Kitap 13
BOOK XIII. PROPOSITIONS.
PROPOSITION 1.
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If a straight line be cut in extreme and mean ratio, the square on the greater segment added to the half of the whole is five times the square on the half.
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For let the straight line AB be cut in extreme and mean ratio at the point C, and let AC be the greater segment; let the straight line AD be produced in a straight line with CA, and let AD be made half of AB; I say that the square on CD is five times the square on AD.
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For let the squares AE, DF be described on AB, DC, and let the figure in DF be drawn; let FC be carried through to G.
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Now, since AB has been cut in extreme and mean ratio at C, therefore the rectangle AB, BC is equal to the square on AC. [VI. Def. 3, VI. 17]
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And CE is the rectangle AB, BC, and FH the square on AC; therefore CE is equal to FH.
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And, since BA is double of AD, while BA is equal to KA, and AD to AH, therefore KA is also double of AH.
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But, as KA is to AH, so is CK to CH; [VI. 1] therefore CK is double of CH.
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But LH, HC are also double of CH.
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Therefore KC is equal to LH, HC.
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But CE was also proved equal to HF; therefore the whole square AE is equal to the gnomon MNO.
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And, since BA is double of AD, the square on BA is quadruple of the square on AD, that is, AE is quadruple of DH.
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But AE is equal to the gnomon MNO; therefore the gnomon MNO is also quadruple of AP; therefore the whole DF is five times AP.
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And DF is the square on DC, and AP the square on DA; therefore the square on CD is five times the square on DA.
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Therefore etc. Q. E. D.
PROPOSITION 2.
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If the square on a straight line be five times the square on a segment of it, then, when the double of the said segment is cut in extreme and mean ratio, the greater segment is the remaining part of the original straight line.
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For let the square on the straight line AB be five times the square on the segment AC of it, and let CD be double of AC; I say that, when CD is cut in extreme and mean ratio, the greater segment is CB.
