Then also the sum of all the antecedents to all the consequents would be as 3 to 2.

Now by prolonging the arc DO until it meets AK at X, KX is the sum of the antecedents.

And by prolonging the arc KQ till it meets AD at Y, the sum of the consequents is DY.

Then KX ought to be to DY as 3 to 2.
Whence it would appear that the curve KDE was of such a nature that having drawn from some point which had been assumed, such as K, the straight lines KA, KB, the excess by which AK surpasses AD should be to the excess of DB over KB, as 3 to 2.

For it can similarly be demonstrated, by taking any other point in the curve, such as G, that the excess of AG over AD, namely VG, is to the excess of BD over DG, namely DP, in this same ratio of 3 to 2.