 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 ... - Page 13by Dennis M'Curdy - 1846 - 138 pagesFull view - About this book
 | Robert Simson - Trigonometry - 1762 - 488 pages
...therefore becaufe in the triangles DBC, ACB, DB is equal to AC, and 1>C common to bath, the two fides DB, BC are equal to the two AC, CB, each to each ; and the angle DBC is equal to the ang'c ^CB ; therefore the bafe DC is equal to the bafe AB, and the triangle... | |
 | Robert Simson - Trigonometry - 1775 - 534 pages
...therefore, becaufe in the triangles DBC, ACBi D3 is equal to AC, and BC common to both, the two fides DB, BC are equal to the two AC, CB, each to each ; and the angle DBC is equal to the angle ACB ; therefore the bale DC is eqtial to the bafe AB, and the triangle... | |
 | Euclid - 1781 - 552 pages
...therefore, becaule in the triangles DBC, ACB, 1 >I> it equal to AC, and BC common to both, the two fides DB, BC are equal to the two AC, CB, each to each ; and the angle DBC is equal to the angle ACB ; therefore the bafe DC is equal to the bare AB, and the triangle... | |
 | Euclid, James Williamson - Euclid's Elements - 1781 - 324 pages
...Jlraight line DB is equat to theJlraight line AC, but the Jlralght line BC common; certainly the two DB, BC are equal to the two AC, CB, each to each; and an angle, the angle contained by DBC is equal to the angle contained by ACB (by fupp,); wherefore (by... | |
 | Robert Simson - Trigonometry - 1781 - 534 pages
...let AB be the greater, and from it cut 'off DB equal to AC, . the lefs,and join DC. therefore becaufe in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two fides DB, BC are equal to the two AC, CB, each to each; and the angle DBC is... | |
 | John Playfair - Euclid's Elements - 1795 - 462 pages
...AB be the greater, and from it cut a off DB equal to AC, the lefe, and join DC ; therefore, becaufe in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two fides DB, BC are equal to the two AC, CB, each to each ; but the angle DBC... | |
 | Alexander Ingram - Trigonometry - 1799 - 374 pages
...AB be the greater, and from it cut a off DB equal to AC, the lefs, and join DC ; therefore, becaufe in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two fides DB, BC are equal to the two AC, CB, each to each ; and the angle DBC... | |
 | Robert Simson - Trigonometry - 1804 - 528 pages
...the greater, and from it cut * oft'DB equal to AC, a. 3. x. the lefs, and join DC. therefore becaufe in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two fides DB, BC are equal to the two AC, CB, each to each ; and the angle DBC... | |
 | Robert Simson - Trigonometry - 1806 - 548 pages
...the greater, and from it cut a off DB a. 3. 1. equal to AC, the less, and join DC ; there- A fore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two sides DB, BC are equal to to the two AC, CB, each to each ; and the angle DBC... | |
 | John Playfair - Mathematics - 1806 - 320 pages
...other : let AB be the greater, 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 ; but the angle DBC... | |
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