 | Euclides - 1883 - 176 pages
...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.... | |
 | Euclides - 1885 - 322 pages
...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,... | |
 | 1884 - 434 pages
...(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... | |
 | Association for the Improvement of Geometrical Teaching - Euclid's Elements - 1888 - 208 pages
...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... | |
 | Edward Wyllys Hyde - Ausdehnungslehre - 1890 - 288 pages
...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*... | |
 | John Maximilian Dyer - Plane trigonometry - 1891 - 306 pages
...= -. (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... | |
 | Robert Lachlan - Geometry - 1893 - 312 pages
...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,... | |
 | Edinburgh Mathematical Society - Mathematics - 1894 - 284 pages
...CAB + ABC ; 2 (D AE + ED A) = ACB + DAC + D AE ; 2(D AE + D AC) = ACB + DAC + D AE ; .•.EAC=ECA. . •.BAC = ABC + BCA. § 20. The perpendicular bisectors...perpendicular to BC, AB. Draw AB', CB' perpendicular to AD, OF. Make AC' = AB' ; C A' = CB' ; and join C'A'; bisect C'A' in E. Then, since A, C, E, are the middle... | |
 | Education - 1915 - 412 pages
...Find the condition that the vectors x — a, b—c may be perpendicular, and hence show that the three perpendiculars from the vertices of a triangle on the opposite sides are concurrent. 18119. (VV SATYANARAYAN.) — The ratio compounded of the ratios in which any two i»ogonal conjugates... | |
 | Wallace Alvin Wilson - Geometry, Analytic - 1915 - 232 pages
...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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The perpendiculars from the vertices of a triangle on the opposite sides are concurrent. 
