How to find triangle vertices given midpoints?

A parallelogram forms between the midpoints and each vertex; the lengths of opposite sides are equal, so equate the x and y values of the vertices from there:

You can see that the "rises" and "runs" are the same, so in this example we're finding B visually, the difference between $x_{E}$ and $x_{D}$ is $3$, so we go to F: $3-3=0$ which gives $x_{B}$. You do the same for the y value: $3-1=2$, go to F; $-4-2=-6$. $$B(0,-6)$$

The matrix equation giving the midpoints from the vertices is $$ \begin{bmatrix} \frac12&\frac12&0\\ 0&\frac12&\frac12\\ \frac12&0&\frac12 \end{bmatrix} \begin{bmatrix} a\vphantom{\frac12}\\b\vphantom{\frac12}\\c\vphantom{\frac12} \end{bmatrix} = \begin{bmatrix} \frac{a+b}2\\\frac{b+c}2\\\frac{c+a}2 \end{bmatrix} $$ Inverting the equation above yields $$ \begin{bmatrix} 1&-1&1\vphantom{\frac12}\\ 1&1&-1\vphantom{\frac12}\\ -1&1&1\vphantom{\frac12} \end{bmatrix} \begin{bmatrix} \frac{a+b}2\\\frac{b+c}2\\\frac{c+a}2 \end{bmatrix} = \begin{bmatrix} a\vphantom{\frac12}\\b\vphantom{\frac12}\\c\vphantom{\frac12} \end{bmatrix} $$ The last equation simply says $$ a=\frac{a+b}2+\frac{c+a}2-\frac{b+c}2\\ b=\frac{b+c}2+\frac{a+b}2-\frac{c+a}2\\ c=\frac{b+c}2+\frac{c+a}2-\frac{a+b}2 $$ This amounts to reflecting each midpoint across the midpoint of the line connecting the other two midpoints.

Anoher elementary construction is this: Guess a point $A_0$, reflect it to find $B_0$, reflact that to find $C_0$, reflect that to find $A_1$, which will differ from $A_0$. But the midpoint of $A_0$ and $A_1$ is the true $A$.