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 Math Help > Geometry > Polygons and Triangles > Cross Ratio Theorem

## Cross Ratio Theorem

Cross Ratio Theorem: Given four collinear points A,B,C,D and four coplanar and concurrent lines a,b,c,d, the point cross ratio (ABCD) = CA/CB : DA/DB equals the line cross-ratio (abcd) = sin(cMa)/sin(cMb) : sin(dMa)/sin(dMb), where line segments and angles are considered "signed".
Corollary: The cross-ratio is invariant under a central projection. (ABCD)=(abcd)=(A'B'C'D')
Additional factoid: The cross-ratio is invariant under inversion.

### Permutataions of points A, B, C, D

If you define (ABCD) = CA/CB : DA/DB, then there are 24 permutations of ABCD which give six distinct values of the cross-ratio. If you let m=(ABCD), then here are the 24 permutations, along with the six distinct cross-ratios:

m = (ABCD) = CA/CB : DA/DB
m = (BADC) = DB/DA : CB/CA
m = (CDAB) = AC/AD : BC/BD
m = (DCBA) = BD/BC : AD/AC

1/m = (ABDC) = DA/DB : CA/CB
1/m = (BACD) = CB/CA : DB/DA
1/m = (DCAB) = AD/AC : BD/BC
1/m = (CDBA) = BC/BD : AC/AD

1/(1-m) = (ACDB) = DA/DC : BA/BC
1/(1-m) = (CABD) = BC/BA : DC/DA
1/(1-m) = (DBAC) = AD/AB : CD/CB
1/(1-m) = (BDCA) = CB/CD : AB/AD

1-m = (ACBD) = BA/BC : DA/DC
1-m = (CADB) = DC/DA : BC/BA
1-m = (BDAC) = AB/AD : CB/CD
1-m = (DBCA) = CD/CB : AD/AB

(m-1)/m = (ADBC) = BA/BD : CA/CD
(m-1)/m = (DACB) = CD/CA : BD/BA
(m-1)/m = (BCAD) = AB/AC : DB/DC
(m-1)/m = (CBDA) = DC/DB : AC/AB

m/(m-1) = (ADCB) = CA/CD : BA/BD
m/(m-1) = (DABC) = BD/BA : CD/CA
m/(m-1) = (CBAD) = AC/AB : DC/DB
m/(m-1) = (BCDA) = DB/DC : AB/AC

Proofs of all these identities are done by algebraic manipulation of the double ratios.  Most of them are quite simple, but one strikes me as harder than the others.  This one follows:

### Proof that if m = (ABCD) = CA/CB : DA/DB then 1-m = (ACBD) = BA/BC : DA/DC

The statement of this proof can be restated,

BA/BC : DA/DC = 1 - (CA/CB : DA/DB)

The proof begins with lemma 1, proved below, which states:

(AB DC) + (BC DA) + (CA DB) = 0

This interesting identity is restated, then divided through by (BC DA), as follows:

(BA DC) = (BC DA) + (CA DB)
(BA DC)/(BC DA) = (BC DA)/(BC DA) + (CA DB)/(BC DA)
(BA DC)/(BC DA) = 1 + (CA DB)/(BC DA)
(BA DC)/(BC DA) = 1 - (CA DB)/(CB DA)
BA/BC : DA/DC = 1 - (CA/CB : DA/DB),

which completes the proof.

Lemma 1: (AB DC) + (BC DA) + (CA DB) = 0

Let real numbers a, b, c, d be defined such that AB=(a-b), BC=(b-c), CD=(c-d), and DA=(d-a), etc.

Then (AB DC) + (BC DA) + (CA DB) =
((a-b)(d-c)) + ((b-c)(d-a)) + ((c-a)(d-b)) =
0

### Internet references

Cut-the-knot: Cross-Ratio

### Related pages in this website:

Summary of geometrical theorems

Inversion circle

The webmaster and author of this Math Help site is Graeme McRae.