| Euclides - 1855 - 262 Seiten
...EDF. Therefore, if two triangles, &c. QED The enunciation of this proposition may be thus simplif'ed: **If two triangles have two angles of the one, equal to two angles of the other, each to each, and** a side of the one equal to a side of the other similarly situated as to the equal angles, the two triangles... | |
| Euclides - 1855 - 230 Seiten
...the angle EBC (4): and the angle AEG is equal to the angle BEH (a); therefore the triangles AEG, BEH **have two angles of the one, equal to two angles of the other, each to each, and** the sides AE, EB, adjacent to the equal angles, equal to one another; wherefore they have their other... | |
| Robert Potts - 1855 - 1050 Seiten
...drawn to intersect one another, the greater segments will be equal to the sides of the pentagon. 3. **If two triangles have two angles of the one equal to two angles of the other,** and one side equal to one side, viz. either the sides adjacent to the equal angles in each, or the... | |
| John Playfair - 1855 - 334 Seiten
...it is not equal to it: therefore the angle BAC is greater than the angle EDF. PROP. XXVI. THLOR. Jf **two triangles have two angles of the one equal to two angles of the** otIirr, each to each; and one side equal to one side, viz. either the sides adjacent to the equal anglrs,... | |
| Cambridge univ, exam. papers - 1856 - 200 Seiten
...on the same side of it, are either two right angles, or are together equal to two right angles. 3. **If two triangles have two angles of the one equal to two angles of the other, each to each, and one** si ie equal to one side, via. the sides opposite to equal angles in each, then shall the other sides... | |
| Peter Nicholson - 1856 - 482 Seiten
...parallel to CD, the alternate angles, GFE, FGH, are also equal ; therefore the two triangles GEF, HFG, **have two angles of the one equal to two angles of the other, each to each ; and** the side FG, adjacent to the equal angles, common ; the triangles are therefore equal (theorem 6) ;... | |
| Euclides - 1856 - 168 Seiten
...BAC, and the angle ABE is equal to the angle ABC (being both right angles), the triangles ABC, ABE **have two angles of the one equal to two angles of the other,** and the side AB common to the two. Therefore the triangles ABC, ABE are equal, and the side AE is equal... | |
| Elias Loomis - 1857 - 242 Seiten
...is parallel to CD, the alternate angles GHE, HEF are also equal. Therefore, the triangles HEF, EHG **have two angles of the one equal to two angles of the other, each to each, and** the side Eli included between the equal angles, common ; hence the triangles are equal (Prop. VII.)... | |
| Adrien Marie Legendre, Charles Davies - 1857 - 442 Seiten
...consequently, the two equiangular triangles BA C, CUD, are similar figures. Cor. Two triangles which **have two angles of the one equal to two angles of the other,** are similar; for, the third angles are then equal, and the two triangles are equian gular (BI, p. 25,... | |
| Elias Loomis - 1858 - 256 Seiten
...is parallel to CD, the alternate angles GHE, HEF are also equal. Therefore, the triangles HEF, EHG **have two angles of the one equal to two angles of the other, each to each, and** the side Eli included between the equal angles, common ; hence the triangles are equal (Prop. VII.)... | |
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