 
Summary: 62 2. Elementary geometrical figures and their properties
Proof. Clearly, the second and third inequalities can be derived from
the first by a renumbering of points, so we show only the first one.
Both sides of the first inequality are translation invariant. Because of
this we can, without loss of generality, assume z1 = 0. The triangle
inequality gives
z2z3  z2z4 z2z3  z3z4 + z3z4  z2z4
and can be written equivalently as
z2z3  z4 z3z2  z4 + z4z2  z3.
This is precisely the inequality d12d34 d13d24 + d23d14. If we have
equality then there exists a real number with
(z2z3  z3z4) = (z3z4  z2z4).
Since z1 = 0 the number  is the cross ratio of the four points. In
other words, equality (in any of the three inequalities) occurs if and
only if the cross ratio of the four points is real. The assertion about
the case of equality is a consequence of Corollary 9.
2.4. The conic sections
2.4.1. The section by a plane of a cone. Consider a cone C with
vertex S = (0, 0, 0), the origin, and half opening angle /2  . The
equation for C in R3
