Trapezoid and Parallelogram

We note that $ABCE$ is a parallelogram and $\triangle DEF$ and $\triangle BCF$ are similar. Therefore, $\dfrac {DE}{BC} = \dfrac {AE-AD}{BC} \\ = 1 - \dfrac {AD}{BC} = \dfrac {EF}{BF} \implies \dfrac {AD}{BC} = 1 - \dfrac {EF}{BF}$ . We need to find $BE$ and $EF$ to solve the problem.

Since $BC=2AB$ , let $AB=CE=1$ and $BC=EA=2$ . Let $\angle EBC = \theta$ . Then

$\begin{aligned} \tan \theta & = \frac {EG}{BG} = \frac {EG}{BC+CG} = \frac {CE\sin 60^\circ}{BC+CE\cos 60^\circ} = \frac {\frac {\sqrt 3}2}{2+\frac 12} = \frac {\sqrt 3}5 \\ \implies \cos \theta & = \frac 5{2\sqrt 7} \\ \implies BF & = BC\cos \theta = 2 \times \frac 5{2\sqrt 7} = \frac 5{\sqrt 7} \end{aligned}$

We note that $BE = \dfrac {BG}{\cos \theta} = \dfrac {\frac 52}{\frac 5{2\sqrt 7}} = \sqrt 7$ .

Therefore, $\dfrac {AD}{BC} = 1 - \dfrac {EF}{BF} = 1 - \dfrac {BE-BF}{BF} = 2 - \dfrac {BE}{BF} = 2 - \dfrac {\sqrt 7}{\frac 5{\sqrt 7}} = \dfrac 35 = \boxed{0.6}$

Without loss of generality, let $AB = EC = 1$ , which would make $BC = AE = 2$ . Place the diagram on a coordinate grid such that $B$ is at $B(0, 0)$ and $C$ is at $C(2, 0)$ .

Since $\angle B = 60°$ and $AB = 1$ , $A$ is at $A(\frac{1}{2}, \frac{\sqrt{3}}{2})$ and since $AD || BC$ , $AD$ has an equation of $y = \frac{\sqrt{3}}{2}$ . Also, since $EC || AB$ , $E$ is at $E(\frac{5}{2}, \frac{\sqrt{3}}{2})$ .

That means $BE$ has a slope of $\frac{\sqrt{3}}{5}$ , and since $CD$ is perpendicular to $BE$ and goes through $C(2, 0)$ , $CD$ has an equation of $y = -\frac{5}{\sqrt{3}}(x - 2)$ .

$CD$ and $AD$ intersect at the solution of $y = -\frac{5}{\sqrt{3}}(x - 2)$ and $y = \frac{\sqrt{3}}{2}$ , which is $D(\frac{17}{10}, \frac{\sqrt{3}}{2})$ . $AD$ then has a distance of $\frac{17}{10} - \frac{1}{2} = \frac{6}{5}$ .

The ratio of $AD$ to $BC$ is then $\frac{\frac{6}{5}}{2} = \frac{3}{5} = \boxed{0.6}$