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; and the two mean proportionals to be found between a b and b c are a d and f g. Let the rectangle b d be completed, and let the diagonal a c be drawn in a c. Let the semicircle a d c e be described, and through the point d let the straight line f g be drawn so that f d is equal to c g. I say then that the two lines c g and a f, which are [means] of a b and b c, are the mean proportionals. For f d is equal to e g, and with the common d e, 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 that under d g, g e is equal to that under b g, g c, as has been shown for semicircles. Wherefore that under e f, f d is equal to that under b f and f a. And because it has been shown in the 6th [book] of the Elements that in equilateral and equiangular parallelograms the sides about the equal angles are mutually equal, so as f b is to b g, so is c g to a f; and 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 between the two given straight lines a b, b c the two mean proportionals c g and f a have been found.

Otherwise (aliter), after Parmenion, disciple of Apollonius of Perga. Let the two given straight lines be a b, b c, such that a b is double b c, between which the two mean proportionals must be found. Let the rectangular parallelogram d b be completed, and let the diagonals a c, b d be drawn, and let a b, b c be produced to f g, and through the point d let f g be fitted,

...a 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 let the parallel straight line e h be drawn; and because the triangle b e c is isosceles and equal to e h, therefore b h is equal to h c; and so, because b c is cut at h in two ways and approaches the mean straight line c g, that which is under b c, c g together with that on h c is equal to that on h e. For let that on e h be taken as common. Therefore that which is under b g, g e together with that on h c, h e is equal to that on e h, h g; but to that on h c, h e is equal that on c e, and to that on e h, h g is equal that on e g and that under b g, g c. Therefore together with that on c e it is equal to that on e g; this again because under b f, f a together with that on a e is equal to that on e f, but e f is equal to e g, and that under b g, g e. Therefore together with that on c e is equal that under b f, f a and that on a e. It remains therefore that with that under b g, g c is equal that on b f, f a. And because, as has been shown in the 6th of the Elements, in equal and equiangular parallelograms the sides about the equal angles are mutually equal, therefore as b f is to b g, so e g to a f. But as b f to b g so to a d; but as b f to b g, so f a to a d and c d to c g. And so as d c 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.