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: Origin at $A$

Let $\vec u=\overrightarrow{AB}$, $\vec v=\overrightarrow{AC}$, $\vec w=\overrightarrow{AD}$. Then $\overrightarrow{CD}=\vec w-\vec v$ and $\overrightarrow{BC}=\vec v-\vec u$, and the condition reads

\[ \vec u^{\,2}+(\vec w-\vec v)^2=\vec w^{\,2}+(\vec v-\vec u)^2. \]

Step 2: Simplification

\[ \vec u^{\,2}+\vec w^{\,2}-2\vec v\cdot\vec w+\vec v^{\,2}=\vec w^{\,2}+\vec v^{\,2}-2\vec u\cdot\vec v+\vec u^{\,2}\ \Longrightarrow\ \vec v\cdot\vec w=\vec u\cdot\vec v, \]

\[ \vec v\cdot(\vec w-\vec u)=0\ \Longleftrightarrow\ \overrightarrow{AC}\cdot\overrightarrow{BD}=0.\qquad\blacksquare \]

\[\overrightarrow{BD}\cdot\overrightarrow{AC}=0\]

Every step is reversible, so the converse holds as well (see Problem 126).