 | William Frothingham Bradbury - Geometry - 1880 - 260 pages
...angle DBC—CDB; therefore the whole angle ABC = ADC; therefore the triangles ABC and ADC, having two sides and the included angle of the one equal to two sides and the included angle of the other, are equal (80). 89. Scholium. In equal triangles the equal angles are opposite the equal... | |
 | Simon Newcomb - Geometry - 1881 - 418 pages
...QU= QW (I.), TQ+QU= TU> TW. 6. Comparing with (4), TU> TV. QED 45 THEOREM XIX. 108. If two triangles .have two sides and the included angle of the one equal to two sides and the included angle of the other, they are identically equal. Hypothesis. ABC&n&MNP, two triangles in which PM = CA. PN... | |
 | Alfred Hix Welsh - Geometry - 1883 - 326 pages
...direction BC; then EF will take the direction BH, and ABH = ABC + DEF. THEOREM IV. If two triangles have two sides and the included angle of the one equal to two sides and the included angle of the other, each to each, the triangles will be equal in all their parts. " Let ABC and DEF be two... | |
 | Edward Olney - Geometry - 1883 - 344 pages
...will fall in AD or AD produced, as E'H'. Now, the two triangles AE'B and DH'C are equal, since they have two sides and the included angle of the one equal to two sides and the included angle of the other; viz., AB = DC, being opposite sides of a parallelogram; and fora like reason BE' = CH'.... | |
 | Edward Olney - Geometry - 1883 - 352 pages
...PROPOSITION XXI. 704. Theorem.—On the same sphere, or on equal spheres, two spherical triangles having two sides and the included angle of the one equal to two sides and the included angle of the other, each to each, are equal, or symmetrical and equivalent. DEMONSTRATION. Let ABC and A'B'C'... | |
 | Edward Olney - Geometry - 1883 - 352 pages
...falls between CA and CB, and CH will meet AB in some point, as H. Draw HE. The triangles HCB and HCE have two sides and the included angle of the one, equal to the corresponding parts of the other, whence HE = HB (?). Now AH + HE > AE but AH + HE = AH + HB =... | |
 | Evan Wilhelm Evans - Geometry - 1884 - 170 pages
...similar. Take AG equal to DE, and AH to DF; also, join GH. Then the triangles AGH, DEF, having two sides and the included angle of the one equal to two sides and the included angle of the other, are equal throughout (Theo. XII, Book I). Now, by hypothesis, Hence, it follows that... | |
 | William John M'Clelland - 1885 - 182 pages
...(any two of which being supposed greater than the third}. (Cf. Euc. I. xxii.) (7). If two triangles have two sides and the included angle of the one equal to two sides and the included angle of the other, each to each, the triangles are equal in every respect. For, since two sides of given... | |
 | Charles Davies, Adrien Marie Legendre - Geometry - 1885 - 538 pages
...CAG equal to the angle D (Post. 7) ; make AG equal to DE, and draw GC. Then the triangles AGC and DEF have two sides and the included angle of the one equal to two sides and the included angle of the other, each to each ; consequently, GC is equal to EF (PV). Now, the point G may be without... | |
 | William Chauvenet - Geometry - 1887 - 346 pages
...ABC and A' B'C' have now been proved to have two sides and the included angle of the one respectively equal to two sides and the included angle of the other, and are equal, by Proposition VI. PROPOSITION XI.—THEOREM. 27. If two angles of a triangle are equal,... | |
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If two triangles have two sides, and the included angle of the one equal to two sides and the included angle of the other, they are equal in all their parts. 
