If a line cuts the lines of a pencil with vertex P at the points A,B,C,D,

then (PA,PB,PC,PD) = sin(<APC)sin(<BPD)/sin(<BPC)sin(<APD).

**proof**

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**=α**a**+β**b**, **d**=γ**a**+δ**b** where **a,b,c,d** are the position vectors of A,B,C,D.

It is gven by

(*) {**a**x**c**/**b**x**c**}/{**a**x**d**/**b**x**d**}.

As A,B,C,D are collinear, α+β=γ+δ=1.
Now **c**=α**a**+β**b**, **d**=γ**a**+δ**b**, so

**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

We usually define **u**x**v** by its magnitude |**u**||**v**|sinθ, where θ is the angle

between **u** and **v**,
measured in the range [0,π], and direction perpendicular

to **u**,**v**, with
{**u**,**v**,**u**x**v**} right-handed.

If we suppose that **u** and **v** lie in the xy-plane, then **u**x**v** is parallel
to **z**, the

unit vector in the
z-direction. Then **u**x**v** is (|**u**||**v**|sinφ)**z**, where φ
is the angle

between
**u** and **v** measured in the
*anticlockwise *direction.

When we use the vectors PA,PB,PC,PD, the lengths cancel, giving the result.