Find the ratio

We note that $ABC$ is an isosceles right triangle, therefore $\alpha+\beta = 45^\circ$ . Therefore, $\angle ADC = 180^\circ - \alpha - \beta = 135^\circ$ . Let $AB=BC = 1$ . Then $AC = \sqrt 2$ . By sine rule,

$\begin{aligned} \frac {AD}{\sin \beta} & = \frac {AC}{\sin 135^\circ} = \frac {\sqrt 2}{\frac 1{\sqrt 2}} = 2 \\ \implies AD & = 2\sin \beta \\ \frac {AD}{BE} & = \frac {2\sin \beta}{\sin \beta} = \boxed 2 \end{aligned}$