| Robert Simson - Trigonometry - 1835 - 513 pages
...equal to KCF, and the right angle FHC equal to the right angle FKC; in the triangles FHC, FKC there are **two angles of the one equal to two angles of the other,** and the side FC, which is opposite to one of the equal angles in each, is common to both : therefore... | |
| Mathematics - 1836 - 472 pages
...which has the greater base, shall be greater than the angle contained by the sides of the other. XXVI. **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, or the sides opposite... | |
| John Playfair - Geometry - 1836 - 114 pages
...bisected by BD, and that the right angle BED is equal to the right angle BFD, the two Iriangles EBD, FBD **have two angles of the one equal to two angles of the other,** and the side BD, which is opposite to one of the equal angles in each, is common to both ; therefore... | |
| Charles Reiner - Geometry - 1837 - 215 pages
...SECTION IV. two angles of the one is equal to the sum of the remaining two angles of the other. 2. **If two triangles have two angles of the one equal to two angles of the other, each to each,** the third angle of the one is equal to the third angle of the other ; that is, the triangles are equiangular.... | |
| Charles Reiner - Geometry - 1837 - 215 pages
...and if the equal angles be subtracted from these equals, the remaining angles must be equal. M.—If **two triangles have two angles of the one equal to two angles of the other, each to each,** what may be said of the remaining third angles ? P.—They must be equal,—for the reason alleged... | |
| William Whewell - Mechanics - 1837 - 182 pages
...angle; therefore MLN is equal to LKH; and the angles at H and at N are right angles. Therefore the **triangles have two angles of the one equal to two angles of the other** ; and the side KL is equal to LM. Therefore the triangles are equal, and HL is equal to MN; that is,... | |
| Andrew Bell - Euclid's Elements - 1837 - 240 pages
...by BD ; and because the right angle BED is equal to the right angle BFD, the two triangles EBD, FBD, **have two angles of the one equal to two angles of the other** ; and the side BD, which is opposite to one of the equal angles in each, is common to both ; therefore... | |
| Euclid, James Thomson - Geometry - 1837 - 390 pages
...is equal (const.) to FBD, and that the right angles BED, BFD are equal, the two triangles EBD, FBD **have two angles of the one equal to two angles of the other,** and the side BD, which is opposite to one of the equal angles in each, is common to both ; therefore... | |
| A. Bell - Conic sections - 1837 - 164 pages
...Def. 7)i and therefore the angles AFG, AEG, are also equal. The triangles AGE, AGF, have therefore **two angles of the one equal to two angles of the other,** and they have also the side AG common ; wherefore they are equal, and the side AF is equal to the side... | |
| Euclides - Euclid's Elements - 1837 - 88 pages
...be > Z EOF. PROPOSITION XXVI. (Argument ad absurdum). Theorem. If two triangles have two angles of **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, or opposite to the... | |
| |