CB ; wherefore the four figures HF, CK, AG, GE are equal to the squares of AC, CB, and to twice the rectangle AC, CB : but HF, CK, AG, GE make up the whole figure ADEB, which is the square of AB: therefore the square of AB is equal to the squares of AC,... Elements of geometry: consisting of the first four,and the sixth, books of ... - Page 47by Euclides - 1842Full view - About this book
| Euclid, John Playfair - Euclid's Elements - 1826 - 326 pages
...AB» =AO+CB3+2AC.CB. Wherefore, if a straight line be divided &e. QED COR. From thedemonstration,it is manifest that the parallelograms about the diameter of a square are likewise squares. •f PROP. V. THEOR. If a straight line be divided into two equal parts, and also into two unequal... | |
| Robert Simson - Trigonometry - 1827 - 546 pages
...HF, CK, AG, GE are equal to the squares of AC, CB, and to twice the rectangle AC, CB : but HF, CK, AG, GE make up the whole figure ADEB, which is the square of AB: therefore the square of AB is equaJf to the squares of AC. t J Ax. CB, and twice the rectangle AC, CB. Wherefore, if a straight line,... | |
| Euclid, Dionysius Lardner - Euclid's Elements - 1828 - 542 pages
...rectangle under DC and CB ; but the square of AC is equal to the squares of AD and DC (XLVII, Book I.); and therefore the square of AB is equal to the squares of AC and CB together with double the rectangle under BC and CD, therefore the square of AB exceeds the sum... | |
| Euclid, Robert Simson - Geometry - 1829 - 548 pages
...HF, CK, AG, GE are equal to the squares of AC, CB, and to twice the rectangle AC, CB ; but HF, CK, AG, GE make up the whole figure ADEB, which is the...rectangle AC, CB. Wherefore, if a straight line, &c. QED COB. From the demonstration it is manifest, that the parallelograms about the diameter of a square... | |
| John Playfair - Euclid's Elements - 1832 - 358 pages
...CK+AG+GE=the figure AE, or AB'; therefore AB' = ACa + CB2+2AC.CB. Wherefore, if a straight line be divided &c QED COR. From the demonstration, it is manifest...THEOR. If a straight line be divided into two equal parls, and also into two unequal parts; the rectangle contained by the unequal parts, together with... | |
| Euclid - Euclid's Elements - 1833 - 216 pages
...the rectangle under DC and CB ; but the square of AC is equal to the squares of AD and DC (1), and therefore the square of AB is equal to the squares of AC and CB, together with double the rectangle under BC and CD ; therefore the square of AB exceeds the... | |
| Euclides - 1834 - 518 pages
...figures HF, CK, AG, GE are equal to the squares of AC, CB, and to twice the rectangle AC, CB: but HF, CK, AG, GE make up the whole figure ADEB, which is the square of AB : t 1 Ax. therefore the square of AB is equal f to the squares of AC, CB, and twice the rectangle AC,... | |
| Robert Simson - Trigonometry - 1835 - 544 pages
...figures HF, CK, AG, GE are equal to the squares of AC, CB, and to twice the rectangle AC, CB: But HF, CK, AG, GE make up the whole figure ADEB, which is the...COR. From the demonstration, it is manifest, that tlie parallelograms about the diameter of a square are likewise squares. a 46. 1. b 31. 1. c 43. 1.... | |
| John Playfair - Euclid's Elements - 1835 - 336 pages
...HF+CK-fAG -t-GE=AC2+CB2+2AC.CB. But HF-f-CK+AG+GE=the figure AE, or AB2 ; therefore AB2= AC2+CB2-f2AC.CB. COR. From the demonstration, it is manifest that the...about the diameter of a square are likewise squares. SCHOLIUM. This property is derived from the square of a binomial. For, let the two parts into which... | |
| Euclid - 1835 - 540 pages
...the rectangle contained by the parts. Let the straight line AB be divided into any two parts in C; the square of AB is equal to the squares of AC, CB, and to twice the rectangle contained by AC, CB. Upon AB describe a the square ADEB, and join BD, and through... | |
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