... 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
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...which joins the points of section will bt 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. Book VI. Join BE, CD ; then the triangle BDE is equal to the triangle CDEa.... | |
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...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 CF. to E A. Ioin BE, CD; then the triangle BDE is equal to the triangle CD E*, because... | |
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...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 CDEa, because... | |
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...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.),... | |
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...tvhich joins ¡he points of section shall be parallel to the remaining tide 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 RUH U equal to the triangle CDE, (87.) because... | |
| Euclid, Phillips - 1826 - 180 pages
...вс, and through D draw DF parallel to вс. And because FD is drawn parallel to one of the sides вс **of the triangle ABC ; therefore, proportionally as CD is to DA so is BF to FA."** But CD is double of DA ; therefore / \K • 4. 6. also BF is double of FA ; hence BA is triple of AF.... | |
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...points of section shall be parallel to the remaining side of the triangle. a \ •-, " • Let DE be **drawn parallel to BC, one of the sides of the triangle ABC** : BD shall be to DA, as CE to EA. * 37. 1. Join BE, CD ; then the triangle BDE is equal* to the triangle... | |
| Euclid, John Davidson - 1835 - 513 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** ; then BD is to DA, as CE to EA. Join BE, CD ; then the triangle BDE is equal to the triangle CDE a,... | |
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