jmc

Let $\overline{DE}$ have length $x$, so $\overline{BD}$, the median, has length $2x$. In a right triangle, the median to the hypotenuse has half the length of the hypotenuse, so $AD=DC=2x$ as well. Then, $$EC=DC-DE=2x-x=x.$$We can find $BE$ by using the Pythagorean theorem on right triangle $△BDE$, which gives $$BE=\sqrt{B{D}^2-D{E}^2}=\sqrt{(2x{)}^2-x^2}=x\sqrt{3}.$$We have $AE=AD+DE=2x+x=3x$. Now, we use the Pythagorean theorem on right triangle $△ABE$, which gives $$AB=\sqrt{A{E}^2+B{E}^2}=\sqrt{(3x{)}^2+(x\sqrt{3}{)}^2}=2x\sqrt{3}.$$(Triangles $△BDE$ and $△ABE$ have sides in a $1:\sqrt{3}:2$ ratio, so they are $30^{\circ }-60^{\circ }-90^{\circ }$ triangles; there are others, too.)

Finally, we have $$\frac{AB}{EC}=\frac{2x\sqrt{3}}{x}=\boxed{2\sqrt{3}}.$$