10.1.19 ** Given isosceles $\triangle A B C$, with base angles $\angle A B C=\angle A C B=50^{\circ}$, points $D$ and $E$ are on $B C$ and $A C$ respectively, $A D$ and $B E$ intersect at point $P, \angle A B E=30^{\circ}, \angle B A D=50^{\circ}$, connect $E D$, find $\angle B E D$.
As shown in the figure, let's assume $A B=A C=1$, then $B C=$ $2 \cos 50^{\circ}$. And $A D=B D=\frac{1}{2 \cos 50^{\circ}}$. From the area, we know $\frac{P D}{\Lambda P}=$ $\frac{B D \sin 20^{\circ}}{A B \sin 30^{\circ}}=\frac{\sin 20^{\circ}}{\cos 50^{\circ}}$, and by the Law of Sines, we get $\frac{P E}{A P}=$ $\frac{\sin 30^{\circ}}{\sin 70^{\circ}}$.

If we can prove these two ratios are equal, then we have $P D=P E$, and since $\angle A P E=80^{\circ}$, we get $\angle B E D=40^{\circ}$. Therefore, our task becomes proving $\cos 50^{\circ} \sin 30^{\circ}-\sin 20^{\circ} \sin 70^{\circ}$ or $\cos 50^{\circ}=2 \sin 20^{\circ} \sin 70^{\circ}$. Using the product-to-sum identities, the right side $=\cos 50^{\circ}-\cos 90^{\circ}=$ the left side, hence the conclusion holds.