14. In the triangle $A B C, \angle B=90^{\circ}, \angle C=20^{\circ}, D$ and $E$ are points on $B C$ such that $\angle A D C=$ $140^{\circ}$ and $\angle A E C=150^{\circ}$. Suppose $A D=10$. Find $B D \cdot C E$.
14. Answer. 50
Direct calculation gives $\angle D A C=20^{\circ}$ and $\angle B A D=50^{\circ}$. Thus $A D=C D=10$. Also $B D=10 \sin 50^{\circ}$. By sine rule applied to the triangle $A E C$, we have $\frac{C E}{\sin 10^{\circ}}=\frac{A C}{\sin 150^{\circ}}=$ $\frac{2 \times 10 \cos 20^{\circ}}{\sin 150}=40 \cos 20^{\circ}$. (Note that $A D=D C$.)

Therefore, $B D \cdot C E=400 \cos 20^{\circ} \sin 10^{\circ} \sin 50^{\circ}$.
Direct calculation shows that $\cos 20^{\circ} \sin 10^{\circ} \sin 50^{\circ}=\frac{1}{4}$ so that $B D \cdot C E=50$.