# Affine transformation implies parallelogram

Let $A,B,C,D$ be four points in the affine plane. Suppose there is an affine transformation $F$ that maps those points cyclically ($A$ to $B$, $B$ to $C$ and so on). Prove that $ABCD$ is a parallelogram. What if $F$ swapped $A$ with $C$ and $B$ with $D$, instead?

My attempt. My goal is to prove that $\overrightarrow{AB}=-\overrightarrow{CD}$ and $\overrightarrow{BC}=-\overrightarrow{DA}$. I know that $F$ has order $4$, but I'm not able to prove that $dF^2$ is $-Id$. I also tried to write a matrix $$\begin{pmatrix} 0&a \\ 1&b \end{pmatrix}$$ with respect to the basis $(\overrightarrow{AB},\overrightarrow{BC})$, raised to the 4th power, but I don't get a unique solution (a=-1,b=0).

Wlog, we can place the origin at $A$ and use $\vec{AB}$ and $\vec{AD}$ as our basis. In that basis, the homogeneous coordinates of the four points are \begin{align}A&=(0,0,1)^T\\B&=(1,0,1)^T\\C&=(x_C,y_C,1)^T\\D&=(0,1,1)^T\end{align} and the problem reduces to showing that $x_C=y_C=1$.

Let $M$ be the matrix of the affine transformation $F$ relative to this basis. We can see immediately that its last column must be equal to $B$. The first column of the matrix is also fairly easy to deduce: Since $\vec{AB}\mapsto\vec{BC}$ and $\vec{AB}$ lies along the first coordinate axis, this column of $M$ must be the direction vector that corresponds to $\vec{BC}$, i.e., $(x_c-1,y_c,0)^T$. Similarly, since $\vec{AD}\mapsto\vec{BA}$, the second column of $M$ must be $(-1,0,0)^T$ (note the sign change). Thus, we have $$M=\pmatrix{x_c-1&-1&1\\y_c&0&0\\0&0&1}.$$ Multiplying $C$ by this matrix and setting the result equal to $D$ yields the following system of equations: $$x_C^2-x_C-y_C+1=0\\x_Cy_C=1$$ for which the solution is $x_C=y_C=1$, as desired. Alternatively, you could set $M^2B=D$, which results in the same equations.

For the second part of the problem, similar considerations yield the transformation matrix $$\pmatrix{-1&0&x_C\\0&-1&y_C\\0&0&1}$$ and applying this matrix to either $B$ or $D$ eventually leads to $x_C=y_C=1$ again.

• +1. The use of homogeneous coordinates makes the problem much easier. Feb 25 '17 at 18:33