$A`B=\mathrm{AB}=\sqrt{2}\Rightarrow A`C=1$ (by Pythagoras)

$\Delta A`\mathrm{BC}$ is therefore a right-angled isosceles triangle.

$\Rightarrow \angle \mathrm{BA}`C={45}^{\circ }\Rightarrow \angle \mathrm{EA}`D={45}^{\circ }$

$\Rightarrow \mathrm{ED}=A`D=\sqrt{2}-1$