Find ED

Theorem: In a right triangle, let the length of the altitude to the hypotenuse be $x$ and let the altitude divide the hypotenuse into two parts of lengths $p$ and $q$ . Then, we have $x^2=pq$ .

Proof: The result follows from a straight forward application of similar triangles.

Using the above fact on $\triangle ABC$ , $BE^2 = AE \times EC=4 \times 1$ $\implies BE=2$

Similarly applying the theorem on $\triangle BAD$ , $AE^2= BE \times ED$ $\implies 4^2=2 \times ED$ $\implies \boxed{ED=8}$

$\cos \theta=\dfrac{AB}{5}=\dfrac{4}{AB}$ $\implies$ $AB^2=20$ $\implies$ $AB=2\sqrt{5}$

By pythagorean theorem, $BC=\sqrt{5^2-\left(2\sqrt{5}\right)^2}=\sqrt{5}$

$\tan \alpha =\dfrac{ED}{4}=\dfrac{2\sqrt{5}}{\sqrt{5}}$ $\implies$ $\dfrac{ED}{4}=2$ $\implies$ $ED=8$