Book 10
91
Now let the square LM be made equal to AI, and let there be subtracted the square NO having a common angle with it, the angle LPM, and equal to FK; therefore the squares LM, NO are about the same diameter. [VI. 26]
91
Let PR be their diameter, and let the figure be drawn.
91
Since then the rectangle contained by AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG. [VI. 17]
91
But, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to KF; [VI. 1] therefore EK is a mean proportional between AI, KF. [V. 11]
91
But MN is also a mean proportional between LM, NO, as was before proved, [Lemma after X. 53] and AI is equal to the square LM, and KF to NO; therefore MN is also equal to EK.
91
But EK is equal to DH, and MN to LO; therefore DK is equal to the gnomon UVW and NO.
91
But AK is also equal to the squares LM, NO; therefore the remainder AB is equal to ST.
91
But ST is the square on LN; therefore the square on LN is equal to AB; therefore LN is the side of AB.
91
I say next that LN is an apotome.
91
For, since each of the rectangles AI, FK is rational, and they are equal to LM, NO, therefore each of the squares LM, NO, that is, the squares on LP, PN respectively, is also rational; therefore each of the straight lines LP, PN is also rational.
91
Again, since DH is medial and is equal to LO, therefore LO is also medial.
91
Since then LO is medial, while NO is rational, therefore LO is incommensurable with NO.
91
But, as LO is to NO, so is LP to PN; [VI. 1] therefore LP is incommensurable in length with PN. [X. 11]
91
And both are rational; therefore LP, PN are rational straight lines commensurable in square only; therefore LN is an apotome. [X. 73]
91
And it is the side of the area AB; therefore the side of the area AB is an apotome.
91
Therefore etc.
PROPOSITION 92.
92
If an area be contained by a rational straight line and a second apotome, the side of the area is a first apotome of a medial straight line.
