Let's construct a triangle, given two heights $m_{a}$ and $m_{b}$, and the median to the third side $s_{c}$.
Consider the task as solved and reflect the $A B C \triangle$ over the midpoint $C_{1}$ of side $A B$. As a result, we obtain the parallelogram $A C B C^{*}$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_b33a6e7051fdefbe05a9g-1.jpg?height=583&width=596&top_left_y=229&top_left_x=754)

The heights $m_{a}$ and $m_{b}$ from point $C$ of this parallelogram end at points $A^{\prime}$ and $B^{\prime}$, which lie on the Thales circle drawn over $C^{*} C$. The cyclic quadrilateral $C A^{\prime} C^{*} B^{\prime}$ is thus easily constructible, and the construction of the parallelogram $C A C^{*} B$ (with $C A \| B^{\prime} C^{*}$ and $C B \| A^{\prime} C^{*}$) is trivial. The diagonal $A B$ of the parallelogram provides the sought vertices $A$ and $B$ of the triangle. The solvability condition is $m_{a}<2 s_{c}, m_{b}<2 s_{c}$.

Zsombok Zoltán (Bp., IV. Könyves Kálmán g. II. o. t.)

Note: If the circle is halved from point $C^{*}$, then point $C$ moves to $C_{1}$, and $A^{\prime}$ and $B^{\prime}$ move to the perpendiculars of length $m a / 2$ and $m_{b} / 2$ dropped from $C_{1}$ to $C^{*} A$ and $C^{*} B$, respectively, providing a second solution that is not significantly different from the first. (We can write $C$ instead of $C^{*}$. )