Kitap 13
5
Therefore the whole DK is equal to the whole AE.
5
And DK is the rectangle BD, DA, for AD is equal to DL; and AE is the square on AB; therefore the rectangle BD, DA is equal to the square on AB.
5
Therefore, as DB is to BA, so is BA to AD. [VI. 17]
5
And DB is greater than BA; therefore BA is also greater than AD. [V. 14]
5
Therefore DB has been cut in extreme and mean ratio at A, and AB is the greater segment. Q. E. D.
PROPOSITION 6.
6
If a rational straight line be cut in extreme and mean ratio, each of the segments is the irrational straight line called apotome.
6
Let AB be a rational straight line, let it be cut in extreme and mean ratio at C, and let AC be the greater segment; I say that each of the straight lines AC, CB is the irrational straight line called apotome.
6
For let BA be produced, and let AD be made half of BA.
6
Since then the straight line AB has been cut in extreme and mean ratio, and to the greater segment AC is added AD which is half of AB, therefore the square on CD is five times the square on DA. [XIII. 1]
6
Therefore the square on CD has to the square on DA the ratio which a number has to a number; therefore the square on CD is commensurable with the square on DA. [X. 6]
6
But the square on DA is rational, for DA is rational, being half of AB which is rational; therefore the square on CD is also rational; [X. Def. 4] therefore CD is also rational.
