Book 10
53
For, since DB is equal to BF, and BE to BG, therefore the whole DE is equal to the whole FG.
53
But DE is equal to each of the straight lines AH, KC, and FG is equal to each of the straight lines AK, HC; [I. 34] therefore each of the straight lines AH, KC is also equal to each of the straight lines AK, HC.
53
Therefore the parallelogram AC is equilateral.
53
And it is also rectangular; therefore AC is a square.
53
And since, as FB is to BG, so is DB to BE, while, as FB is to BG, so is AB to DG, and, as DB is to BE, so is DG to BC, [VI. 1] therefore also, as AB is to DG, so is DG to BC. [V. 11]
53
Therefore DG is a mean proportional between AB, BC.
53
I say next that DC is also a mean proportional between AC, CB.
53
For since, as AD is to DK, so is KG to GC— for they are equal respectively— and, componendo, as AK is to KD, so is KC to CG, [V. 18] while, as AK is to KD, so is AC to CD, and, as KC is to CG, so is DC to CB, [VI. 1] therefore also, as AC is to DC, so is DC to BC. [V. 11]
53
Therefore DC is a mean proportional between AC, CB. Being what it was proposed to prove.
PROPOSITION 54.
54
If an area be contained by a rational straight line and the first binomial, the side of the area is the irrational straight line which is called binomial.
54
For let the area AC be contained by the rational straight line AB and the first binomial AD; I say that the side of the area AC is the irrational straight line which is called binomial.
54
For, since AD is a first binomial straight line, let it be divided into its terms at E, and let AE be the greater term.
54
It is then manifest that AE, ED are rational straight lines commensurable in square only, the square on AE is greater than the square on ED by the square on a straight line commensurable with AE, and AE is commensurable in length with the rational straight line AB set out. [X. Deff. II. 1]
54
Let ED be bisected at the point F.
