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" CB, BA, by twice the rectangle CB, BD. Secondly, Let AD fall without the triangle ABC. Then, because the angle at D is a right angle, the angle ACB is greater than a right angle ; (i. "
Euclid's Elements of plane geometry [book 1-6] explicitly enunciated, by J ... - Page 60
by Euclides - 1860
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Woolwich Mathematical Papers for Admission Into the Royal Military Academy ...

Eldred John Brooksmith - Mathematics - 1901 - 368 pages
...indefinitely in both directions, must meet. 4. C is a point in a finite straight line AB. Prove that the square on AB is equal to the squares on AC and CB, together with twice the rectangle contained by AC and CB. Express this result algebraically. 5. ABC...
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Books 10-13 and appendix

Euclid - Mathematics, Greek - 1908 - 576 pages
...Since then the angle ACB is an angle in the semicircle ACB, therefore the angle ACB is right. [in. 31] Therefore the square on AB is equal to the squares on AC, CB. [i. 47] Hence the square on AB is greater than the square on AC by the square on CB. But AC is...
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Euclid's Elements in Greek: Vol. I: Books 1-4

Richard Fitzpatrick - Mathematics - 2005 - 298 pages
...(figures) HF, CK, AC, and GE are (equivalent to) the whole of ADEB, which is the square on AB. Thus, the square on AB is equal to the squares on AC and CB, and twice the rectangle contained by AC and CB. Thus, if a straight-line is cut at random, then the square on the whole (straight-line)...
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The Thirteen Books of Euclib's Elements.

Euclid - 454 pages
...rectangle contained by AC, CB. But HF, CK, AG, GE are the whole A DEB, 50 which is the square on AB, Therefore the square on AB is equal to the squares on AC, CB and twice the rectangle contained by AC, CB. Therefore etc. QED i. twice the rectangle contained...
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