| Robert Simson - 1806 - 546 páginas
...which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise... | |
| John Playfair - 1806 - 320 páginas
...which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore, upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise... | |
| John Mason Good - 1813 - 714 páginas
...subtend, or arc. opposite to» the equal angles, shall be equal to one another. Prop. VII. Theor. Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise... | |
| Euclides - 1816 - 588 páginas
...which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore, upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise... | |
| John Playfair - 1819 - 350 páginas
...which the vertex of one triangle is upon a side of the other, needs no demonstration.. Therefore, upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise... | |
| Euclides - 1821 - 294 páginas
...every equiangular triangle is equilateral ; vide, Elrington. PROP. 7. THEOR. i On the same right line and on the same side of it there cannot be two triangles formed whose conterminous sides are equal. If it be possible that there can, 1st, let the vertex of... | |
| Rev. John Allen - 1822 - 508 páginas
...it are equal, and therefore the sides opposite to them. PROP. VII. THEOR. Upon the same base (AB), and on the same side of it, there cannot be two triangles (ACB, ADB), whose conterminous sides are equal, (namely AC to AD, and BC to BD). For, if possible,... | |
| Peter Nicholson - 1825 - 1046 páginas
...which the vertex of one triangle is upon a side of the other, needs uo demonstration. Therefore, upon the same base, and on the same side of it, there cannot be two triangles, that have their sides which are terminated >n one extremity of the base equal to one another, and likewise... | |
| Robert Simson - 1827 - 546 páginas
...which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore, upon the same base, and on the same side of it, there cannot be two triangles that have thtir sides, which are terminated in one extremity of the base, equal to another, and likewise... | |
| Jared Sparks, Edward Everett, James Russell Lowell, Henry Cabot Lodge - 1828 - 598 páginas
...angles, the lines AI, BD, produced, will meet.' The other is from Simson's Euclid, prop. 7, b. 1 . ' Upon the same base and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise... | |
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