Book 13
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I say next that MB is also a fourth apotome.
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Let the square on N be equal to that by which the square on BK is greater than the square on KM; therefore the square on BK is greater than the square on KM by the square on N.
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And, since KF is commensurable with FB, componendo also, KB is commensurable with FB. [X. 15]
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But BF is commensurable with BH; therefore BK is also commensurable with BH. [X. 12]
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And, since the square on BK is five times the square on KM, therefore the square on BK has to the square on KM the ratio which 5 has to 1.
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Therefore, convertendo, the square on BK has to the square on N the ratio which 5 has to 4 [V. 19, Por.], and this is not the ratio which a square number has to a square number; therefore BK is incommensurable with N; [X. 9] therefore the square on BK is greater than the square on KM by the square on a straight line incommensurable with BK.
11
Since then the square on the whole BK is greater than the square on the annex KM by the square on a straight line incommensurable with BK, and the whole BK is commensurable with the rational straight line, BH, set out, therefore MB is a fourth apotome. [X. Deff. III. 4]
11
But the rectangle contained by a rational straight line and a fourth apotome is irrational, and its square root is irrational, and is called minor. [X. 94]
11
But the square on AB is equal to the rectangle HB, BM, because, when AH is joined, the triangle ABH is equiangular with the triangle ABM, and, as HB is to BA, so is AB to BM.
11
Therefore the side AB of the pentagon is the irrational straight line called minor. Q. E. D.
PROPOSITION 12.
12
If an equilateral triangle be inscribed in a circle, the square on the side of the triangle is triple of the square on the radius of the circle.
12
Let ABC be a circle, and let the equilateral triangle ABC be inscribed in it; I say that the square on one side of the triangle ABC is triple of the square on the radius of the circle.
