 In any obtuse triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides, increased by twice the product of one of these sides and the projection of the other side upon it.  Elements of Geometry: Plane geometry - Page 148by Andrew Wheeler Phillips, Irving Fisher - 1896Full view - About this book
 | Charles Davies - Geometry - 1872 - 464 pages
...x CD. Bat, JB& + Aff = AH\ and CD 1 + A& = AC* : hence, ~ X CD ; PROPOSITION XIII. THEOREM. In any obtuse-angled triangle, the square of the side opposite the obtuse angle is equal to the sum of the sqitares of the base and the other side, increased by twice the rectangle of the base and the distance... | |
 | Adrien Marie Legendre - Geometry - 1874 - 500 pages
...and CD2 + AJ? = hence, X CD ; AB2 = BC2 + which was to be proved. PROPOSITION Xm. THEOREM. N In any obtuse.angled triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of t/>e base and the other side, increased by twice the rect> angle of the base and the distance... | |
 | William Guy Peck - Conic sections - 1876 - 376 pages
...(P. 8), we have, CD 8 = DA 8 +AC 8 - 2AC x AE, which was to be proved. PROPOSITION X, THEOREM. In any obtuse-angled triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides, increased by twice the rectangle of the base and the distance from... | |
 | Richard Wormell - 1876 - 268 pages
...Then the rectangle AG = the square on С D. THEOREM LIV. In any obtuse-angled triangle, tl1e square on the side opposite the obtuse angle is equal to the sum of the squares on the sides containing the obtuse angle, together with twice the rectangle contained by either... | |
 | William Frothingham Bradbury - Geometry - 1877 - 262 pages
...XXIX. 70. In a triangle the square of the side opposite the obtuse angle is equivalent to the sum of the squares of the other two sides plus twice the product of one of these sides and the distance from the vertex of the obtuse angle to the foot of the perpendicular let fall upon this side... | |
 | Elias Loomis - Conic sections - 1877 - 458 pages
...AD falls upon AB,this proposition reduces to the same as Pr. 11, Cor. 1. PROPOSITION XIII. THEOREM. In an obtuse-angled triangle, the square of the side opposite the obtuse angle is greater than the squares of the base and the other side by twice the rectangle contained by the base,... | |
 | Thomas Hunter - Geometry, Plane - 1878 - 142 pages
...(AB-AC)=AB 2 -AC 2 . Algebraically: Let AB=a, and AC=5y then (a-\-b) PROPOSITION XIX. — THEOREM. In an obtuse-angled triangle the square of the side...opposite the obtuse angle is equal to the sum of the squares of the other two sides and twice the rectangle contained by the base and the distance from... | |
 | William Frothingham Bradbury - Geometry - 1880 - 260 pages
...obtuse-angled trianr/le the square of the side opposite the obtuse anyle is equivalent to the sum of the squares of the other two sides plus twice the product of one of these sides and the distance from the vertex of the obtuse angle to the foot of the perpendicular let fall upon this side... | |
 | Franklin Ibach - Geometry - 1882 - 208 pages
...260. In any obtuse-angled triangle, the square on the side opposite the obtuse angle equals the sum of the squares of the other two sides plus twice the product of one of those sides and the projection of the other upon that side. In the A ABC, let c be the obtuse Z., and... | |
 | Richard Pears Wright - 1882 - 136 pages
...these sides and the projection of the other upon it. Rule 2. In obtuse-angled triangles, the square on the side opposite the obtuse angle is equal to the sum of the squares on the sides which contain it increased by twice the product of either of these sides and the... | |
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