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" The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. A D A' Hyp. In triangles ABC and A'B'C', To prove AABC A A'B'C' A'B' x A'C ' Proof. Draw... "
Elements of Geometry and Trigonometry Translated from the French of A.M ... - Page 86
by Charles Davies - 1849 - 359 pages
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Schultze and Sevenoak's Plane Geometry

Arthur Schultze, Frank Louis Sevenoak - Geometry, Plane - 1913 - 328 pages
...two given squares. [The solution is left to the student.] PROPOSITION XIII. THEOREM 378. The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the product of the sides including the equal angles. Given A ABC and A'B'C', Z...
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Schultze and Sevenoak's Plane and Solid Geometry

Arthur Schultze, Frank Louis Sevenoak - Geometry - 1913 - 490 pages
...vertices of an inscribed rectangle inclose a rhombus. Ex. 1067. Two parallelograms are similar when they have an angle of the one equal to an angle of the other, and the including sides proportional. Ex. 1068. Two rectangles are similar if two adjacent sides are proportional....
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Plane and Solid Geometry

George Albert Wentworth, David Eugene Smith - Geometry - 1913 - 491 pages
...Given .'.ZA = ZA'. §282 Then AABC AB * AC 111611 AA'B'C' ~ AW X A'C" (The areas of two triangles that have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles.) AABC AB AC But f§ = I^'...
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Plane and Solid Geometry

George Albert Wentworth, David Eugene Smith - Geometry - 1913 - 496 pages
...triangles are similar, Given ^A'. §282 §282 A A'B'C' A'B' X A'C' (The areas of two triangles that have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles.) 4 AABC AB AC 1S> A A'R'r'...
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Connecticut School Document, Issues 1-13

Education - 1913 - 396 pages
...only one If two triangles have their homologous sides proportional they are similar If two triangles have an angle of the one equal to an angle of the other their areas are to each other as the products of the sides including the equal angles The area of a...
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Plane and Solid Geometry

George Albert Wentworth, David Eugene Smith - Geometry - 1913 - 496 pages
...• 1 AA'B'C' AW Proof. Since the triangles are similar, Given §282 (The areas of two triangles that have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles.) AABC AB AC AC (Similar...
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Military Education in the United States

Ira Louis Reeves - Military education - 1914 - 508 pages
...5, find the area of the segment subtended by the side of a regular hexagon. 8. Theorem: The areas of two triangles which have an angle of the one equal to an angle of the other, are to each other as the products of the sides including those angles. 9. Problem: Through a given...
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Report of the Secretary for Public Instruction ...

Queensland. Department of Public Instruction - Education - 1914 - 284 pages
...circle is half that of the square oircumBCribed about the same circle. 8. Prove that if two triangles have an angle of the one equal to an angle of the other arxcl the sides about these angles proportional the triangles will be similar. 9. Two circles intersect...
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Plane Geometry

John Wesley Young, Albert John Schwartz - Geometry, Modern - 1915 - 248 pages
...equal to an acute angle of the other. PLANE GEOMETRY 410. THEOREM. Two triangles are similar, if they have an angle of the one equal to an angle of the other and the including sides are proportional. , FIG. 186. Given the A ABC and A'B'C', with ZA = Z A', and AB =...
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Plane Geometry

Edith Long, William Charles Brenke - Geometry, Plane - 1916 - 292 pages
...to their corresponding sides. Suggestion. Apply Art. 220, Cor. 3. 222. Theorem XVI. // two triangles have an angle of the one equal to an angle of the other, and the including sides proportional, the triangles are similar. Given the triangles ABC and AiBiCi with angle...
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