If four points A,B,C,D in space satisfy…

If four points $A,B,C,D$ in space satisfy $\overline{AB}^2+\overline{CD}^2=\overline{AD}^2+\overline{BC}^2$, then $\overrightarrow{BD}\cdot\overrightarrow{AC}=0$. Prove this.

Step 1: Expansion

With position vectors $\vec a,\vec b,\vec c,\vec d$ the squares $|\vec a|^2,\dots,|\vec d|^2$ appear once on each side and cancel: \[|\vec b-\vec a|^2+|\vec d-\vec c|^2-|\vec d-\vec a|^2-|\vec c-\vec b|^2=-2\big(\vec a\cdot\vec b+\vec c\cdot\vec d-\vec a\cdot\vec d-\vec b\cdot\vec c\big).\]

Step 2: Factorisation

\[ \vec a\cdot\vec b+\vec c\cdot\vec d-\vec a\cdot\vec d-\vec b\cdot\vec c=\vec a\cdot(\vec b-\vec d)-\vec c\cdot(\vec b-\vec d)=(\vec a-\vec c)\cdot(\vec b-\vec d)=\overrightarrow{CA}\cdot\overrightarrow{DB}. \]

Hence the condition is equivalent to $\overrightarrow{CA}\cdot\overrightarrow{DB}=0$, i.e.

\[\overline{AB}^2+\overline{CD}^2=\overline{AD}^2+\overline{BC}^2\iff\overrightarrow{BD}\cdot\overrightarrow{AC}=0\]

The implication even holds in both directions. For a (possibly skew) quadrilateral $ABCD$: the diagonals are perpendicular exactly when the sums of squares of opposite sides are equal.