 | Joe Garner Estill - 1896 - 214 pages
...and the exterior bisector of the angle B of a triangle ABC meet at D, prove that angle BDC = i A. 7. In any triangle the product of two sides is equal to the diameter of the circumscribed circle multiplied by the perpendicular to the third side from its opposite... | |
 | George Washington Hull - Geometry - 1897 - 408 pages
...into segments whose product is equal to the square of the radius. PROPOSITION XXXIII. THEOREM. 282. In any triangle the product of two sides is equal to the product of the diameter of the circum* scribed circle by tlie altitude upon the third side. Given—CE the altitude of the triangle... | |
 | George Albert Wentworth - Geometry - 1899 - 500 pages
...NP = NO 2 + MO X 0I\ Whence , !f(? = NM X NP - MO X OP. Ax. .,5 QED PROPOSITION XXXV. THEOREM. 384. In any triangle the product of two sides is equal...of the diameter of the circumscribed circle by the altitude upon the third side. Let NMQ be a triangle, NO the altitude, and QNMP the circle circumscribed... | |
 | George Albert Wentworth - Geometry, Modern - 1899 - 272 pages
....'. MN X NP = NO* +MOX OP. Whence NV* = NM X NP - MO X OP. Ax. 3 QBD PROPOSITION XXXV. THBOREM. 384. In any triangle the product of two sides is equal...of the diameter of the circumscribed circle by the altitude upon the third side. Let NMQ be a triangle, NO the altitude, and QNMP the circle circumscribed... | |
 | Charles Austin Hobbs - Geometry, Plane - 1899 - 266 pages
...mean ratio, to construct the line. Proposition 16O. Theorem. 195. In any triangle, the product of any two sides is equal to the product of the diameter of the circumscribed circle and the perpendicular drawn to the third side from the vertex of the opposite angle. Hypothesis. AD... | |
 | Edward Brooks - Geometry, Modern - 1901 - 278 pages
...external segment. For, from the above proportion we have = PC x PD. PROPOSITION XXXIV. — THEOREM. In any triangle the product of two sides is equal...of the diameter of the circumscribed circle by the altitude upon the third side. PLANE GEOMETRY.— BOOK IV. To Prove. — Then we are to prove that ABx... | |
 | Arthur Schultze, Frank Louis Sevenoak - Geometry - 1901 - 394 pages
...7, respectively. Find the length of the bisector corresponding with 7. PROPOSITION XLI. THEOREM 330. In any triangle the product of two sides is equal to the altitude upon the third side, multiplied by the diameter of the circumscribed circle. Hyp. rf is a... | |
 | Arthur Schultze, Frank Louis Sevenoak - Geometry - 1902 - 394 pages
...7, respectively. Find the length of the bisector corresponding with 7. PKOPOSITION XLI. THEOREM 330. In any triangle the product of two sides is equal to the altitude upon the third side, multiplied by the diameter of the circumscribed circle. or Hyp. d is... | |
 | Arthur Schultze - 1901 - 260 pages
...7, respectively. Find the length of the bisector corresponding with 7. PROPOSITION XLI. THEOREM 330. In any triangle the product of two sides is equal to the altitude upon the third side, multiplied by the diameter of the circumscribed circle. Hyp. d is a diameter... | |
 | Arthur Schultze, Frank Louis Sevenoak - Geometry - 1901 - 396 pages
...7, respectively. Find the length of the bisector corresponding with 7. PROPOSITION XLI. THEOREM 330. In any triangle the product of two sides is equal to the altitude upon the third side, multiplied by the diameter of the circumscribed circle. J7 Hyp. d is... | |
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In any triangle the product of two sides is equal to the product of the diameter... 
