Kitap 2
13
In acute-angled triangles the square on the side subtending the acute angle is less than the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle, namely that on which the perpendicular falls, and the straight line cut off within by the perpendicular towards the acutc angle.
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Let ABC be an acute-angled triangle having the angle at B acute, and let AD be drawn from the point A perpendicular to BC;
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I say that the square on AC is less than the squares on CB, BA by twice the rectangle contained by CB, BD.
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For, since the straight line CB has been cut at random at D, the squares on CB, BD are equal to twice the rectangle contained by CB, BD and the square on DC. [II. 7]
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Let the square on DA be added to each; therefore the squares on CB, BD, DA are equal to twice the rectangle contained by CB, BD and the squares on AD, DC.
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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.
