 | Robert Simson - 1762 - 488 ページ
...ftraight line AC making any angle with AB ; and in AC take any point D, and take AC which is the fam« multiple of AD that AB is of the part which is to...join BC, and draw DE parallel to it. then AE is the fame part of AB that AD is of AC ; that is, AE is the part required to be cut off. Becaufe ED is parallel... | |
 | Robert Simson - 1775 - 534 ページ
...requi- gee N. fed. Let AB be the given ftraight line ; it is required to cut off any part from it. Fiom the point A draw a ftraight line AC making any angle...parallel to one of the fides of the triangle ABC, viz.toBC,asCD is to DA, fois'BE to EA ; and, by compofition b, CA is to AD, as BA to AE : But CA is... | |
 | Euclid, John Playfair - 1795 - 462 ページ
...and in AC take any point D, and take AC M 4 the Book VI. the fame multiple of AD, that AB i? - f - of the part which is to be cut off from it ; join...one of the fides of the triangle ABC, viz. to BC, a i. 6. as CD is to DA, fo is» BE to EA; and, 6.18. 5. by compofition b, CA is to AD, as BA to, AE... | |
 | John Playfair, Euclid - 1804 - 468 ページ
...take AC M 4 fnch Book VI «t. '. b i«. 5. fuch that it fhall contain AD, as oft as AB js to contain the part, which is to be cut off from it ; join BC,...one of the fides of the triangle ABC, viz. to BC, CD : DA : : BE : EA » ; and by compofition b, CA : AD : : BA : AE : But CA is a multiple of AD ; therefore... | |
 | Robert Simson - 1804 - 530 ページ
...ftraight line AC making any angle with AB; and in AC take any point D, and take AC which is the ftme multiple of AD that AB is of the part which is to...join BC, and draw DE parallel to it. then AE is the fame part of AB that AD is of AC ; that is, AE is the part required to be cut off. Becaufe ED is parallel... | |
 | John Playfair - 1806 - 320 ページ
...; in AC take any point D, and take AC sucli that it shall contain AD as" often as AB is to contain the part which is to be cut off from it ; join BC,...to it ; then AE is the part required to be cut off. Because ED is parallel to BC, one of the sides of the triangle ABC, CD : DA : : BE : EAa ; therefore,... | |
 | Robert Simson - 1806 - 546 ページ
...draw a straight line AC making any angle with AB ; and in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to be cut off from it ; join DC, and draw DE parallel to it : then AE is the part required to be cut off. Because ED is parallel... | |
 | Euclid - 1810 - 554 ページ
...draw a straight line AC making any angle with AB ; and in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to...to it : then AE is the part required to be cut off. Because ED is parallel to one of the sides of the triangle ABC, viz. to BC, as CD is to / \ DA, so... | |
 | Euclides - 1816 - 588 ページ
...draw a straight line AC, making any angle with AB ; and in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to...to it: Then AE is the part required to be cut off. . Because ED is parallel to one of the sides of the triangle ABC, viz. to BC, as CD is to DA, so isa... | |
 | John Playfair - 1819 - 350 ページ
...; and in AC take any point D, and take AC such that it shall contain AD, as oft as AB is to contain the part, which is to be cut off from it ; join BC,...to it : then AE is the part required to be cut off. Because ED is parallel to one of the sides of the triangle ABC, viz. to BC, CD : DA : : DK : 1^(2^6.)... | |
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AC the same multiple of AD, that AB is of the part which is to be cut off from it : join BC, and draw DE parallel to it : then AE is the part required to be cut off. 
