 | William Chauvenet - 1905 - 336 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... | |
 | International Correspondence Schools - Building - 1906 - 634 pages
...each triangle, and adding the results. 53. Comparison of the Areas of Similar Polygons. The areas of two similar triangles are to each other as the squares of their homologous sides. In Fig. 35, An*. ABC =\AB*CD (1) '£'C' = \A'B<y. O Z> (2) Dividing equation (1) by equa_ A'.B CD n(... | |
 | Wisconsin. Department of Public Instruction - Education - 1906 - 124 pages
...each other as their bases; triangles having equal bases are to each other as their altitudes. 108. Similar triangles are to each other as the squares of their homologous sides. 109. Similar polygons are to each other as the squares of the homologous sides. 110. To construct a... | |
 | Webster Wells - Geometry - 1908 - 336 pages
...they intercept, the areas of the figures may be found by §§ 284 and 286. PROP. VII. THEOREM 288. Two similar triangles are to each other as the squares of their homologous sides. Draw A ABC ; and construct A A'B' C' similar to A ABC, AB (c) and A'B' (c') being homologous sides.... | |
 | Webster Wells - Geometry, Plane - 1908 - 208 pages
...they intercept, the areas of the figures may be found by §§ 284 and 286. PROP. VII. THEOREM 288. Two similar triangles are to each other as the squares of their homologous sides. Draw A ABC ; and construct A A'B' C' similar to A ABC, AB (c) and A'B' (c') being homologous sides.... | |
 | Geometry, Plane - 1911 - 192 pages
...whose diagonals intersect at E. If F is the middle point of BC, prove that EF produced bisects AD. 8. Two similar triangles are to each other as the squares of their homologous sides. SEPTEMBER, 1893 1. If two sides of a triangle are unequal, the angles opposite are unequal, and the... | |
 | Arthur Schultze, Frank Louis Sevenoak - Geometry - 1913 - 490 pages
...and ZA = ZA'. Find AC if AB = 2 in., 4'J3' = 3 in., and 4'C' = 4 in. PROPOSITION XIV. THEOREM 379. Similar triangles are to each other as the squares of their homologous sides. Given To prove Proof. A ABC ~ AA'B'C A ABC ~AB L AA'B'C' ZA A ABC AB X AC AA'B'C' ''A'B'xA'C" (303)... | |
 | Arthur Schultze, Frank Louis Sevenoak - Geometry - 1913 - 486 pages
...are to each other as the products of their diagonals. 381. METHOD XXI. As similar polygons (including triangles) are to each other as the squares of their homologous sides, Prop. IX may be used to draw a polygon that shall be any given part of a given polygon, and similar... | |
 | Arthur Schultze, Frank Louis Sevenoak - Geometry, Plane - 1913 - 328 pages
...are to each other as the products of their diagonals. 381. METHOD XXI. As similar polygons (including triangles) are to each other as the squares of their homologous sides, Prop. IX may be used to draw a polygon that shall be any given part of a given polygon, and similar... | |
 | Arthur Schultze, Frank Louis Sevenoak - Geometry - 1918 - 486 pages
...Ex. 1158. If A ABC = A A'B'C', and £A = £ A', then AB:A'B' =A'C' :AC. PROPOSITION XIV. THEOREM 379. Similar triangles are to each other as the squares of their homologous sides. "AA'B'C' A'B'XA'C'' AB ., AC C', (303) (378) AB AB In like manner A'B' A ABC ACT AA'B'C' QED Ex. 1159.... | |
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Similar triangles are to each other as the squares of their homologous sides. 
