How to find triangle vertices given midpoints? I have a task to find vertices if midpoints are given: $M1(2;1)$, $M2(5;3)$, $M3(3;-4)$. I know one way to solve it through making a system of equations with three variables.
My teacher says there is faster way by using the midline of a triangle, and I can`t find this way of solving it on the Internet. How can I do it?
 A: 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 $A$.
A: 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; $-4-2=-6$. $$B(0,-6)$$
A: 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.
A: By a construction procedure: Draw a line parallel to $M_2 M_3$ through $M_1$ and same for other two pairs, to find three  intersection  points $A_1, A_2, A_3.$
To translate this into vectors:There are 3 vectors. $ A, B, B-A $. Find their position vectors. The other 3 vector triangles drawn around them are labelled same between $A_1, A_2, A_3$ to fix their coordinates with parallel displacement. 
A: The vertex opposite to $M_1$ is the reflection of $M_1$ in the midpoint of $M_2M_3$.
