 | Robert Gibson, James Ryan - 1839 - 452 páginas
...may be proved that the square ACGF is equal to the parallelogram KCLM. So ABDE+ACGF the sum of the squares =BKLH+KCML, the sum of the two parallelograms or square BCMH; therefore the sum of ihe squares on AB and AC is equal to the square on BC. QED* Cor. 1. Hence the hypothenuso of a right-angled... | |
 | Nathaniel Bowditch - 1846 - 854 páginas
...JMrallelograms BKLH and KCML; but the sum of these parallelograms is equal to the square ВСЛ1Н ; therefore the sum of the squares on AB and AC is equal to the square on BC. Cor. Hence, in any right-angled triangle, if we have the hypotenuse and one of the legs, we may easily... | |
 | Sir Henry Edward Landor Thuillier - 1851 - 826 páginas
...square ACGF, is equal to the parallelogram KCLM. So ABDE + ACGF the sum of the squares, = BKLH = KCML, the sum of the two parallelograms or square BCMH ;...the squares on AB and AC is equal to the square on EC. QED Cor. 1; The hypothenuse of a right-angled triangle may be found by having the other two sides... | |
 | Euclid - 1853 - 176 páginas
...square on AE is equal in area to twice the square on AD together with twice the square on DE (rf); therefore the sum of the squares on AB and AC is equal in area to twice the ^ square on BE, twice the square on DE, and twice the square on AD taken together.... | |
 | James McDowell - 1878 - 310 páginas
...the sides of the triangle ABC meet in <?. Since AD bisects the side BC in D, therefore by (41), twice the sum of the squares on AB and AC is equal to four times the squares on AD and DB, that is, to four tunes the square on AD to- . gether with the... | |
 | 1880 - 880 páginas
...to the sum of the parallelograms BKLH and KCML; but the sum of these parallelograms is equal to the square BCMH ; therefore the sum of the squares on AB and AC is equal to the square on BC. Cor. Hence, in any right-angled triangle, if we have the hypotenuse and one of the legs, we may easily... | |
 | Nathaniel Bowditch - 1888 - 704 páginas
...to the sum of the parallelograms BKLH and KCML; but the sum of these parallelogram? is equal to the square BCMH; therefore the sum of the squares on AB and AC is equal to the square on BC. Cor. Hence, in any right angled triangle, if we have the hypotenuse and one of the legs, we may easily... | |
 | University of Calcutta - 1908 - 562 páginas
...CB, and twice the rectangle AC, CD. ABC is a triangle whose side BC is bisected at D ; show that 4 the sum of the squares on AB and AC is equal to twice the sum of the squares on AD and DB. 9. Show how to bisect an arc of a circle. 4 10. Inscribe... | |
 | Harold Ordway Rugg, John Roscoe Clark - 1919 - 392 páginas
...same unit 5 times. By constructing squares on the sides of the triangle, you can see by counting that the sum of the squares on AB and AC is equal to the square on BC. To test this further, the pupil should construct a right triangle with the base 12 units and the altitude... | |
 | S. N. Forrest - 1947 - 444 páginas
...hypotenuse. Now figures 1, 2, 3, 4 and 5 will fit together exactly to make the square on BC, showing that the sum of the squares on AB and AC is equal to the square onBC. Fig. 128 shows another right-angled triangle ABC. Again, figures 1, 2, 3, 4 and 5 fit exactly... | |
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ABDE+ACGF the sum of the squares —BKLH-\-KCML, the sum of the two parallelograms or square BCMH; therefore the sum of the squares on AB and AC is equal to the square on BC. 
