Let's construct an isosceles triangle, given the height to the base: $m_{a}$ and the heights to the legs: $m_{b}=m_{c}$.
Consider the task as solved. The labeling is shown in Figure 1.

![](https://cdn.mathpix.com/cropped/2024_05_02_9e712873dbf9782e45cdg-1.jpg?height=599&width=588&top_left_y=259&top_left_x=736)

From the right triangle $A A_{0} A_{1}$, it is evident that the hypotenuse $A A_{1}$ is always greater than the leg $A_{1} A_{0}$, which means a necessary condition for solvability is $m_{a} > \frac{m_{b}}{2}$.

This condition is also sufficient, because for any $m_{a}$ and $m_{b}$ that satisfy this condition, the right triangle $A A_{0} A_{1}$ can always be constructed, and from this triangle, the construction of $\triangle ABC$ is trivial.

The construction process: The right triangle $A A_{0} A_{1}$ can be constructed in various ways. Perhaps the most convenient way is to construct the isosceles triangle $A A_{1}^{\prime} A_{1}$ (Figure 1) from the base $A_{1} A_{1}^{\prime} = m_{b}$ and the legs $A_{1} A = A_{1}^{\prime} A = m_{a}$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_9e712873dbf9782e45cdg-1.jpg?height=632&width=645&top_left_y=1256&top_left_x=726)

Figure 2

The altitude $A A_{0}$ of this triangle to its base will contain the point $C$. The line $BC$ passes through $A_{1}$ and $BC \perp A A_{1}$, and $A_{1} B = A_{1} C$.

There is always one and only one solution if $m_{a} > \frac{m_{b}}{2}$.