Given two straight lines, to find the two mean proportionals.

Let the two straight lines be a b and b c, and let a b be a multiple of b c; the two mean proportionals to be found from a b and b c are a d and f g. Complete the rectangle b d and draw the diameter a c along a c. Describe the semicircle a d c e, and through the point d draw the straight line f g so that f d is equal to c g. I say, then, that the two lines c g and a f, belonging to a b and b c, are the mean proportionals. For f d is equal to e g, and d e is common; therefore f e is equal to d g. Therefore the rectangle under d g and g e is equal to that under e f and f d. But the rectangle under d g, g e is equal to that under b g, g c, as has been shown in the semicircles; for the rectangle under e f, f d is equal to that under b f and f a. And because it has been shown in the sixth book of the Elements that in equilateral and equiangular parallelograms the sides about the equal angles are reciprocally equal, as f b is to b g so is c g to a f; but as d f to b g, so f a to a d, and e d to c g, so c g to a f, and b a to d c. Thus, to the two given straight lines a b, b c, the two mean proportionals c g and f a have been found.

Another method, by Parmenion, disciple of Apollonius of Perga. Let the two given straight lines be a b and b c, so that a b is double b c, of which the two mean proportionals must be found. Complete the rectangular parallelogram d b and draw the diagonals a c, b d, and extend a b, b c to f g, and through the point d fit the line f g,

the straight line, so that e f is equal to e g. Then I say that of the straight lines a b, b c the two mean proportionals are c g and a f. For from a b to b c draw the parallel straight line e h; and because the triangle b e c is isosceles and equal to e h, therefore b h is equal to h c; and thus, because b c is cut at h in two ways and approaches the mean straight line c g, the rectangle under b c, c g together with the square on h c is equal to the square on h e. For let the square on e h be taken as common. Therefore the rectangle under b g, g e together with the squares on h c, h e is equal to the squares on e h, h g; but to the squares on h c, h e is equal the square on c e, and to the squares on e h, h g is equal the square on e g and the rectangle under b g, g c. Therefore, together with the square on c e it is equal to the square on e g; and this also because under b f, f a together with the square on a e is equal to the square on e f, while e f is equal to e g and to the rectangle under b g, g e. Therefore, together with the square on c e is equal the rectangle under b f, f a and the square on a e. It remains, then, that together with the rectangle under b g, g c is equal the square on b f, f a. And because, as shown in the sixth book of the Elements, in equal and equiangular parallelograms the sides about the equal angles are reciprocally equal, therefore as b f is to b g, so is e g to a f. But as b f to b g, so to a d; and as b f to b g, so f a to a d and c d to c g. And so as d c is to c g, so c g to a f and a f to a d; and d c is equal to a b, but a d is equal to b c; and so as a b is to c g, so c g to a f and a f to c d.