Book 13
4
And, since AF is equal to FE, let CK be added to each; therefore the whole AK is equal to the whole CE; therefore AK, CE are double of AK.
4
But AK, CE are the gnomon LMN and the square CK; therefore the gnomon LMN and the square CK are double of AK.
4
But, further, AK was also proved equal to HG; therefore the gnomon LMN and the squares CK, HG are triple of the square HG.
4
And the gnomon LMN and the squares CK, HG are the whole square AE and CK, which are the squares on AB, BC, while HG is the square on AC.
4
Therefore the squares on AB, BC are triple of the square on AC. Q. E. D.
PROPOSITION 5.
5
If a straight line be cut in extreme and mean ratio, and there be added to it a straight line equal to the greater segment, the whole straight line has been cut in extreme and mean ratio, and the original straight line is the greater segment.
5
For let the straight line AB be cut in extreme and mean ratio at the point C, let AC be the greater segment, and let AD be equal to AC.
5
I say that the straight line DB has been cut in extreme and mean ratio at A, and the original straight line AB is the greater segment.
5
For let the square AE be described on AB, and let the figure be drawn.
5
Since AB has been cut in extreme and mean ratio at C, therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17]
5
And CE is the rectangle AB, BC, and CH the square on AC; therefore CE is equal to HC.
5
But HE is equal to CE, and DH is equal to HC; therefore DH is also equal to HE.
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.
