| Thomas Fowler - 1895 - 365 sider
...which our deductive reasoning proceeds. The proposition proved in Euclid, Book i. Prop. 38, that ' **Triangles upon equal bases, and between the same parallels, are equal to one another,'** is derived from, or is the total result of, the previous deductions (i) that ' Parallelograms upon... | |
| Northwest Territories Council of Public Instruction - 1897
...right angles. I. 32. Cor. 2. (6) Divide a right angle into five equal parts. 10. ('/) Parallelograms on **equal bases and between the same parallels are equal to one another.** I. 36. (6) Extend the proof of proposition (a) to any number of parallelograms. (c) Distinguish "equal... | |
| Seymour Eaton - 1899 - 340 sider
...square shall be less than that of the parallelogram. Lesson No. 17 PROPOSITION 38. THEOREM Triangles on **equal bases, and between the same parallels, are equal to one another.** Let the triangles ABC, DEF be on equal bases BC, EF, and between the same parallels BF, AD : then the... | |
| Manitoba. Department of Education - 1900
...so that ED is equal to twice BA. Prove that the angle DBC is equal to onethird of the angle ABC. 5. **Triangles upon equal bases and between the same parallels are equal to one another.** If E and D are the points of trisection (nearest to A) of the sides AB, AC of a triangle, and F the... | |
| 1907
...form of an equation : 2DOA + 2DOC .-= 2DOB. Now the triangle DOA equals triangle BCO (for triangles on **equal bases and between the same parallels are equal to one another—** Euclid, 1.38). Therefore, 2DOA + 2DOC = 2BCO + 2DOC = 2DOB. Therefore the sum of the moments of P and... | |
| Joseph Gregory Horner - 1907
...form of an equation : 2DOA + 2DOC - 2DOB. Now the triangle DOA equals triangle BCO (for triangles on **equal bases and between the same parallels are equal to one another—** Euclid, I. 38). Therefore, 2DOA + 2DOC = 2BCO + 2DOC = 2DOB. Therefore the sum of the moments of P... | |
| Paul Carus - 1909
...of the article in a universal sense is regular in Greek. Euclid does not say "All parallelograms on **equal bases and between the same parallels are equal to one another"** but "the parallelograms" (TO. TOpoAAiyAoypo/t/aa) ; so in the famous 47th it is not "in all" but "In... | |
| R. H. Warn, John G. Horner - 2002 - 284 sider
...parallels, are equal to one another. (Euc. I. 37, and I. 38.) (jr) Parallelograms upon the same base, or **upon equal bases, and between the same parallels, are equal to one another.** (Enc. I. 35, and I. 36.) (7i) If a parallelogram and a triangle be upon the same base, and between... | |
| Richard Fitzpatrick - 2005 - 296 sider
...^apaXXr]Xoiç ïaa àXXrjXoiç èaiiv OTisp sSsi ELEMENTS BOOK l Proposition 38 В С EF Triangles which are on **equal bases and between the same parallels are equal to one another.** Let ABC and DEF be triangles on the equal bases BC and EF, and between the same parallels BF and AD.... | |
| Popular educator - 1860
...angle at the base), the perpendicular will fall within the triangle, upon the base itself. PROPOSITION **XXXVIII. THEOREM. Triangles upon equal bases, and...between the same parallels, are equal to one another.** ng. 38. Inflg.38,letthetrianglesABC Ç AD н and D в F, be upon equal bases в о and EF, and .between... | |
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