It's All Ratios!

Let $\theta = \angle EAF = \angle FAD$ . Then by alternate interior angles of parallel lines, $\angle BEA = 2\theta$ .

From $\triangle ADF$ , $FD = \tan \theta$ and $AF = \sec \theta$ .

From $\triangle ABE$ , $AE = \csc 2\theta$ .

Therefore, $\cfrac{AF^2}{FD \cdot AE} = \cfrac{\sec^2 \theta}{\tan \theta \cdot \csc 2\theta} = \cfrac{\frac{1}{\cos^2 \theta}}{\frac{\sin \theta}{\cos \theta} \cdot \frac{1}{2 \sin \theta \cos \theta}} = \boxed{2}$ .

$\tan(2\theta) = \dfrac{x}{a}$ and $\tan(\theta) = \dfrac{b}{x} \implies$ $\dfrac{x}{a} = \tan(2\theta) = \dfrac{2\tan(\theta)}{1 - \tan^2(\theta)} = \dfrac{2bx}{x^2 - b^2}$

$\implies 2ab = x^2 - b^2 \implies x^2 = b^2 + 2ab = (b + a)^2 - a^2 \implies \overline{AE} = b + a$

and

$\overline{AF}^2 = (a + b)^2 - a^2 + b^2 = (a + b)^2 - (a - b)(a + b) = 2(a + b)(b) =$

$2 * (\overline{AE}) * (\overline{FD}) \implies \dfrac{\overline{AF}^2}{\overline{FD} * \overline{AE}} = \boxed{2}$ .