Kitap 13
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Therefore, when the straight line CD is cut in extreme and mean ratio, CB is the greater segment.
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Therefore etc. Q. E. D.
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Lemma. That the double of AC is greater than BC is to be proved thus.
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If not, let BC be, if possible, double of CA.
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Therefore the square on BC is quadruple of the square on CA; therefore the squares on BC, CA are five times the square on CA.
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But, by hypothesis, the square on BA is also five times the square on CA; therefore the square on BA is equal to the squares on BC, CA: which is impossible. [II. 4]
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Therefore CB is not double of AC.
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Similarly we can prove that neither is a straight line less than CB double of CA; for the absurdity is much greater.
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Therefore the double of AC is greater than CB. Q. E. D.
PROPOSITION 3.
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If a straight line be cut in extreme and mean ratio, the square on the lesser segment added to the half of the greater segment is five times the square on the half of the greater segment.
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For let any straight line AB be cut in extreme and mean ratio at the point C, let AC be the greater segment, and let AC be bisected at D; I say that the square on BD is five times the square on DC.
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For let the square AE be described on AB, and let the figure be drawn double.
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Since AC is double of DC, therefore the square on AC is quadruple of the square on DC, that is, RS is quadruple of FG.
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And, since the rectangle AB, BC is equal to the square on AC, and CE is the rectangle AB, BC, therefore CE is equal to RS.
