Let's construct a rhombus, given one side and the difference between the two diagonals.
Let $a$ be the given side, $2d$ the difference of the two diagonals. We draw a triangle $ABE$ where $AB=a$, $AE=d$, and $\angle AEB=135^{\circ}$. On the extension of $AE$, we determine point $C$ such that $AB=CB$. Then $A$, $B$, and $C$ are three vertices of the desired rhombus. We denote the foot of the perpendicular from $B$ to $AC$ by $O$.

$$
\angle BEO=180^{\circ}-135^{\circ}=45^{\circ},
$$

thus

$$
AO-BO=AO-EO=AE-d
$$

(László Szántó, Pécs.)