 | James Wallace MacDonald - Geometry - 1894 - 76 pages
...triangle. Proposition XXXI. A Theorem. 68. If two triangles have two sides of the one respectively equal to two sides of the other and the included angles unequal, the third side of the one having the greater angle will be longer than the third side of the other. SCHOLIUM.... | |
 | James Wallace MacDonald - Geometry - 1889 - 80 pages
...Proposition XXXII. A Theorem. 69. Conversely, if two triangles have two sides of the one respectively equal to two sides of the other and the included angles unequal, the angle opposite the longer third side will be greater than the angle opposite the shorter. Proposition... | |
 | William Chauvenet - 1893 - 340 pages
...greater than A. PROPOSITION XIV.—THEOREM. 30. If two triangles have two sides of the one respectively equal to two sides of the other, and the included angles unequal, the triangle which has the greater included angle has the greater third side. Let ABC and ABD be the two... | |
 | George Cunningham Edwards - Geometry - 1895 - 324 pages
...difference of the interior angles at P and C will equal the sum of the A PAC and PBC. ' 3. Show that if two triangles have two sides of one equal to two sides of the other, and their included angles unequal, the triangle having the greater included angle will have the greater... | |
 | Andrew Wheeler Phillips, Irving Fisher - Geometry - 1896 - 570 pages
...points.] Substituting XC for its equal XC, BC'<BX+XC. Or BC'<BC. QED PROPOSITION XXVI. THEOREM 93. If two triangles have two sides of one equal to two sides of the other but the third side of the first greater than thc third side of the second, then the angle opposite... | |
 | Andrew Wheeler Phillips, Irving Fisher - Geometry, Modern - 1896 - 276 pages
...points.] Substituting XC for its equal XC', BC'<BX+XC. Or BC'<BC. QED PROPOSITION XXVI. THEOREM 93. If two triangles have two sides of one equal to two sides of the other but the third side of the first greater than the third side of the second, then the angle opposite... | |
 | Andrew Wheeler Phillips, Irving Fisher - Geometry - 1897 - 374 pages
...A TD are equal. § 78 Hence AB=AD. But AB + BC>AC. By subtraction BC>DC. The triangles BTC and DTC have two sides of one equal to two sides of the other, and the third side BC of one greater than the third side DC of the other. Therefore angle BTC>angle DTC. §92... | |
 | Andrew Wheeler Phillips, Irving Fisher - Geometry - 1896 - 574 pages
...TD are equal. § 79 Hence AB = AD. But AB + BC>AC. By subtraction BC>DC. The triangles ETC and DTC have two sides of one equal to two sides of the other, and the third side-BC of one greater than the third side DC of the other. Therefore BTC>DTC. §93 By construction... | |
 | Harvard University - Geometry - 1899 - 39 pages
...than the side opposite the less angle. THEOREM V. If two triangles have two sides of one respectively equal to two sides of the other, and the included angles unequal, the triangle which has the greater included angle has the greater third side. If two triangles have two... | |
 | William James Milne - Geometry, Modern - 1899 - 258 pages
...ADC? In angle CDB ? In angle ACD ? In angle DCS ? Proposition XXX 129. Construct two triangles having two sides of one equal to two sides of the other, and the angles included between the sides unequal. How do the third sides compare in length? Which triangle... | |
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If two triangles have two sides of one respectively equal to two sides of the other, and the included angles unequal, the triangle which has the greater included angle has the greater third side. 
