I found two triangles. Is this correct? Well, suppose we have constructed the triangle $ABC$ with the altitudes $AD=h_a, BE=h_b$ and the median $AM=m_a$ The perpendicular from $M$ to $AC$ intersects the line $AC$ at a point $K$. From the triangle $BCE$ we can see that $MK = \dfrac{h_b}{2}$. So, let's start with the construction. We construct first the triangle $ADM$. It's easy to construct it, since we know that it is right with hypotenuse $AM$ and one perpendicular side $AD$. (so it should be $m_a > h_a$) Then, we construct another right triangle $AMK$ with the same hypotenuse $AM$ and one perpendicular side $MK$ equal to $\dfrac{h_b}{2}$. So the other restriction is $m_a > \dfrac{h_b}{2}$ There is not just one point $K$ with the above properties. There are two points $K_1, K_2$ on opposite sides of $AM$. For each $K$ ($K_1$ or $K_2$) , the ray $AK$ intersects the line $DM$ at the point $C$. Now it's easy to find the point $B$, which is the symmetric of $C$ w.r.t. $M$. The triangle $ABC$ has the altitude $h_a=AD$, and the median $m_a=AM$. If we bring the altitude $BE$ from $B$, then $BE=2 \cdot MK = h_b$ In my scetch there are two triangles, one for each place of $K$

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I have discussed this problem on my YouTube channel: Video link: Video Solution