 | Adrien Marie Legendre - 1819 - 574 sider
...AE. we have But 4JiE is the square of 2.AE or of .-1C ; and A In. is the square BD ; therefore the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals. THEOREM. Fig 114. 196. The line DE (fig. 114), drawn parallel to the base... | |
 | Adrien Marie Legendre, John Farrar - 1825 - 280 sider
...BC = 4AE + IDE. But 4AE is the square of 2AE or of AC; and 4DE is the square of BD ; therefore the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals. THEOREM. Fig. 114. 196. The line DE (fig. 114), drawn parallel to the base... | |
 | Adrien Marie Legendre, John Farrar - 1825 - 294 sider
...BC- 4AE + 4DE. But 4.4JU is the square of <2JLE or of AC; and 4DE is the square of BD ; therefore the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals. THEORKM. Fig. 114. 1 96. The line DE (fig. 1 1 4), drawn parallel to the... | |
 | Adrien Marie Legendre - 1825 - 276 sider
...IDE. 2 2 • But 4AE is the square of 2.4£ or of AC; and 4DE is the square of BD ; therefore the su;n of the squares of the sides of a parallelogram is equal to the sum of the squares- f the diagonals. THEOREM. Fig. 114. 1 96. Tlie line DE (fig. 1 1 4), drawn parallel to the... | |
 | Euclid, Dionysius Lardner - 1828 - 542 sider
...squares of BF and DF is equal to twice that of EF and D E. Hence the proposition is manifest. (301) The sum of the squares of the sides of a parallelogram is equal to that of the diagonals. For in that case the line EF vanishes, since the diagonals bisect each other... | |
 | Euclid - 1833 - 216 sider
...with twice the square of BC ; because CP and BC are equal. Fig: 7. Schol. Hence it is evident that the sum of the squares of the sides of a parallelogram is equal to the (i)Prcp.29. sum of the squares of the diagonals. B- 1- Because the angles DEC and BDA, ACB and CAD... | |
 | Adrien Marie Legendre - 1841 - 288 sider
...+ BC = 4AE + 4DE. But 4AE is the square of 2AE or ofAC; and 4DEis the square of BD ; therefore the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals. THEOREM. Fig 114. 196. The line DE (fig. 114), drawn parallel to the lose... | |
 | Euclid, Isaac Todhunter - 1867 - 426 sider
...is equal to the square on AB, together with twice the square on BC. 147. The sum of the squares on the sides of a parallelogram is equal to the sum of the squares on the diagonals. 148. The base of a triangle is given and is bisected by the centre of a given... | |
 | Euclid, Isaac Todhunter - 1867 - 424 sider
...is equal to the square on AB, together with twice the square on BC. 147. The sum of the squares on the sides of a parallelogram is equal to the sum of the squares on the diagonals. 148. The base of a triangle is given and is bisected by the centre of a given... | |
 | Adrien Marie Legendre - 1871 - 490 sider
...equal to AC', and 4.DE2 to BD' (P. Vin., C.) : hence, AB2 + BC2 + CD2 + DA2 = AC2 + BD'. That is, the sum of the squares of the sides of a parallelogram, is equal to the sum of the squares of its diagonals. / ^ PROPOSITION XV. THEOREM. In any triangle, a line drawn parallel to the... | |
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The sum of the squares on the sides of a parallelogram is equal to the sum of the squares on the diagonals. 
