1. In rectangle $A B C D, E$ and $F$ are chosen on $\overline{A B}$ and $\overline{C D}$, respectively, so that $A E F D$ is a square. If $\frac{A B}{B E}=\frac{B E}{B C}$, determine the value of $\frac{A B}{B C}$.
1. Let $x$ be $B E$ and $y$ be $A E$. Note that $A E F D$ is a square so $A E=B C=y$. Also, $A B=B E+A E$ so $A B=x+y$. Since $\frac{A B}{B E}=\frac{B E}{B C}$ then $\frac{x+y}{x}=\frac{x}{y}$. Thus, we have, $x y+y^{2}=x^{2}$ which yields to $x^{2}-x y-y^{2}=0$. Solving for $x$ using the quadratic formula gives us, $x=\frac{y \pm \sqrt{y^{2}-4(1)\left(-y^{2}\right)}}{2}=\left(\frac{1 \pm \sqrt{5}}{2}\right) y$. However, we will only take $x=$ $\left(\frac{1+\sqrt{5}}{2}\right) y$ since the other solution will mean that $x<0$ which is absurd since $x$ is a measure of length. Thus, $\frac{A B}{B C}=\frac{x+y}{y}=\frac{\left(\frac{1+\sqrt{5}}{2}\right) y+y}{y}=\frac{1+\sqrt{5}+2}{2}=\frac{3+\sqrt{5}}{2}$. Therefore, the answer is $\frac{3+\sqrt{5}}{2}$.