 | Timothy Walker - Geometry - 1829 - 156 pages
...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 - Geometry - 1831 - 166 pages
...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 - Geometry - 1836 - 394 pages
...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, equal to the 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... | |
 | Charles Davies - Trigonometry - 1849 - 372 pages
...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, equal to the 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... | |
 | Elias Loomis - Conic sections - 1849 - 252 pages
...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 one, equal to 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... | |
 | Charles Davies - Logic - 1850 - 400 pages
...of this method, Proposition XVII. of the First Book of Legendre : " When two right-angled triangles have the hypothenuse and a side of the one equal to the hypothenuse and a side of the other, each Enunciation, to each, the remaining parts will be equal, each to each, and the triangles themselves... | |
 | Charles Davies - Geometry - 1850 - 218 pages
...perpendicular is the shortest distance from a point to a line. THEOREM XIX. If two right angled triangles have the hypothenuse and a side of the one equal to the hypothenuse and a side of the other, the remaining parts will also be equal, each to each. * Let the two right angled triangles AI) ABC... | |
 | Charles Davies - Geometry - 1850 - 238 pages
...perpendicular is the shortest distance from a point to a line. THEOREM XIX. If two right angled triangles have the hypothenuse and a side of the one equal to the hypothenuse and a side of the other, the remaining parts will also be eqnal, each to eack. Let the two right angled triangles A ABC and... | |
 | Adrien Marie Legendre - Geometry - 1852 - 436 pages
...perpendicular to AB at the middle point 0. PROPOSITION XVII. THEOEEM. If two right-angled triangles have the hypothenuse and a side of the one equal to the hypothenuse and a side of the other y each to each, the triangles are equal. Let BAC and EDF be two right-angled triangles, having the... | |
 | Charles Davies, William Guy Peck - Mathematics - 1855 - 628 pages
...of this method, proposition XVII. of the first book of Legendre. " When two right angled triangles have, the hypothenuse and a side of the one equal to the hypothenuse and a side of the other, euch to each, the remaining parts will be equal each to each, and the triangles themselvue will be... | |
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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... 
