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Euler's Distribution Theorem

​
A
1
B
2
C
5
D
9
AB = 1
AC = 4
AD = 8
BC = 3
BD = 7
CD = 4
AB×CD + AC×DB + AD×BC =
(1) × (4) + (4) × (-7) + (8) × (3) =
(4) + (-28) + (24) = 0
For signed distances on a line segment (so that XY = -YX), AB×CD + AC×DB + AD×BC = 0. If
a
,
b
,
c
, and
d
are the coordinates of the four points on the line, this follows from the algebraic identity
(b-a)(d-c)+(c-a)(b-d)+(d-a)(c-b)=0
.​
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