infinitely little altitudes to be quantity, what need you this limitation of yours, “_so far forth as that by multiplication they may be made equal to the altitude of the whole figure_?” May not the half, the third, the fourth, or the fifth part, &c. be made equal to the whole by multiplication? Why could you not have said plainly, _so far forth as that every one of those infinitely little altitudes be not only something but an aliquot part of the whole_? So you will have an _infinitely little_ altitude, that is to say, _a point to be both nothing and something and an aliquot part_. And all this proceeds from not understanding the ground of your profession. Well, the lemma is true. Let us see the theorems you draw from it. The first is (p. 3) “_that a triangle to a parallelogram of equal base and altitude is as one to two_.” The conclusion is true, but how know you that? “_Because_,” say you, “_the triangle consists as it were_ [_as it were_, is no phrase of a geometrician] _of an infinite number of straight parallel lines_.” Does it so? Then by your own doctrine, which is, that “_lines have no breadth_,” the altitude of your triangle consisteth of an infinite number of no altitudes, that is of an infinite number of nothings, and consequently the area of your triangle has no quantity. If you say that by the parallels you mean infinitely little parallelograms, you are never the better; for if infinitely little, either they are nothing, or if somewhat, yet seeing that no two sides of a triangle are parallel, those parallels cannot be parallelograms. I see they may be counted for parallelograms by not considering the quantity of their altitudes in the demonstration. But you are barred of that plea, by your spiteful arguing against it in your _Elenchus_ . Therefore this third proposition, and with it the fourth, is undemonstrated.