| Euclid, Robert Simson - 1806 - 518 sider
...in 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 - 311 sider
...in 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
...which 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... | |
| Charles Butler - 1814
..." for if -4EB do not coincide with CFD, it must fall otherwise (as in the figure to prop. 23.) then **upon the same base, and on the same side of it, there** will be two similar segments of circles not coinciding with one another, but this has been shewn (in... | |
| Euclides - 1816 - 528 sider
...in 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 - 317 sider
...two angles, &c. QED II C COR. Hence every equiangular triangle is also equilateral. PROP. VII. THEOR. **Upon the. same base, and on the same side of it, there** caitnot be two triangles, that have their sides which are terminated in one extremity of the base equal... | |
| Euclides - 1821
...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 - 494 sider
...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 - 372 sider
...&c. QED COR. Hence every equiangular triangle is also equal equilátera. Proposition Vll. Theorem. **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... | |
| Robert Simson - 1827 - 513 sider
...two angles, &c. QED COR. — Hence every equiangular triangle is also equilateral. PROP. VII. THEOR. **Upon the same base, and on the same side of it, there** can- See N. not be two triangles that have their sides which are terminated in one extremity of the... | |
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