| William Mitchell Gillespie - Surveying - 1880 - 540 pages
...AD is derived from the area of a triangle being equal to its base by half its altitude. (52 T) Since similar triangles are to each other as the squares of their homologous sides, ABC : DBE : : AB" : BD« ; whence BD = AB J 5|? = AB <J — ^— . f ADU fn .f. 1fc The construction... | |
| Simon Newcomb - Geometry - 1881 - 418 pages
...:: PB : BC. 4. Comparing with (1) and (2), AB : BR :: PB : BC. QED THEOREM XXII. 423. The areas of similar triangles are to each other as the squares of their homologous sides. Hypothesis. ABC and PQR, two triangles in which AB : BC : CA :: PQ : QR : RP. Conclusion. Area ABC... | |
| Franklin Ibach - Geometry - 1882 - 208 pages
...BD X DA for its equal DE X CD. ELEMENTS OF PLANE GEOMETRY. RELATION OF POLYGONS. THEOREM XXX. 290. Similar triangles are to each other as the squares of their homologous sides. Let the As ABC and EFG be similar. DH To prove that A ABC : A EFG :: A~ÏÏ : ËF\ Draw the altitudes AD and... | |
| Charles Davies, Adrien Marie Legendre - Geometry - 1885 - 538 pages
...: : AE : AC ; hence (B. II, P. IV.), we 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 F : then... | |
| Webster Wells - Geometry - 1886 - 392 pages
...prove that the parallelogram ABFG is equivalent to tlie tnipezoid. 160 PROPOSITION VII. THEOREM. 334. Two similar triangles are to each other as the squares of their homologous sides. Let AB and A'B' be homologous sides of the similar triangles ABC and A'B'C'. To prove that -A^- = -^MA'B'C'... | |
| William Chauvenet, William Elwood Byerly - Geometry - 1887 - 331 pages
...trapezoid is equal to the product of its altitude byhalf the sum of its parallel bases. PROPOSITION VIII. Similar triangles are to each other as the squares of their homologous sides. PROPOSITION IX. Similar polygons are to each other as the squares of their homologous sides. PROPOSITION... | |
| William Chauvenet, William Elwood Byerly - Geometry - 1887 - 342 pages
...trapezoid is equal to the product of its altitude by half the sum of its parallel bases. PROPOSITION VIII. Similar triangles are to each other as the squares of their homologous sides. PROPOSITION IX. Similar polygons are to each other as the squares of their homologous sides. PROPOSITION... | |
| James Wallace MacDonald - Geometry - 1889 - 80 pages
...construct a parallelogram equivalent to a given square. Proposition XVIII. A Theorem. 253. The areas of similar triangles are to each other as the squares of their homologous sides. Proposition XIX. A Theorem. 254. The areas of any similar polygons are proportional to the squares... | |
| James Wallace MacDonald - Geometry - 1894 - 76 pages
...construct a parallelogram equivalent to a given square. Proposition XVIII. A Theorem. 253. The areas of similar triangles are to each other as the squares of their homologous sides. See Proposition VII. Proposition XIX. A Theorem. 254. The areas of any similar polygons are proportional... | |
| Edward Albert Bowser - Geometry - 1890 - 418 pages
...other as the products of the sides including these angles. Proposition 9. Theorem. 377. The areas of similar triangles are to each other as the squares of their homologous sides. BC Hyp. Let ABC, A'B'C' be similar AS. A ABC BC' To prove __U7 = __t. Proof. Since ZB = / B', (Hyp.)... | |
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