proof
As A,B,C,D are collinear, α+β=γ+δ=1.
Now c=αa+βb, d=γa+δb, so
We usually define uxv by its magnitude |u||v|sinθ, where θ is the angle
When we use the vectors PA,PB,PC,PD, the lengths cancel, giving the result.
We know that the cross-ratio is that defined by the points A,B,C,D.
The proof of lemma 1 indicates that this depends on numbers such that
c=αa+βb, d=γa+δb where a,b,c,d are the position vectors of A,B,C,D.
It is gven by
(*) {axc/bxc}/{axd/bxd}.
c-p=α(a-p)+β(b-p), d-p=γ(a-p)+δ(b-p).
Thus we can replace a,b,c,d in (*) by the vectors PA,PB,PC,PD
between u and v,
measured in the range [0,π], and direction perpendicular
to u,v, with
{u,v,uxv} right-handed.
If we suppose that u and v lie in the xy-plane, then uxv is parallel
to z, the
unit vector in the
z-direction. Then uxv is (|u||v|sinφ)z, where φ
is the angle
between
u and v measured in the
anticlockwise direction.