 | Adrien Marie Legendre - 1825 - 570 σελίδες
...; THEOREM. 56. Two right-angled triangles are equal, when the hypothenuse and a side of the one are equal to the hypothenuse and a side of the other, each to each. Demonstration. Let the hypothenuse AC = DF (fig. 33), and Fig. 33, the side AB = DE ; the right-angled... | |
 | Adrien Marie Legendre, John Farrar - 1825 - 294 σελίδες
...and THEOREM. 56. Two right-angled triangles are equal, when the kypothenuse and a side of the one are equal to the hypothenuse and a side of the other, each to each. Demonstration. Let the hypothenuse AC = DF (Jig. 33), and *V the side AB = DE ; the right-angled triangle... | |
 | Adrien Marie Legendre, John Farrar - 1825 - 280 σελίδες
...THEOREM. ' a 56. Two right-angled triangles are equal, when the hypothenuse and a side of the one are equal to the hypothenuse and a side of the other, each to each. Demonstration. Let the hypothenuse AC = DF (fig. 33), and Fig. 33. the side AB = DE ; the right-angled... | |
 | James Hayward - 1829 - 218 σελίδες
...learner show that no different triangle could be constructed with these things given.] We say then — If two right-angled triangles have the hypothenuse and a side of the one, equal respectively to the hypothcnuse and a side of the other, the two triangles will be equal in all their... | |
 | James Hayward - 1829 - 228 σελίδες
...learner show that no different triangle could be constructed with these things given.] We say then—If two right-angled triangles have the hypothenuse and a side of the one, equal respectively to the hypothenuse and a side of the other, the two triangles will be equal in all their... | |
 | Timothy Walker - 1829 - 156 σελίδες
...right triangle A F'D=BEC (39), since AF=BE and AD=BC from the nature of parallelograms, and the two triangles have the hypothenuse and a side of the one, equal to the hvpothenuse and a side of the other. Now if from the whole figure ABCF, we take AFD, there will remain... | |
 | Timothy Walker - 1831 - 166 σελίδες
...The right triangle AFD=BEC (59), since AF=BE and AD=BC from the nature of parallelograms, and the two triangles have the hypothenuse and a side of the one,...equal to the hypothenuse and a side of the other. Now if from the whole figure ABCF, we take AFD, there will remain ABC D. Again if from the whole figure... | |
 | Adrien Marie Legendre - 1836 - 394 σελίδες
...it will be perpendicular to AB at the middle point C. PROPOSITION XVII. THEOREM. If two right angled triangles have the hypothenuse and a side of the one,...hypothenuse and a side of the other, each to each, the remaining parts will also be equal, each to each, and the triangles themselves will be equal. In the... | |
 | Adrien Marie Legendre - 1841 - 288 σελίδες
...THEOREM. , 56. Two right-angled triangles are equal, when the hypothenuse and a side of the one are equal to the hypothenuse and a side of the other, each to each. Demonstration. Let the hypothenuse AC — DF (fig. 33), and Fig- 33 the side AB = DE ; the right-angled... | |
 | Elias Loomis - 1849 - 252 σελίδες
...second line, it will be perpendicular to the second line at its middle point PROPOSITION XIX. THEOREM. If two right-angled triangles have the hypothenuse...and a side of the other, each to each, the triangles are equal. Let ABC, DEF be two right-angled triangles, having the hypothenuse AC and the side AB of... | |
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If two right-angled triangles have the hypothenuse and a side of the one, equal to the hypothenuse and a side of the other, each to each, the triangles are equal. Let... 
