Kitap 6
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Let AB be the given uncut straight line, and AC the straight line cut at the points D, E; and let them be so placed as to contain any angle; let CB be joined, and through D, E let DF, EG be drawn parallel to BC, and through D let DHK be drawn parallel to AB. [I. 31]
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Therefore each of the figures FH, HB is a parallelogram; therefore DH is equal to FG and HK to GB. [I. 34]
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Now, since the straight line HE has been drawn parallel to KC, one of the sides of the triangle DKC, therefore, proportionally, as CE is to ED, so is KH to HD. [VI. 2]
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But KH is equal to BG, and HD to GF; therefore, as CE is to ED, so is BG to GF.
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Again, since FD has been drawn parallel to GE, one of the sides of the triangle AGE, therefore, proportionally, as ED is to DA, so is GF to FA. [VI. 2]
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But it was also proved that, as CE is to ED, so is BG to GF; therefore, as CE is to ED, so is BG to GF, and, as ED is to DA, so is GF to FA.
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Therefore the given uncut straight line AB has been cut similarly to the given cut straight line AC. Q. E. F.
PROPOSITION 11.
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To two given straight lines to find a third proportional.
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Let BA, AC be the two given straight lines, and let them be placed so as to contain any angle; thus it is required to find a third proportional to BA, AC.
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For let them be produced to the points D, E, and let BD be made equal to AC; [I. 3] let BC be joined, and through D let DE be drawn parallel to it. [I. 31]
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Since, then, BC has been drawn parallel to DE, one of the sides of the triangle ADE, proportionally, as AB is to BD, so is AC to CE. [VI. 2]
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But BD is equal to AC; therefore, as AB is to AC, so is AC to CE.
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Therefore to two given straight lines AB, AC a third proportional to them, CE, has been found. Q. E. F. to find. The Greek word, here and in the next two propositions, is προσευρεῖν, literally to find in addition.
PROPOSITION 12.
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To three given straight lines to find a fourth proportional.
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Let A, B, C be the three given straight lines; thus it is required to find a fourth proportional to A, B, C.
