_A._ Hitherto all is evident. Proceed. _B._ From the point Y raise a perpendicular cutting V R wheresoever in _t._, and then (because P Y and V R are parallel) the angle Y _t_ V will be a right angle. And the figure _u_ Y _t_ V a rectangle, and _u t_ equal to Y V. But Y V is equal to Z X; and therefore Z X is equal to _u t_. And _u t_ must pass through the point T (for the diameters of any rectangle divide each other in the middle), therefore Z and _u_ are the same point, and X and _t_ the same point. Therefore Y Z produced falls upon P. And D X is the lesser of the two means between A D and D V. And the cube of D X double to the cube of D V, which was to be demonstated _A._ I cannot imagine what fault there can be in this demonstration, and yet there is one thing which seems a little strange to me. And it is this. You take B R, which is half the diagonal, and which is the sine of forty-five degrees, and which is also the mean proportional between the two extremes; and yet you bring none of these proprieties into your demonstration. So that though you argue from the construction, yet you do not argue from the cause. And this perhaps your adversaries will object, at least, against the art of your demonstration, or enquire by what luck you pitched upon half the diagonal for your foundation. _B._ I see you let nothing pass. But for answer you must know, that if a man argue from the negative of the truth, though he know not that it is the truth which is denied, yet he will fall at last, after many consequences, into one absurdity or another. For though false do often produce truth, yet it produces also absurdity, as it hath done here. But truth produceth nothing but truth. Therefore in demonstrations that tend to absurdity, it is no good logic to require all along the operation of the cause. _A._ Have you drawn from hence no corollaries? _B._ No. I leave that for others that will; unless you take this for a corollary, that, what arithmetical calculation soever contradicts it, is false. _A._ Let me see now the algebraical demonstration against it. _B._ Here it is: Let A B or A D be equal to 2 Then D F or D V is equal to 1 And B R or A S is equal to the square root of 2 And D Y equal to 3 want the square root of 2 The cube of A B is equal to 8 The cube of D Y is equal to 45 want the square root of 1682 that is almost equal to 4 For 45 want the square root of 1681 is equal to 4 Therefore D Y is a little less then the greater of the two means between A D and D V.