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The perpendiculars from the vertices of a triangle on the opposite sides are concurrent.
Proceedings of the Edinburgh Mathematical Society - Página 13
por Edinburgh Mathematical Society - 1894
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Geometry for schools, comprising books i. and ii. of Euclid, with some ...

Euclides - 1883 - 96 páginas
...trisection of each of them. 217. To construct a triangle having given the three medians. 218. The three perpendiculars from the vertices of a triangle on the opposite sides are concurrent. 219. The lines perpendicular to the sides of a triangle through their middle points are concurrent....
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The Elements of Euclid, books i. to vi., with deductions, appendices and ...

Euclides - 1885
...position between AD and AF ; AE lies in magnitude between AD and AF. I. 19, Cor. 20. Since the sum of the perpendiculars from the vertices of a triangle on the opposite sides is greater than the semiperimeter, by the fifteenth deduction ; and since, by the preceding deduction,...
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Mathematical Questions with Their Solutions: From the ..., Volúmenes40-42

1884
...(Professor Malet, FR8.)— If A = I—**2, prove that * loc1 A -^7332. (The Editor.) — If plt p,, p, be the perpendiculars from the vertices of a triangle on the opposite sides ; rf,, dj, rfa the distances from the vertices to the points of contact of the escribed circles with...
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The Elements of Plane Geometry:pPart I(corresponding to Euclid Books I.-II ...

...from B or C on the opposite side of the triangle meet AD at O ; shew that OD is equal to DP. *i66. The perpendiculars from the vertices of a triangle on the opposite sides meet in a point (called the orthocentre of the triangle). *i67. If O is the orthocentre of the triangle...
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The Directional Calculus: Based Upon the Methods of Hermann Grassmann

Edward Wyllys Hyde - 1890 - 247 páginas
...the mean point of eu e2, e3, and trisects the distance from el to p^ (2) To find the common point of the perpendiculars from the vertices of a triangle on the opposite sides. Let l, m, n be the ratios of the sides of the triangle to the cosines of the opposite angles ; ie T>f*...
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Elementary Trigonometry

John Maximilian Dyer - 1891 - 266 páginas
...= -. (3) If л/o2 — x¡i cos ß + o sin a = x sin /3, find a. (4) If Z, m, и, be the lengths of the perpendiculars from the vertices of a triangle on the opposite sides, prove that the area of the triangle is - Z' »»' и* (cosec A cosec В cosec C)*. ¿ * (5) An object...
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An Elementary Treatise on Modern Pure Geometry

Robert Lachlan - 1893 - 288 páginas
...PQR divides the line joining the median points of the triangles ABC, A'B'C' in the ratio m : n. 115. The perpendiculars from the vertices of a triangle on the opposite sides meet in a point (§ 9G, Ex. 2), which is called the orthocentre of the triangle. If ABC be the triangle,...
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Analytic Geometry

Wallace Alvin Wilson - 1915
...10. Show that the perpendicular bisectors of the sides of a triangle meet in a point. 11. Show that the perpendiculars from the vertices of a triangle on the opposite sides meet in a point. 12. Show that the three points of intersection found in Problems 9, 10, and 11 lie...
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Proceedings of the Edinburgh Mathematical Society, Volumen2

...+ CAB + ABC ; 2 (D AE + ED A) = ACB + D AC + D AE ; 2(DAE + DAC) = ACB + DAC + DAE ; .•.EAO=ECA. .•.BAC = ABC + BCA. § 20. The perpendicular bisectors...perpendicular to BC, AB. Draw AB', CB' perpendicular to AD, CF. Make AC'=AB'; CA' = CB'; and join C'A'; bisect C'A' in E. Then, since A, C, E, are the middle points...
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A Course of Pure Geometry

...a conic. Reciprocate with respect to a circle the theorems contained in Exx. 5 — 12 inclusive. 5. The perpendiculars from the vertices of a triangle on the opposite sides are concurrent. 6. The tangent to a circle is perpendicular to the radius through the point of contact. 7. Angles in...
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