KL is drawn parallel to AZ. Then KJ : JL :: (KJ : AG & AG : JL) :: (JD : GD & BG : JB). These proportions might be written today as equations: KJ/JL = (KJ/AG)(AG/JL) = (JD/GD)(BG/JB). The last compound ratio (namely JD : GD & BG : JB) is what is known today as the cross ratio of the collinear points J, G, D, and B in that order; it is denoted today by (J, G; D, B). So we have shown that this is independent of the choice of the particular straight line JD that crosses the three straight lines that concur at A. In particular (J, G; D, B) = (J, Z; H, E). It does not matter on which side of A the straight line JE falls. In particular, the situation may be as in the next diagram, which is the diagram for Lemma X. Just as before, we have (J, G; D, B) = (J, Z; H, E). Pappus does not explicitly prove this; but Lemma X is a converse, namely that if these two cross ratios are the same, and the straight lines BE and DH cross at A, then the points G, A, and Z must be collinear. What we showed originally can be written as (J, ∞; K, L) = (J, G; D, B), with ∞ taking the place of the (nonexistent) intersection of JK and AG. Pappus shows this, in effect, in Lemma XI, whose diagram, however, has different lettering: What Pappus shows is DE.ZH : EZ.HD :: GB : BE, which we may write as (D, Z; E, H) = (∞, B; E, G). The diagram for Lemma XII is: The diagram for Lemma XIII is the same, but BA and DG, extended, meet at N. In any case, considering straight lines through G as cut by the three straight lines through A, (and accepting that equations of cross ratios remain valid after permutation of the entries,) we have by Lemma III or XI (G, J; E, H) = (G, D; ∞ Z). Considering straight lines through D as cut by the three straight lines through B, we have (L, D; E, K) = (G, D; ∞ Z). Thus (E, H; J, G) = (E, K; D, L), so by Lemma X, the points H, M, and K are collinear. That is, the points of intersection of the