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.

Let it be proposed to find the surface generated by the revolution of the curve KDE, which, receiving the incident rays coming to it from the point A, shall deviate them toward the point B. Then considering this other curve as already known, and that its apex D is in the straight line AB, let us divide it up into an infinitude of small pieces by the points G, C, F; and having drawn from each of these points, straight lines towards A to represent the incident rays, and other straight lines towards B, let there also be described with centre A the arcs GL, CM, FN, DO, cutting the rays that come from A at L, M, N, O; and from the points K, G, C, F, let there be described the arcs KQ, GR, CS, FT cutting the rays towards B at Q, R, S, T; and let us suppose that the straight line HKZ cuts the curve at K at right-angles.

And this operates so that all the parts of the wave of light DN, coming to meet the surface KDE, will advance thence along parallels to KS and will arrive at this straight line at the same time; of which the proof is again the same as that which served for the first oval.