14. In the figure below, $B A D, B C E, A C F$ and $D E F$ are straight lines. It is given that $B A=B C, A D=A F, E B=E D$. If $\angle B E D=x^{\circ}$, find the value of $x$.
14. Answer: 108
Let $\angle A B C=\alpha$ and $\angle B A C=\beta$. Since $B A=B C$, we have $\angle B C A=\angle B A C=\beta$. As $E B=E D$, it follows that $\angle E D B=\angle E B D=\angle A B C=\alpha$. Then $\angle A F D=\angle A D F=$ $\angle E D B=\alpha$ since $A D=A F$. Note that $\angle D A F=180^{\circ}-\beta$. In $\triangle A B C$, we have $\alpha+2 \beta=180^{\circ}$; and in $\triangle A D F$, we have $2 \alpha+180^{\circ}-\beta=180^{\circ}$. From the two equations, we obtain $\alpha=36^{\circ}$. By considering $\triangle B D E$, we obtain $x=180^{\circ}-2 \alpha=108^{\circ}$.