Kitap 10
32
For the same reason the rectangle BC, CD is also equal to the square on AC.
32
And since, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the perpendicular so drawn is a mean proportional between the segments of the base, [VI. 8, Por.] therefore, as BD is to DA, so is AD to DC; therefore the rectangle BD, DC is equal to the square on AD. [VI. 17]
32
I say that the rectangle BC, AD is also equal to the rectangle BA, AC.
32
For since, as we said, ABC is similar to ABD, therefore, as BC is to CA, so is BA to AD. [VI. 4]
32
Therefore the rectangle BC, AD is equal to the rectangle BA, AC. [VI. 16] Q. E. D.
PROPOSITION 33.
33
To find two straight lines incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial.
33
Let there be set out two rational straight lines AB, BC commensurable in square only and such that the square on the greater AB is greater than the square on the less BC by the square on a straight line incommensurable with AB, [X. 30] let BC be bisected at D, let there be applied to AB a parallelogram equal to the square on either of the straight lines BD, DC and deficient by a square figure, and let it be the rectangle AE, EB; [VI. 28] let the semicircle AFB be described on AB, let EF be drawn at right angles to AB, and let AF, FB be joined.
