_A._ Is their calculation so inconstant, or rather so foolish as you make it? _B._ Yes. For the same number is sometimes so many lines, sometimes so many planes, and sometimes so many solids; as you shall plainly see, if you will take the pains to examine first a demonstration I have to prove the said duplication, and after that, the algebraic calculation which is pretended to confute it. And not only that this one is false, but also any other arithmetical account used in geometry, unless the numbers be always so many lines, or always so many superficies, or always so many solids. _A._ Let me see the geometrical demonstration. _B._ There it is. Read it. TO FIND A CUBE DOUBLE TO A CUBE GIVEN: Let the side of the cube given be V D. Produce V D to A, till A D be double to D V. Then make the square of A D, namely A B C D. Divide A B and C D in the middle at E and F. Draw E F. Draw also A C cutting E F in I. Then in the sides B C and A D take B R and A S, each of them equal to A I or I C. Lastly, divide S D in the middle at T, and upon the centre T, with the distance T V, describe a semi-circle cutting A D in Y, and D C in X. I say the cube of D X is double to the cube of D V. For the three lines D Y, D X, D V are in continual proportion. And continuing the semi-circle V X Y till it cut the line R S, drawn and produced in Z, the line S Z will be equal to D X. And drawing X Z it will pass through T. And the four lines T V, T X, T Y and T Z will be equal. And therefore joining Y X and Y Z, the figure V X Y Z will be a rectangle. [Illustration: _Delphic Problem. Vol. VII. Eng. p. 60._ ] Produce C D to P so as D P be equal to A D. Now if Y Z produced fall on P, there will be three rectangle equiangled triangles, D P Y, D Y X, and D X V; and consequently four continual proportionals, D P, D Y, D X, and D V, whereof D X is the least of the means. And therefore the cube of D X will be double to the cube of D V. _A._ That is true; and the cube of D Y will be double to the cube of D X; and the cube of D P double to the cube of D Y. But that Y Z produced, falls upon P, is the thing they deny, and which you ought to demonstrate. _B._ If Y Z produced fall not on P, then draw P Y, and from V let fall a perpendicular upon P Y, suppose at _u._. Divide P V in the midst at _a._, and join _a u._; which done _a u._ will be equal to _a._ V or _a._ P. For because V _u._ P is a right angle, the point _u._ will be in the semi-circle whereof P V is the diameter. Therefore drawing V R, the angle _u._ V R will be a right angle. _A._ Why so? _B._ Because T V and T Y are equal; and T D, T S equal; S Y will also be equal to D V. And because D P and R S are equal and parallel, R Y will be equal and parallel to P V. And therefore V R and P Y that join them will be equal and parallel. And the angles P _u._ V, R V _u._ will be alternate, and consequently equal. But P _u._ V is a right angle; therefore also R V _u._ will be a right angle.