Book 10
53
Let the parallelogram AC be completed; I say that AC is a square, that DG is a mean proportional between AB, BC, and further that DC is a mean proportional between AC, CB.
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]
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Therefore DG is a mean proportional between AB, BC.
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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.
