Euler's Distribution Theorem

For signed distances,

$$\overline{AB}\cdot\overline{CD}+\overline{AC}\cdot\overline{DB}+\overline{AD}\cdot\overline{BC}=0,$$

since

$$(b-a)(d-c)+(c-a)(b-d)+(d-a)(c-b)=0.$$

References

Johnson, R. A. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, p. 3, 1929.