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ABCD altitude axis base becomes bisected chord circle circumference circumscribed common cone constructed contained cylinder described determine diagonal diameter difference direction distance divided draw drawn edges equal equation equiangular equivalent expressed extremities faces fall feet figure follows formed four frustum give given greater half Hence the theorem homologous hypotenuse included inscribed intersect joining length less magnitudes mean measured meet multiplied number of sides opposite parallel parallelogram parallelopipedon pass perpendicular placed plane pole polygon portion PROBLEM produced proportion PROPOSITION prove pyramid radius ratio rectangle regular represent respectively right angles segment sides similar solid sphere spherical square straight line suppose surface taken third triangle triangular prisms unit vertex vertices volume whole
Page 56 - If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.
Page 28 - Therefore all the interior angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Page 253 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.
Page 121 - In a given circle to inscribe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle ; it is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF.
Page 111 - From a given point, to draw a line parallel to a given line. Let A be the given point, and BC the given line.
Page 172 - A cylinder is conceived to be generated by the revolution of a rectangle about one of its sides as an axis.
Page 94 - The angle in a semicircle is a right angle ; the angle in a segment greater than a semicircle is less than a right angle ; and the angle in a segment less than a semicircle is greater than a right angle.
Page 75 - FGL ; (vi. 6.) and therefore similar to it ; (vi. 4.) wherefore the angle ABE is equal to the angle FGL: and, because the polygons are similar, the whole angle ABC is equal to the whole angle FGH ; (vi.