} \sin ^{2} 80^{\circ}=\frac{\sqrt{3}}{4} \sin 50^{\circ}$ or $\frac{1+\cos 20^{\circ}}{2} \cdot \sin 20^{\circ}=$ $\frac{\sqrt{3}}{4} \sin 50^{\circ}$, which is equivalent to $2 \sin 20^{\circ}+\sin 40^{\circ}=\sqrt{3} \sin 50^{\circ}$.

This equation is easy to prove, because $\frac{\sqrt{3}}{2} \sin 50^{\circ}-\frac{1}{2} \sin 40^{\circ}=\sin 50^{\circ} \cos 30^{\circ}-\cos 50^{\circ} \sin 30^{\circ}=$ $\sin 20^{\circ}$. Therefore, $A B^{2}=B E \cdot B D \quad$ or $\frac{A B}{B D}=\frac{B E}{A B}$. Hence, $\triangle A B E \sim \triangle D B A$, and thus $\angle B A E=$ $60^{\circ}$.