Problem source

Let $ABCD$ be a parallelogram. By using vectors, or otherwise, prove that: 
\begin{questionparts}
\item $AB^{2}+BC^{2}+CD^{2}+DA^{2}=AC^{2}+BD^{2}$; 
\item $|AC^{2}-BD^{2}|$ is 4 times the area of the rectangle whose sides are any side of the parallelogram and the projection of an adjacent side on that side. 
\end{questionparts}
State and prove a result like $\textbf{(ii)}$ about $|AB^{2}-AD^{2}|$ and the diagonals.

Solution source

Set up coordinates such that $A$ at the origin and $\vec{AB} = \mathbf{x}$ and $\vec{AD} = \mathbf{y}$ and so $\vec{AC} = \mathbf{x}+\mathbf{y}$

\begin{questionparts}
\item \begin{align*}
AC^2 + BD^2 &= (\mathbf{x}+\mathbf{y})\cdot (\mathbf{x}+\mathbf{y}) + (\mathbf{y}-\mathbf{x})\cdot(\mathbf{y}-\mathbf{x}) \\
&= 2\mathbf{x}\cdot\mathbf{x} + 2\mathbf{y}\cdot\mathbf{y} \\
&= AB^2 + CD^2 +AD^2 + BC^2
\end{align*}

\item \begin{align*}
AC^2 -BD^2 &=  (\mathbf{x}+\mathbf{y})\cdot (\mathbf{x}+\mathbf{y}) - (\mathbf{y}-\mathbf{x})\cdot(\mathbf{y}-\mathbf{x}) \\
&= 4 \mathbf{x}\cdot \mathbf{y}
\end{align*}

$\mathbf{x}\cdot\mathbf{y} = |\mathbf{x}||\mathbf{y}|\cos \theta$ which is exactly the lenth of one side mutliplied by the length of the projection to that same side.
\end{questionparts}
\begin{align*}
AB^2 - AD^2 &= \mathbf{x}\cdot\mathbf{x} - \mathbf{y}\cdot \mathbf{y} \\
&= (\mathbf{x}+\mathbf{y})\cdot(\mathbf{x}-\mathbf{y}) \\
&= AC \cdot BD
\end{align*}
So this is the area of the rectangle formed by the length of one diagonal and the projection of the other diagonal onto it.