What is the geometric meaning of this vector equality? $\vec{BC}\cdot\vec{AD}+\vec{CA}\cdot\vec{BD}+\vec{AB}\cdot\vec{CD}=0$

I was doing some exercises for linear algebra. One of them was to prove that for any four points $A, B, C, D \in \mathbb{R}^3$ the following equality holds:
$$\overrightarrow{BC} \cdot \overrightarrow{AD}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}\ = 0$$
The proof is easy; you just make three vectors starting in $A$ and then see that all the terms cancel out.

My question is: what is the geometric interpretation of this equality? How can I visualize it or understand its deeper meaning? Does this equality have a name or where can I read more about it?

I’m asking this because it turns out that it is not just a random equality and is rather useful. For example, if we want to prove the existence of orthocenter, we can do it surprisingly easily and quickly using this equality.


Let $O$ be the orthocenter $O$ of $\triangle ABC$. Then
\ +\ \overrightarrow{BC}\cdot\overrightarrow{AD}
\ +\ \overrightarrow{CA}\cdot\overrightarrow{BD}\\
=\ &\left(\overrightarrow{AB}\cdot\overrightarrow{CO}
\ +\ \overrightarrow{BC}\cdot\overrightarrow{AO}
\ +\ \overrightarrow{CA}\cdot\overrightarrow{BO}\right)
+ \left(\overrightarrow{AB}\cdot\overrightarrow{OD}
\ +\ \overrightarrow{BC}\cdot\overrightarrow{OD}
\ +\ \overrightarrow{CA}\cdot\overrightarrow{OD}\right)\\
=\ &\left(\overrightarrow{AB}\cdot\overrightarrow{CO}
\ +\ \overrightarrow{BC}\cdot\overrightarrow{AO}
\ +\ \overrightarrow{CA}\cdot\overrightarrow{BO}\right)
+ \left(\overrightarrow{AB}\ +\ \overrightarrow{BC}\ +\ \overrightarrow{CA}\right)\cdot\overrightarrow{OD}\tag{$\dagger$}\\
=\ &0+0=0.\\

The first bracket on line $(\dagger)$ is zero because every side of $\triangle ABC$ is perpendicular to the altitude dropped from the opposite vertex. The second bracket is zero because it is the sum of directed edges of a closed circuit.

In short, the identity is basically a cyclic sum of expressions of the form “side dot altitude” on $\mathbb R^2$, but another cyclic sum of the form “side dot $\overrightarrow{OD}$” has been added to conceal the significance of the orthocenter and make the identity present in $\mathbb R^3$.

Source : Link , Question Author : Matthew , Answer Author : user1551

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