Let's construct an isosceles triangle, given the base $a$, and the difference between the leg and the height corresponding to $a$ is $b-m$.
We are constructing a right-angled triangle $CDE$ where

$$
C D=\frac{a}{2} \quad \text{and} \quad D E=b-m
$$

On the hypotenuse $CD$, we erect a perpendicular at its midpoint. This perpendicular intersects $DE$ at $A$. If we choose point $B$ on the line $CE$ such that $BE=EC$, then $ABC$ is the desired triangle, because

$$
A C-A E=A D-A E=D E=b-m
$$

and

$$
B C=2 C E=a
$$

(Ferenc Grünhut, Budapest.)