 ... 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 - Σελίδα 212των Euclid - 1908Πλήρης προβολή - Σχετικά με αυτό το βιβλίο
 | John Playfair - 1806 - 320 σελίδες
...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.... | |
 | Euclid - 1810 - 554 σελίδες
...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... | |
 | Euclides - 1816 - 588 σελίδες
...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... | |
 | John Playfair - 1819 - 354 σελίδες
...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, W. Little - 1822 - 916 σελίδες
...that semicircle, is a right angle. (Euc. ///. 31. Simp. III. 13. Em. VI. 14.J 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 EA, ( Euc, VI. 2, Simp. IV. 12. Em. II. 12.} :i the preceding figure, DE... | |
 | Anthony Nesbit - 1824 - 476 σελίδες
...that semicircle, is a right angle. (Euc. III. 31. Simp. III. 13. Em. vI. 14.; B 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 EA. (Euc. vI. 2. Simp. IV. 12. Em. II. 12J DB THEOREM XI. In the preceding... | |
 | Peter Nicholson - 1825 - 1046 σελίδες
...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 - 1826 - 234 σελίδες
...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.... | |
 | Robert Simson - 1827 - 546 σελίδες
...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 - 1835 - 540 σελίδες
...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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