Exercise 1.1.7

Proof. Let the four vertices of the parallelogram be $A$, $B$, $C$, $D$ counterclockwise. Say $x=\overrightarrow{\mathit{AB}}$ and $y=\overrightarrow{\mathit{AD}}$. Then the line joining points $B$ and $D$ should be $x+s(y-x)$, where $s$ is in $𝔽$. The line joining points $A$ and $C$ should be $t(x+y)$, where $t$ is in $𝔽$. To find the intersection of the two lines we should solve $x+s(y-x)=t(x+y)$ for $s$ and $t$. By doing so, we obtain $(1-s-t)x=(t-s)y$. But since $x$ and $y$ can not be parallel, we have $1-s-t=0$ and $t-s=0$. So $s=t=\frac{1}{2}$ and the midpoint would be the head of the vector $\frac{1}{2}(x+y)$ emanating from $A$ and by the previous exercise we know it to be the midpoint of segment $\mathit{AC}$ or segment $\mathit{BD}$. □