Book 13
1
Therefore etc. Q. E. D.
PROPOSITION 2.
2
If the square on a straight line be five times the square on a segment of it, then, when the double of the said segment is cut in extreme and mean ratio, the greater segment is the remaining part of the original straight line.
2
For let the square on the straight line AB be five times the square on the segment AC of it, and let CD be double of AC; I say that, when CD is cut in extreme and mean ratio, the greater segment is CB.
2
Let the squares AF, CG be described on AB, CD respectively, let the figure in AF be drawn, and let BE be drawn through.
2
Now, since the square on BA is five times the square on AC, AF is five times AH.
2
Therefore the gnomon MNO is quadruple of AH.
2
And, since DC is double of CA, therefore the square on DC is quadruple of the square on CA, that is, CG is quadruple of AH.
2
But the gnomon MNO was also proved quadruple of AH; therefore the gnomon MNO is equal to CG.
2
And, since DC is double of CA, while DC is equal to CK, and AC to CH, therefore KB is also double of BH. [VI. 1]
2
But LH, HB are also double of HB; therefore KB is equal to LH, HB.
2
But the whole gnomon MNO was also proved equal to the whole CG; therefore the remainder HF is equal to BG.
2
And BG is the rectangle CD, DB, for CD is equal to DG; and HF is the square on CB; therefore the rectangle CD, DB is equal to the square on CB.
2
Therefore, as DC is to CB, so is CB to BD.
2
But DC is greater than CB; therefore CB is also greater than BD.
