on the lines carrying A, A0, C and C
0*respectively, if the maps* {sA 7→ sA0 , eA 7→
eA0 , SA,C 7→ SA0,C0 , pA 7→ pA0} and {sC 7→ sC0 , eC 7→ eC0 , SA,C 7→ SA0,C0 , pC 7→
pC0} *preserve betweenness and equality, then the angles* ∠(
−−−−−→ SA,C pA,
−−−−−→ SA,C pC )
and ∠(
−−−−−→ SA,C pA,
−−−−−→ SA,C pC ) *have the same sign.*
Proof. Since *A R C* and A0 R C0, the angles ∠(−−−→ sA eA, *−−−→* sC eC ) and
∠(−−−−→ sA0 eA0 , *−−−−→* sC0 eC0 ) have the same sign. By the assumption of the preservation of betweenness and equality, this carries over to angles ∠(
−−−−−→ SA,C pA,
−−−−−→ SA,C pC )
and ∠(
−−−−−→ SA,C pA,
−−−−−→ SA,C pC ).

Theorem 54. Composition in DRAfp *is strong.*
Proof. Let rac ∈ rab ◦ rbc be an entry in the composition table, with rac, rab and rbc base relations. Given dipoles A and C with AracC, we need to show the existence of a dipole B with ArabB and BrbcC.

Since rac ∈ rab ◦ rbc, we know that there are dipoles A0, B0 and C
0 with A0rabB0, B0rbcC
0 and A0racC
0. Given dipoles X and Y , let SX,Y denote the point of intersection of the lines carrying X and Y ; it is only defined if X
and Y are not parallel. Consider now the three lines carrying A0, B0 and C
0, respectively. According to the results of Section 3.1, for the configuration of these three lines, there are fifteen qualitatively different cases 1, 2, 3a, 3b, 3c, 4a, 4b, 4c, 5a, 5b, 5c, 6a, 6b, 6c and 7:
1. The three points of intersection SA0,B0 , SB0,C0 and SA0,C0 exist and are

![0_image_0.png](0_image_0.png)

different. Since AracC and A0racC
0, by Corollary 51, the point sets
{sA, eA, SA,C } and {sA0 , eA0 , SA0,C0} are ordered in corresponding ways on their lines. Hence, it is possible to choose SA,B in such a way that the point sets {sA, eA, SA,C , SA,B} and {sA0 , eA0 , SA0,C0 , SA0,B0} are ordered in corresponding ways on their lines. In a similar way (interchanging A
and C), SB,C can be chosen.

Since both {SA,B, SA,C , SB,C } and {SA0,B0 , SA0,C0 , SB0,C0} are affine bases, there is a unique affine bijection h: R
2 −→ R
2 with h(SA0,B0 ) = SA,B,
h(SA0,C0 ) = SA,C and h(SB0,C0 ) = SB,C . By Lemma 53, h preserves orientation, and thus by Thm. 28 also the DRAfp relations. Hence, by choosing B = h(B0), we get h(A0)rabB and Brbch(C
0). Since the sets