Therefore CA, AB are commensurable; but they were also, by hypothesis, incommensurable: which is impossible. Therefore no magnitude will measure AB, BC; therefore AB, BC are incommensurable. [X. Def. 1] Therefore etc.
LEMMA.
If to any straight line there be applied a parallelogram deficient by a square figure, the applied parallelogram is equal to the rectangle contained by the segments of the straight line resulting from the application. For let there be applied to the straight line AB the parallelogram AD deficient by the square figure DB; I say that AD is equal to the rectangle contained by AC, CB. This is indeed at once manifest; for, since DB is a square, DC is equal to CB; and AD is the rectangle AC, CD, that is, the rectangle AC, CB. Therefore etc.PROPOSITION 17.
If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into parts which are commensurable in length, thenthe square on the greater will be greater than the square on the less by the square on a straight line commensurable with the greater. And, if the square on the greater be greater than the square on the less by the square on a straight line commensurable with
