 AB be the greater, and from it cut (3. 1.) off DB equal to AC the less, and join DC ; therefore, because A in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two sides DB, BC are equal to the two AC, CB. each to each ; and the angle...  Euclid's Elements: Or, Second Lessons in Geometry,in the Order of Simson's ... - Σελίδα 13των Dennis M'Curdy - 1846 - 138 σελίδεςΠλήρης προβολή - Σχετικά με αυτό το βιβλίο
 | George Gordon Byron Baron Byron - 1847 - 880 σελίδες
...proposition of Euclid : " Because, in the triangles DEC, AC B, DB Is equal to AC, and BC common to both ; cy, in iU excursions, could collect ; and, how little fettered he was b angle DBC is equal to the angle ACS: therefore, the base DC is equal to the base AB. and the triangle... | |
 | Elias Loomis - 1849 - 252 σελίδες
...than the other. Let AB be the greater, and from it cut off DB equal to AC the less, and join CD. Then, because in the triangles DBC, ACB, DB is equal to AC, and BC is common to both triangles, also, by supposition, the angle DBC is equal to the angle ACB; therefore, the triangle DBC... | |
 | Euclid, Thomas Tate - 1849 - 120 σελίδες
...let AB be the greater, and from it cut (i. 3.) off DB equal to AC, the less, and join DC.; therefore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two sides DB, BC are equal to the two AC, CB, each to each; and the angle DBC is... | |
 | Euclides - 1852 - 152 σελίδες
...AB be the greater; " i'i, '. and from it cut" off DB equal to AC, the less, and join DC; Therefore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two sides, DB, BC are equal to the two AC, CB, each to each; and the angle DBC... | |
 | Royal Military Academy, Woolwich - 1853 - 400 σελίδες
...; therefore, because in the triangles DBC, ACB, DB is equal to AC (Constr.), and BC common to both, the two sides, DB, BC are equal to the two AC, CB, each to each ; and the angle DBC is equal to the angle (Hyp.) ACB; therefore the base DC is equal to the base AB, and I he... | |
 | Euclides - 1853 - 146 σελίδες
...Let AB be the greater, and from it cut off (I. 3.) DB equal to AC, the less, and join DC. Therefore because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, 1. The two sides DB, BC, are equal to the two AC, CB, each to each; and the angle DBC... | |
 | George Gordon Byron Baron Byron - 1854 - 1126 σελίδες
...proposition of Euclid : " Because, In EBB triangles DBC, ACB, DB Is equal to AC, anil EC, common to both ; the two sides DB, BC, are equal to the two AC, CB, each to each, and the angle DEC ij equal to the angle ACB : therefore, tí» tuse DC Is equal to the base AB, aüd the triangle... | |
 | George Gordon Byron Baron Byron - 1854 - 1104 σελίδες
...Euclid : " Became, In S» trungVi DBC, ACB, RB is equal to AC, nnd BC, common to both ; the пим^-tDB, BC, are equal to the two AC, CB, each to each, and the angle Dû: л nfxil to the angle ACB : therefore, the baae DC la equal to the bue A3, ind tb, triangle... | |
 | Elias Loomis - 1858 - 256 σελίδες
...greater, and from it cut off DB equal to AC the less, and join CD. Then, because in the triangles DEC, ACB, DB is equal to AC, and BC is common to both triangles, also, by supposition, the angle DBG is equal to the angle ACB ; therefore, the triangle... | |
 | Euclides - 1862 - 172 σελίδες
...Then in the triangles DEC, ACB, because DB is assumed to be equal to AC, and BC is common to both, the two sides DB, BC, are equal to the two AC, CB, each to each -, and the angle DBC is equal to the angle ACB ; (hyp.) therefore the base DC must be equal to the base AB, and... | |
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