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" In any triangle, the product of two sides is equal to the square of the bisector of the included angle plus the product of the segments of the third side. Hyp. In A abc, the bisector t divides c into the segments, p and q. To prove ab = t "
Wentworth & Hill's Exercise Manuals: Geometry - Page 64
by George Albert Wentworth - 1884
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Exercise Manuals, Issue 3

George Albert Wentworth - 1889 - 276 pages
...A, B, C, the other in D, E, F, respectively. Prove that the triangles ABC, D .ELF are similar. 16. In every triangle the product of two sides is equal to the product of the diameter of the circumscribed circle and the altitude upon the third side. If AC, BC...
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Wentworth & Hills's Exercise Manuals: Geometry, Issue 3

George Albert Wentworth - 1889 - 264 pages
...in A, B, C, the other in D, E, F, respectively. Prove that the triangles ABC, DJSFaxe similar. 16. In every triangle the product of two sides is equal to the product of the diameter of the circumscribed circle and the altitude upon the third side. If AC, BCare...
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Plane and Solid Geometry

Arthur Schultze, Frank Louis Sevenoak - Geometry - 1901 - 396 pages
...parallelogram is equal to the sum of the squares of the diagonals. PROPOSITION XL. THEOREM 328. In any triangle, the product of two sides is equal to the...the included angle plus the product of the segments of the third side. Hyp. In A abc, the bisector t divides c into the segments, p and q. To prove ab...
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Plane and Solid Geometry

Arthur Schultze, Frank Louis Sevenoak - Geometry - 1902 - 394 pages
...20, and 18, respectively. Find the length of the median to 18. or PROPOSITION XL. THEOREM 328. In any triangle, the product of two sides is equal to the...the included angle plus the product of the segments of the third side. Hyp. In A abc, the bisector t divides c into the segments, p and q. To prove ab...
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Plane Geometry

Arthur Schultze - 1901 - 260 pages
...20, and 18, respectively. Find the length of the median to 18. PROPOSITION XL. THEOREM 328. In any triangle, the product of two sides is equal to the...the included angle plus the product of the segments of the third side. Hyp. In A abc, the bisector t divides c into the segments, p and q. To prove ab...
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Plane and Solid Geometry

Arthur Schultze, Frank Louis Sevenoak - Geometry - 1901 - 394 pages
...20, and 18, respectively. Find the length of the median to 18. PROPOSITION XL. THEOREM 328. In any triangle, the product of two sides is equal to the...the included angle plus the product of the segments of the third side. Hyp. In A abc, the bisector t divides c into the segments, p and q. To prove ab...
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Original Exercises in Plane and Solid Geometry

Levi Leonard Conant - Geometry - 1905 - 140 pages
...joining their homologous vertices are concurrent. What happens when the triangles are equal ? 510. In any triangle the product of two sides is equal to the...the segments into which it divides the third side. HINT. Produce the bisector to meet the circumscribed circle, and join the point where it meets the...
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Plane Geometry

Edward Rutledge Robbins - Geometry, Plane - 1906 - 268 pages
...(?) (250). .-. these A are similar (?). .-. e: d=h:a(?) (323, 3). QED 1(;2 163 338. THEOREM. In any triangle the product of two sides is equal to the square of the bisector of their included angle, plus the product of the segments of the third side formed by the bisector. Given...
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Plane and Solid Geometry

Isaac Newton Failor - Geometry - 1906 - 440 pages
...If PA = 3, AB = 9, find PT. 669 If PT = 8, PB = 16, find PA. PROPOSITION XXXIII. THEOREM 381 In any triangle, the product of two sides is equal to the square of the bisector of their included angle, increased by the product of the segments of the third side. HYPOTHESIS. In the...
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Plane and Solid Geometry

Isaac Newton Failor - Geometry - 1906 - 431 pages
...If PA = 3, AB = 9, find PT. 669 If PT = 8, PB = 16, find PA. PROPOSITION XXXIII. THEOREM 381 In any triangle, the product of two sides is equal to the square of the bisector of their included angle, increased by the product of the segments of the third side. HYPOTHESIS. In the...
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