Book 6
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Let two straight lines DE, DF be set out containing any angle EDF; let DG be made equal to A, GE equal to B, and further DH equal to C; let GH be joined, and let EF be drawn through E parallel to it. [I. 31]
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Since, then, GH has been drawn parallel to EF, one of the sides of the triangle DEF, therefore, as DG is to GE, so is DH to HF. [VI. 2]
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But DG is equal to A, GE to B, and DH to C; therefore, as A is to B, so is C to HF.
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Therefore to the three given straight lines A, B, C a fourth proportional HF has been found. Q. E. F.
PROPOSITION 13.
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To two given straight lines to find a mean proportional.
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Let AB, BC be the two given straight lines; thus it is required to find a mean proportional to AB, BC.
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Let them be placed in a straight line, and let the semicircle ADC be described on AC; let BD be drawn from the point B at right angles to the straight line AC, and let AD, DC be joined.
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Since the angle ADC is an angle in a semicircle, it is right. [III. 31]
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And, since, in the right-angled triangle ADC, DB has been drawn from the right angle perpendicular to the base, therefore DB is a mean proportional between the segments of the base, AB, BC. [VI. 8, Por.]
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Therefore to the two given straight lines AB, BC a mean proportional DB has been found. Q. E. F.
PROPOSITION 14.
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In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; and equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal.
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Let AB, BC be equal and equiangular parallelograms having the angles at B equal, and let DB, BE be placed in a straight line; therefore FB, BG are also in a straight line. [I. 14]
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I say that, in AB, BC, the sides about the equal angles are reciprocally proportional, that is to say, that, as DB is to BE, so is GB to BF.
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For let the parallelogram FE be completed.
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Since, then, the parallelogram AB is equal to the parallelogram BC, and FE is another area, therefore, as AB is to FE, so is BC to FE. [V. 7]
