Book 2
13
But the square on AB is equal to the squares on BD, DA, for the angle at D is right; [I. 47] and the square on AC is equal to the squares on AD, DC; therefore the squares on CB, BA are equal to the square on AC and twice the rectangle CB, BD,
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so that the square on AC alone is less than the squares on CB, BA by twice the rectangle contained by CB, BD.
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Therefore etc. Q. E. D.
Proposition 14.
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To construct a square equal to a given rectilineal figure.
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Let A be the given rectilineal figure; thus it is required to construct a square equal to the rectilineal figure A.
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For let there be constructed the rectangular parallelogram BD equal to the rectilineal figure A. [I. 45]
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Then, if BE is equal to ED, that which was enjoined will have been done; for a square BD has been constructed equal to the rectilineal figure A.
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But, if not, one of the straight lines BE, ED is greater.
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Let BE be greater, and let it be produced to F; let EF be made equal to ED, and let BF be bisected at G.
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With centre G and distance one of the straight lines GB, GF let the semicircle BHF be described; let DE be produced to H, and let GH be joined.
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Then, since the straight line BF has been cut into equal segments at G, and into unequal segments at E, the rectangle contained by BE, EF together with the square on EG is equal to the square on GF. [II. 5]
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But GF is equal to GH; therefore the rectangle BE, EF together with the square on GE is equal to the square on GH.
