For let BC be bisected at the point E, and let EF be made equal to DE.

Therefore the remainder DC is equal to BF.

And, since the straight line BC has been cut into equal parts at E, and into unequal parts at D, therefore the rectangle contained by BD, DC, together with the square on ED, is equal to the square on EC; [[II. 5](elem.2.5)]

And the same is true of their quadruples; therefore four times the rectangle BD, DC, together with four times the square on DE, is equal to four times the square on EC.

But the square on A is equal to four times the rectangle BD, DC; and the square on DF is equal to four times the square on DE, for DF is double of DE.

And the square on BC is equal to four times the square on EC, for again BC is double of CE.

Therefore the squares on A, DF are equal to the square on BC, so that the square on BC is greater than the square on A by the square on DF.

It is to be proved that BC is also commensurable with DF.

Since BD is commensurable in length with DC, therefore BC is also commensurable in length with CD. [X. 15]

But CD is commensurable in length with CD, BF, for CD is equal to BF. [[X. 6](elem.10.6)]

Therefore BC is also commensurable in length with BF, CD, [[X. 12](elem.10.12)] so that BC is also commensurable in length with the remainder FD; [[X. 15](elem.10.15)] therefore the square on BC is greater than the square on A by the square on a straight line commensurable with BC.

Next, let the square on BC be greater than the square on A by the square on a straight line commensurable with BC, let a parallelogram be applied to BC equal to the fourth part of the square on A and deficient by a square figure, and let it be the rectangle BD, DC.

It is to be proved that BD is commensurable in length with DC.

With the same construction, we can prove similarly that the square on BC is greater than the square on A by the square on FD.

But the square on BC is greater than the square on A by the square on a straight line commensurable with BC.

Therefore BC is commensurable in length with FD, so that BC is also commensurable in length with the remainder, the sum of BF, DC. [[X. 15](elem.10.15)]

But the sum of BF, DC is commensurable with DC, [[X. 6](elem.10.6)] so that BC is also commensurable in length with CD; [[X. 12](elem.10.12)] and therefore, separando, BD is commensurable in length with DC. [[X. 15](elem.10.15)]

Therefore etc.
45. After saying literally that the square on BC is greater than the square on A by the square on DF,

Euclid adds the equivalent expression with δύναται in its technical sense, ἡ ΒΓ ἄρα τῆς Α μεῖζον δύναται τῇ ΔΖ. As this is untranslatable in English except by a paraphrase in practically the same words as have preceded, I have not attempted to reproduce it.