Using only a straight edge and compass, geometrically construct a triangle from the three given segments representing the base (side $c$), and the medians to the other two sides ($m_a$ and $m_b$)

I've looked at numerous websites trying to get help with this construction, but all I could find were constructions of triangles given 2 sides and a median or 2 medians and an altitude. I know I'm supposed to start with side $c$ and draw circles using the medians, but that's about it. How do I go about doing this?

So let's try to understand the problem first, so we can construct the figure.
Let's draw a triangle $\triangle ABC$, with $AB$ the horizontal side. Then the median from $A$ onto $BC$ will intersect $BC$ at $D$, and similarly $E$ is the intersection of the median from $B$ with $AC$. Due to similarity, you have $ED||AB$. You also have $ED=\frac 12 AB$. Now extend $AB$ past the $B$ point (away from $A$) to $B'$, suc that $B'B=\frac 12 AB=ED$. Note that $EBB'D$ is a parallelogram, so $B'D=BE$. In the triangle $\triangle ADB'$ you know that the lengths are $AB'=1.5c$, $AD=m_a$ and $DB'=EB=m_b$.
So now to construct the figure:
The last step can alternatively be calculated by drawing a parallel to $AB$ through $D$ and then use the intersection of this line with the center centered on $B$, and radius $m_b$