$$\begin{align} m^2_a&=\frac{2b^2+2c^2−a^2}4\\[4pt] m^2_b&=\frac{2c^2+2a^2−b^2}4\\[4pt] m^2_c&=\frac{2a^2+2b^2−c^2}4 \end{align}$$
Solving for $a$, $b$, $c$ in terms of $m^2_a$, $m^2_b$, $m^2_c$ gives:
$$\begin{align} a^2&=\frac{8m^2_b+8m^2_c−4m^2_a}9 \\[4pt] b^2&=\frac{8m^2_c+8m^2_a−4m^2_b}9 \\[4pt] c^2&=\frac{8m^2_a+8m^2_b−4m^2_c}9 \end{align}$$
How can I construct a general triangle using this information?
You do not need explicitly find the lengths of the sides. Just build a triangle having side lengths $2m_a,2m_b,2m_c$ (the red one in the picture), find its centroid and draw a few parallel lines to the medians:
The hexagon is made of six congruent triangles with side lengths $a,b,c$.