 | William Alexander Willock - Circle - 1875 - 196 pages
...centres of the four circles which may be described touching three intersecting directives. 17. The three perpendiculars from the vertices of a triangle on the opposite sides, meet in a point. 18. The point of meeting of these perpendiculars is the centre of the circle inscribed in the triangle... | |
 | 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
...other three. *i68. The rectangles contained by the segments into which the orthocentre divides each of the perpendiculars from the vertices of a triangle on the opposite sides are equal to one another. *i69. If three straight lines drawn from the vertices of a triangle to the... | |
 | Mathematics - 1889 - 200 pages
...drawn through a triangle. What chance has each side of escaping uncut 'i 122 9760. (GG Storr, MA)— If the perpendiculars from the vertices of a triangle on the opposite sides meet the circum-circle in А', В', С', prove that a. AA' + b . BB' + c.CC' = 8Д 116 9766. (Rev. T. Roach,... | |
 | 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, and 0 the... | |
 | Edinburgh Mathematical Society - Mathematics - 1894 - 284 pages
...internal bisector of the angles of a triangle are concurrent. A direct proof may also be given. § 22. The perpendiculars from the vertices of a triangle on the opposite sides are concurrent. (Fig. 7). Let ABC be the triangle ; AD, CF perpendicular to BC, AB. Draw AB', CB' perpendicular... | |
 | 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 in a straight line.... | |
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The perpendiculars from the vertices of a triangle on the opposite sides are concurrent. 
