... be parallel to the remaining side of the triangle. Let DE be drawn parallel to BC, one of the sides of the triangle ABC : BD is to DA, as CE to EA. Join BE, CD ; Then the triangle BDE is equal to the triangle CDE*, * «.i. Books 3-9 - Page 212by Euclid, Sir Thomas Little Heath, Johan Ludvig Heiberg - 1908Full view - About this book
| John Playfair - Geometry - 1836 - 114 pages
...which joins the points of section shall be parallel to the remaining side of the triangle. Let DE be **drawn parallel to BC, one of the sides of the triangle ABC** ; BD is to DA, as CE to EA. Join BE, CD ; then the triangle BDE is equal to the triangle CDE (Cor.... | |
| Euclides - 1838
...which joins the points of section shall be parallel to the remaining side of the triangle, LetDE be **drawn parallel to BC, one of the sides of the triangle ABC** : BD shall be to DA, as CE to EA. «3, 1 Join BE, CD ; then the triangle BDE is equal' to the triangle... | |
| Euclides - 1841
...which joins the points of section shall be parallel to the remaining side of the triangle. Let DE be **drawn parallel to BC, one of the sides of the triangle ABC:** then BD shall be to DA, as CE to EA. Join BE, CD; then the triangle BDE is equal* to the triangle CDE,... | |
| John Playfair - Euclid's Elements - 1842 - 317 pages
...which , joins the points of section will be parallel to the remaining side of the triangle. Let DE be **drawn parallel to BC, one of the sides of the triangle ABC** : BD is to DA as CE to EA. Join BE, CD ; then the triangle BDE is equal to the triangle CDE (37. 1.),... | |
| Robert Simson - Geometry - 1845 - 199 pages
...which joins the points of section, shall be parallel to the remaining side of the triangle. Let DE be **drawn parallel to BC, one of the sides of the triangle ABC** : BD is to DA, as CE to EA. Join BE, CD ; Then the triangle BDE is equal to the triangle CDE*, * «.i.... | |
| Robert Potts - 1845
...nhich joins the points of section shall be parallel to the remaining side of the triangle. Let DE be **drawn parallel to BC, one of the sides of the triangle ABC.** Then BD shall be to DA, as CE to EA. Join BE, CD. Then the triangle BDE is equal to the triangle CDE,... | |
| Euclides - 1846
...off from it ; join BC, and draw DE parallel to it : then AE is the part required. For, because ED is **parallel to BC, one of the sides of the triangle ABC, therefore** BE is to EA as CD to DA, and, by composition (5. 18), BA is to AE as CA to AD : But CA is a certain... | |
| John Playfair - Euclid's Elements - 1846 - 317 pages
...which joins the points of section will be parallel to the remaining side of the triangle. Let DE be **drawn parallel to BC, one of the sides of the triangle ABC** : BD is to DA as CE to EA. Join BE, CD ; then the triangle BDE is equal to the triangle CDE (37. 1.),... | |
| Anthony Nesbit - Plane trigonometry - 1847 - 426 pages
...that semi-circle, is a right angle. (Euc. III. 31. Simp. in. 13. Em. VI. 14.) THEOREM X. Let DE be **drawn parallel to BC, one of the sides of the triangle ABC;** then BD is to DA, as CE to E A. (Euc. VI. 2. Simp. IV. 12. Em. II. 12.) THEOREM XI. DB In the preceding... | |
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