| William Mitchell Gillespie - Surveying - 1856 - 478 pages
...AD is derived from the area of a triangle being equal to its base by half its altitude. (527) Since similar triangles are to each other as the squares of their homologous sides, ABC : DBK : : AB' : BD> ; whence BD = AB J ^ = AB The construction of Fig. 363 is founded on the proportion... | |
| William Mitchell Gillespie - Surveying - 1857 - 538 pages
...AD is derived from the area of a triangle being equal to its base by half its altitude. (527) Since similar triangles are to each other as the squares of their homologous sides, ABC : DBE : : AB' : BD3 ; whence BD = AB J ^5| ~ AB A/— ^ — . j A150 f in -f- n The construction... | |
| Adrien Marie Legendre - Geometry - 1863 - 464 pages
...AD : AB : : AE : AC ; hence (B. II., P. IV.), w(j have, ADE : ABE : : ABE : ABC ; PROPOSITION XXV. THEOREM. Similar triangles are to each other as the squares of their homologous sides. Let the triangles ABC and DEF be similar, the angle A being equal to the angle D, B to E, and C to... | |
| Eli Todd Tappan - Geometry, Modern - 1864 - 288 pages
...a I. \ / Substituting these in the equation of the area, it becomes, 391. Theorem. — The areas of similar triangles are to each other as the squares of their homologous lines. HC area BCD. In a similar manner, prove that the areas have the same ratio as the squares of... | |
| Charles Davies - Mathematics - 1867 - 186 pages
...implies a general relation of the magnitudes, which is measured by the Ratio. For example : we say that " Similar triangles are to each other as the squares of their homologous sides." What do we mean by that? Just this : That the area of a trinngle Is to the area of a similar triangle,... | |
| Benjamin Peirce - 1868 - 200 pages
...right-angled at M, we have, by § 260, MH*: MP = HP:PI. But by § 181, Ratio of Similar and lte.;u!ar Polygons. 266. Theorem. Similar triangles are to each...which, multiplied by the proportion gives \AB X CE: lAB ' X CE' = AB Z : A'B 2 , and, by ^251, the area of ABC: the area of A'B'C ' = AB t : A'B'*. 267.... | |
| Edward Brooks - Geometry - 1868 - 284 pages
...and omitting the common factor AHC, we have, or, ABC:DEF::ACXSC:DFXFE. Therefore, etc. THEOREM XVI. Similar triangles are to each other as the squares of their homologous sides. Let ABC and A DE be two similar triangles ; then will they be to each other as the squares of any two... | |
| William Mitchell Gillespie - Surveying - 1869 - 550 pages
...AD is derived from the area of a triangle being equal to its base by half its altitude. (527) Since similar triangles are to each other as the squares of their homologous sides, ABC : DBE : : AB' : BD' ; whence BD = AB J 55? — AB J "* . y ABC ym -f- n The construction of Fig.... | |
| Benjamin Peirce - 1870 - 200 pages
...right-angled at M, we have, by f5 260, M1V: MP = HP: PI. But by § 181, Hatio of Similar and Regular Polyguns. 266. Theorem. Similar triangles are to each other...triangles ABC, ABC ' (fig. 109), we have, by § 199, CE: CE'=AB: AB', which, multiplied by the proportion gives \AB x CE : and, by § 251, the area of ABC:... | |
| William Chauvenet - Geometry - 1871 - 380 pages
...theorems, and the sum of these areas will be the area of the polygon. PROPOSITION VII.— THEOREM. 20. Similar triangles are to each other as the squares of their homologous sides. Let ABC, A'B'C' be similar tri- •* angles; then, ABC BC2 D' C' DC A'B'C' B'C'* Let AD, A'D', be the... | |
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