Let's construct a rhombus, given the side of the rhombus $a$ and the difference of the diagonals $d(=e-f)$.
Consider the task as solved. The construction is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_823be7e7a33b403264fbg-1.jpg?height=556&width=629&top_left_y=183&top_left_x=737)

$M N = M B = \frac{f}{2}, A N = M A - M N = \frac{e}{2} - \frac{f}{2} = \frac{d}{2}$. Therefore, the problem essentially involves constructing the right triangle $A M B$, where the hypotenuse $A B = a$ and the difference between the two legs is $\frac{d}{2}$. Since $B M N$ is an isosceles right triangle, $\angle A N B = 135^{\circ}$. Thus, the geometric locus of the points $N$ is, on one hand, the arc of the angle of view drawn over the distance $A B$, from which the segment $A B$ is seen at an angle of $135^{\circ}$, and on the other hand, the circle with radius $\frac{d}{2}$ centered at $A$.

Accordingly, the construction process is as follows: Draw the Thales circle over $A B$, and one of its intersection points with the diameter perpendicular to $A B$ will be the center $O$ of the angle of view arc. The intersection of the angle of view arc and the circle with radius $\frac{d}{2}$ centered at $A$ provides the point $N$. The other intersection of the line $A N$ with the Thales circle is the point $M$, which is the intersection of the diagonals of the rhombus. The solvability condition is that the circle with radius $\frac{d}{2}$ intersects the angle of view arc, i.e., $\frac{d}{2} < a$, or $d < 2a$.