Problem source

$ABCD$ is a skew (non-planar) quadrilateral, and its pairs of opposite sides are equal, i.e. $AB=CD$ and $BC=AD$. Prove that the line joining the midpoints of the diagonals $AC$ and $BD$ is perpendicular to
each diagonal.

Solution source

Let $\mathbf{a}$ denote the vector position of $A$ and similarly for $B, C, D$. Then we know that $(\mathbf{b}-\mathbf{a})\cdot(\mathbf{b}-\mathbf{a})=(\mathbf{c}-\mathbf{d})\cdot(\mathbf{c}-\mathbf{d})$ and $(\mathbf{b}-\mathbf{c})\cdot(\mathbf{b}-\mathbf{c})=(\mathbf{a}-\mathbf{d})\cdot(\mathbf{a}-\mathbf{d})$.

Subtracting these two equations we see that $|\mathbf{a}|^2 -2\mathbf{a}\cdot\mathbf{b}+2\mathbf{c}\cdot\mathbf{b} - |\mathbf{c}|^2 = |\mathbf{c}|^2-2\mathbf{c}\cdot\mathbf{d}+2\mathbf{a}\cdot\mathbf{d}-|\mathbf{a}|^2$ or 
$2|\mathbf{a}|^2 -2\mathbf{a}\cdot\mathbf{b}+2\mathbf{c}\cdot\mathbf{b} - 2|\mathbf{c}|^2 +2\mathbf{c}\cdot\mathbf{d}-2\mathbf{a}\cdot\mathbf{d}=0$

The midpoints of the diagonals $AC$ and $BD$ are $\frac{\mathbf{a}+\mathbf{c}}{2}$ and $\frac{\mathbf{b}+\mathbf{d}}{2}$, so the line is parallel to: $\mathbf{a}-\mathbf{b}+\mathbf{c}-\mathbf{d}$. The diagonals are parallel to $\mathbf{a}-\mathbf{c}$ and $\mathbf{b}-\mathbf{d}$.

So it suffices to prove that  $(\mathbf{a}-\mathbf{b}+\mathbf{c}-\mathbf{d}) \cdot (\mathbf{a}-\mathbf{c}) = 0$ (since the other will follow by symmetry,

\begin{align*}
(\mathbf{a}-\mathbf{b}+\mathbf{c}-\mathbf{d}) \cdot (\mathbf{a}-\mathbf{c}) &= |\mathbf{a}|^2-\mathbf{a}\cdot\mathbf{b}-\mathbf{a}\cdot \mathbf{d}+\mathbf{b}\cdot \mathbf{c}-|\mathbf{c}|^2+\mathbf{c}\cdot \mathbf{d} \\
\end{align*}

But this is exactly half the equation we determined earlier, so we are done.