 | Robert Simson - 1835 - 544 ページ
...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... | |
 | 1836 - 488 ページ
...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 - 1836 - 148 ページ
...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 - 1837 - 246 ページ
...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 - 1837 - 254 ページ
...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 - 1837 - 226 ページ
...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 - 1837 - 290 ページ
...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 - 1837 - 410 ページ
...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 - 1837 - 180 ページ
...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 - 1837 - 112 ページ
...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... | |
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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. 
