Construct a pentagon from the midpoints of its sides Let $p_{1},p_{2},p_{3},p_{4},p_{5}$ be five points in the euclidean plane such that no set of three of those points lie on the same line. It is easy to prove that there exists a unique pentagon such that $p_{1},p_{2},p_{3},p_{4},p_{5}$ are the midpoints of its sides (In fact there is a more general result saying that the same is true for any odd number n of points as the midpoints of the sides of an n-gon).
The proof uses $\mathbb{C}$ as a model of the euclidean plane and then proves, that the system of linear equations $$\frac{1}{2}(x_{i}+x_{i+1}) = p_{i} \space\space\space \space 1 \leq i \leq 5$$ where $x_{6} = x_{1}$, has unique solutions for $x_{1},x_{2},x_{3},x_{4},x_{5}\in\mathbb{C}$ since the corresponding 5x5 matrix is invertible.
My question is whether there is a way to construct the solution using ruler and  compass. (which is possible in the case with only 3 points)
 A: Consider the transformation on the plane consisting of reflecting a point over $p_1$, then $p_2$, then $p_3,p_4,p_5$. Since it is the composition of $5$ $\Pi$-rotations, it itself is a $\Pi$-rotation or just a reflection. Thus it has a fixed point $A$, which is precisely the vertex adjacent to $p_1$ and $p_5$. To find the fixed point of the transformation, apply it to any other point $X$ in the plane to get a new point $X'$. Since this was a reflection over the fixed point $A$, $A$ is simply the midpoint of $XX'$, so we can construct $A$ and all other vertices similarly.
A: Let $A,B,C,D,E$ be the given midpoints opposite the unknown vertices $V,W,X,Y,Z$ respectively; both sets of points in rotational order. Then quadrilateral $WXYZ$ must have the midpoints of its sides on a parallelogram, and three of those midpoints are given by $E,A,B$. To find the midpoint $M$ of $\overline{ZW}$, construct lines parallel to $\overline{AB}$ through $E$ and parallel to $\overline{EA}$ through $B$. These lines intersect at $M$.
Then $C,D,M$ are midpoints of the sides of $\triangle ZVW$, and the sides of that triangle are parallel to those of $\triangle DMC$. Construct lines parallel to $\overline{MC}$ through $D$ and parallel to $\overline{DM}$ through $C$. These lines intersect at $V$, and the lines contain the sides of the pentagon through $V$.
Once $V$ is obtained, the remaining vertices follow readily. The given midpoint $D$ is the midpoint of $W$, so $W$ is thereby obtained from $V$ and $D$. Similarly $X$ is obtained from $W$ and $E$, and so on around the pentagon.
