 Therefore, in a right-angled, &c. QED COR. From this it is manifest, that the perpendicular drawn from the right angle of a right-angled triangle to the base, is a mean proportional between the segments of the base; and also that each of the sides is...  The Rekhâgaṇita; or, Geometry in Sanskrit - Side 109av Euclid - 1901Uten tilgangsbegrensning - Om denne boken
 | Robert Simson - 1806 - 546 sider
...right angled triangle to the base, is a mean proportional between the segments of the base : and also that each of the sides is a mean proportional between the base, and its segment adjacent to that side: because in the triangles BDA, ADC, BD is to DA, as DA to DC b ;... | |
 | John Playfair - 1806 - 320 sider
...a right angled triangle to the base, is a mean proportional between the segments of the base ; and each of the sides is a mean proportional between the base and its segment adjacent to that side. For, in the triangles BDA, ADC, BD : DA : : DA : DCb ; and in the... | |
 | John Playfair - 1819 - 350 sider
...right angled-triangle, to the base, is a mean proportional between the segments of the base ; and also that each of the sides is a mean proportional between the base, and its segment adjacent to that side. For in the triangles BDA, ADC, BD : DA : : DA : DC (4.-6.) ; and... | |
 | Anthony Nesbit - 1824 - 476 sider
...each other. Also the perpendicular AD is a mean proportional between the segments of the base ; and each of the sides is a mean proportional between the base and its segment adjacent to that side ; therefore, BD is to DA, as DA to DC ; BC is to BA, as BA to BD... | |
 | Euclides - 1826 - 226 sider
...from the right angle at the base, it is a mean proportional between the segments of the base; and also that each of the sides is a mean proportional between the base and its segment adjacent to that side. Deduction. PROPOSITION IX. PROBLEM. From a given right line to cut... | |
 | Euclid - 1826 - 234 sider
...the right angle at the base, it is a mean proportional between the segments of the base ; and also that each of the sides is a mean proportional between the base and its segment adjacent to that side. Deduction. To divide a given finite right line into two parts, such... | |
 | Robert Simson - 1827 - 546 sider
...right angled triangle to the base, is a mean proportional between the segments of the base, and also that each of the sides is a mean proportional between the base, and the segment of it adjacent to that side : because in the triangles BDA, *4.e. ADC, BD is to DA*, as DA to DC; and... | |
 | John Playfair - 1829 - 210 sider
...angled triangle to the hypothenuse or base, is a mean proportional between the segments of the base; and each of the sides is a mean proportional between the base and its segment adjacent to that side. For, m the equiangular triangles BDA, ADC, BD : DA : : DA : DC;... | |
 | Euclides - 1834 - 518 sider
...right-angled triangle to the base, is a mean proportional between the segments of the base, and also that each of the sides is a mean proportional between the base, and the segment of it adjacent to that • 4. 6. side : because in the triangles BDA, ADC, BD is to DA *, as • 4. 6.... | |
 | Euclid - 1835 - 540 sider
...right angled triangle to tlie base, is a mean proportional between the segments of the base : And also, that each of the sides is a mean proportional between the base and its segment adjacent to that side : Because in the triangles BDA, ADC, BD is to DA, as DA to DC b ;... | |
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