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" The square of the bisector of an angle of a triangle is equal to the product of the sides of this angle diminished by the product of the segments made by the bisector upon the third side of the triangle. "
Plane Geometry, with Problems and Application - Page 208
by Herbert Ellsworth Slaught - 1918 - 332 pages
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A Text-book of Geometry

George Albert Wentworth - Geometry - 1888 - 272 pages
...the shortest side of the one, : OM, the shortest side of the other. PROPOSITION XXIII. THEOREM. 349". The square of the bisector of an angle of a triangle is equal to the product of the sides of this angle diminished by the product of the segments determined by the bisector upon the third...
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Syllabus of Geometry

George Albert Wentworth - Mathematics - 1896 - 68 pages
...are drawn, the tangent is a mean proportional between the whole secant and the external segment. 349. The square of the bisector of an angle of a triangle is equal to the product of the sides of this angle diminished by the product of the segments determined by the bisector upon the third...
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Plane and Solid Geometry

George Albert Wentworth - Geometry - 1899 - 500 pages
...external segment is constant in whatever direction the secant is drawn. PROPOSITION XXXIV. THEOREM. 383. The square of the bisector of an angle of a triangle is equal to the product of the sides of this angle diminished by the product of the segments made by the bisector upon the third side...
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Plane Geometry

George Albert Wentworth - Geometry, Modern - 1899 - 272 pages
...external segment is constant in whatever direction the secant is drawn. PROPOSITION XXXIV. THEOREM. 383. The square of the bisector of an angle of a triangle is equal to the product of the sides of this angle diminished by the product of the segments made by the bisector upon the third side...
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Plane and Solid Geometry

George Albert Wentworth - Geometry - 1904 - 496 pages
...external segment is constant in whatever direction the secant is drawn. PROPOSITION XXXIV. THEOREM. 383. The square of the bisector of an angle of a triangle is equal to the product of the sides of this angle diminished by the product of the segments made by the bisector upon the third side...
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Plane and Solid Geometry

Fletcher Durell - Geometry - 1911 - 553 pages
...secants and their external segments are reciprocally proportional. PROPOSITION XXXIV. THEOREM 362. The square of the bisector of an angle of a triangle is equal to the product of the sides forming the angle, diminished l)y the product of the segments of the third side formed by the...
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Plane Geometry

Fletcher Durell - Geometry, Plane - 1904 - 382 pages
...secants and their external segments are reciprocally proportional. PROPOSITION XXXIV. THEOREM 362. The square of the bisector of an angle of a triangle is equal to the product of the sides forming the angle, diminished by the product of the segments of the third side formed by the...
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Exercises in Geometry

Grace Lawrence Edgett - Geometry - 1909 - 104 pages
...sides is equal to twice the product of the side by the projection of the median upon that side. 7. The square of the bisector of an angle of a triangle is equal to the product of the sides forming the angle diminished by the product of the segments of the third side formed by the bisector....
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Plane Geometry: With Problems and Applications

Herbert Ellsworth Slaught, Nels Johann Lennes - Geometry, Plane - 1910 - 304 pages
...— = and hence that A'( A ABC AC AB AB- AC AA'B'C' A'C' A'B' A'B'-A'C' 428. THEOREM. The square on the bisector of an angle of a triangle is equal to...opposite side. Outline of proof: Produce the bisector of the given angle to meet the circumscribed circle. Since ABDC~AAEC AC-BC=CD-CE. (1) But CD • CE=...
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Plane Geometry: With Problems and Applications

Herbert Ellsworth Slaught, Nels Johann Lennes - Geometry, Plane - 1910 - 300 pages
...and hence that , , , — , , . , — , , ,-.AA'B'C' A'C* A'B' A'B'-A1C1 428. THEOREM. The square on the bisector of an angle of a triangle is equal to...opposite side. Outline of proof: Produce the bisector of the given angle to meet the circumscribed circle. Since ABDC~AAKC AC • nc = CD . CE. (1) But CD...
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