 | E. J. Brooksmith - 1889
...this proposition. How many of the exterior angles of any triangle must be obtuse ? 4. Parallelograms upon the same base, and between the same parallels, are equal to one another. Of all triangles that can be drawn upon a given base and between the same parallels, shew that an isosceles... | |
 | Edward Mann Langley, W. Seys Phillips - 1890 - 515 Seiten
...that 'the three medians of a triangle are concurrent.' PROPOSITION 35. THEOREM. Parallelograms on the same base and between the same parallels are equal to one another. Let ||gms A BCD, EBCF be on the same base and between the same ||s AF, BC, ABCD shall be equal to EBCF.... | |
 | Rupert Deakin - 1891 - 79 Seiten
...the diameter bisects the parallelogram, ie divides it into two equal parts. 35. Parallelograms on the same base and between the same parallels are equal to one another. 36. Parallelograms on equal bases and between the same parallels are equal to one another. 37. Triangles... | |
 | Euclid - 1892 - 518 Seiten
...PROPOSITION 37. THEOREM. Triangles on the same base, and between the sanie parallels, are equal in area. Let the triangles ABC, DBC be upon the same base BC, and between the same parallels BC, AD. Then shall the triangle ABC be equal to the triangle DBC. Construction. Through B draw BE parallel... | |
 | Euclid, John Bascombe Lock - 1892 - 167 Seiten
...of equal area. 6. Rectangles having two adjacent sides equal are of equal area. Proposition 37. 139. Triangles upon the same base and between the same parallels are equal, Let ABC, DBC represent triangles upon the same base BC and between the same parallels AD, BC ; it is... | |
 | Seymour Eaton - 1899 - 340 Seiten
...ABC is equal to the triangle BCD (Proposition 4). PROPOSITION 35. THEOREM 323 Parallelograms on the same base and between the same parallels are equal to one another. Let the parallelograms ABCD, EBCF be on the same base BC, and between the same parallels AF, BC; then the parallelogram... | |
 | American Association for the Advancement of Science - 1899
...part a strict treatment of equivalence. Even Euclid, in proving his I. 35, " Parallelograms on the same base, and between the same parallels, are equal to one another," does not show that the parallelograms can be divided into pairs of pieces admitting of superposition... | |
 | American Association for the Advancement of Science - 1899
...third part a strict treatment of equivalence. Even Euclid, in proving his I. 35, "Parallelograms on the same base, and between the same parallels, are equal to one another," does notshowthat the parallelograms can be divided into pairs of pieces admitting of superposition... | |
 | Sidney Herbert Wells - 1900
...irregular figure to a triangle of equal area. The principle of this reduction depends upon the fact that triangles upon the same base and between the same parallels are equal (Euclid, i., 37), and the method consists of converting certain triangles, obtained from the figure,... | |
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TRIANGLES upon the same base, and between the same parallels, are equal to one another. 
