| Euclides - 1855 - 262 pages
...and the right angle BED (I. Ax. 11) to the right angle BF D. Therefore the two triangles EBD and FBD **have two angles of the one equal to two angles of the other, each to each** ; and the side BD, which is opposite to one of the equal angles in each, is common to both. Therefore... | |
| Euclides - 1855 - 230 pages
...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... | |
| Robert Potts - 1855 - 1050 pages
...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 - Geometry - 1855 - 336 pages
...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 pages
...angles are equal, these straight lines are, two and two, in the same straight line. 3. If two triangles **have two angles of the one equal to two angles of the other, each to each,** and one side equal to one side, viz. the sides adjacent to equal angles in each; then shall the other... | |
| Peter Nicholson - Cabinetwork - 1856 - 482 pages
...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... | |
| Euclides - 1856 - 168 pages
...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 - Conic sections - 1857 - 242 pages
...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 - Geometry - 1857 - 444 pages
...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,... | |
| Euclides - 1858 - 248 pages
...demonstration of the following propositions. PROP. 26.— THEOR. — (Important.) If two triangles **have two angles of the one equal to two angles of the other, each to each,** and one side equal to one side, viz., either the sides adjacent to the equal angles in each, or the... | |
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