14. [8] Let $A B C D$ be a trapezoid with $A B \| C D$ and $\angle D=90^{\circ}$. Suppose that there is a point $E$ on $C D$ such that $A E=B E$ and that triangles $A E D$ and $C E B$ are similar, but not congruent. Given that $\frac{C D}{A B}=2014$, find $\frac{B C}{A D}$.
Answer: $\sqrt{4027}$ Let $M$ be the midpoint of $A B$. Let $A M=M B=E D=a, M E=A D=b$, and $A E=B \bar{E}=c$. Since $\triangle B E C \sim \triangle D E A$, but $\triangle B E C$ is not congruent to $\triangle D A E$, we must have $\triangle B E C \sim \triangle D E A$. Thus, $B C / B E=A D / D E=b / a$, so $B C=b c / a$, and $C E / E B=A E / E D=c / a$, so $E C=c^{2} / a$. We are given that $C D / A B=\frac{\frac{c^{2}}{a}+a}{2 a}=\frac{c^{2}}{2 a^{2}}+\frac{1}{2}=2014 \Rightarrow \frac{c^{2}}{a^{2}}=4027$. Thus, $B C / A D=\frac{b c / a}{b}=c / a=\sqrt{4027}$.