 | 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
...being homologous, DE is parallel to BC, and we have, AD : AB : : AE : AC ; hence (B. II, P. IV.), we have, ADE : ABE : : ABE : ABC ; PROPOSITION XXV. THEOREM....angle A being equal to the angle D, B to E, and C to F : then the triangles are to each other as the squares of any two homologous sides. Because the angles... | |
 | Webster Wells - Geometry - 1886 - 392 pages
...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 - 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... | |
 | 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... | |
 | 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... | |
 | 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... | |
 | Education - 1890 - 714 pages
...with an exercise in Geometry. He has proven absolutely and beyond all peradventure that the areas of similar triangles are to each other as the squares of their homologous sides. The proposition admits of no debate, and whoever does not accept the conclusion " is not of sound mind... | |
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Similar triangles are to each other as the squares of their homologous sides. 
