14. Given an equilateral triangle $ABC$, points $D$, $E$, and $F$ are on $BC$, $AC$, and $AB$ respectively, with $BC=3BD$, $BA=3BF$, and $EA=\frac{1}{3}AC$. Find the degree measure of $\angle ADE + \angle FEB$.
【Answer】 $30^{\circ}$
【Analysis】 $\because \angle A D E=\angle F E D, \therefore \angle A D E+\angle F E B=\angle F E D$. Also, $\because \angle F A E=60^{\circ}$, $A E: A F=1: 2, \therefore \angle A E F=90^{\circ}$, and $A B / / D E$, then $\angle B \cdot 4 C=\angle D E C=60^{\circ}$ : Therefore, $\angle A D E+\angle F E B=\angle F E D=180^{\circ}-90^{\circ}-60^{\circ}=30^{\circ}$