In a triangle $ABC$, the point $P$ inside the triangle that minimizes the sum $PA+PB+PC$ is called the Fermat point.  In case there is an angle in the triangle with measure $\geq 120^\circ$, say $\angle A \geq 120^\circ$, then the Fermat point is that vertex $A$.  If every angle is less than $120^\circ$, then the Fermat point is the point in the interior of the triangle such that the three segments $PA$, $PB$, and $PC$ form $120^\circ$ angles with each other. (This is a definition, and there's a lot of ways to find that point).

In this question, we look for the Fermat point of $\triangle ABD$, and calculate the sum of the lengths based on the side length of the square, and solve the equation of the side length.  The method of rotation can be used: rotate the triangle $AED$ around $A$ by $60^\circ$ to the outside of the triangle. (Note that it is $60^\circ$, not $90^\circ$).