56. If the sides of any polygon be cut by a transversal, the product of one set of alternate segments is equal to the product of the remaining set.
57. A transversal being drawn cutting the sides of a triangle, the lines from the angles of the triangle to the middle points of the segments of the transversal intercepted by those angles meet the opposite sides in collinear points.
58. If lines be drawn from any point P to the angles of a triangle, the perpendiculars at P to these lines meet the opposite sides of the triangle in three collinear points.
59. Divide a given semicircle into two parts by a perpendicular to the diameter, so that the radii of the circles inscribed in them may have a given ratio.
60. From a point within a triangle perpendiculars are let fall on the sides; find the locus of the point, when the sum of the squares of the lines joining the feet of the perpendiculars is given.
61. If a circle make given intercepts on two fixed lines, the rectangle contained by the perpendiculars from its centre on the bisectors of the angle formed by the lines is given.
62. If the base and the difference of the base angles of a triangle be given, the rectangle contained by the perpendiculars from the vertex on two lines through the middle point of the base, parallel to the internal and external bisectors of the vertical angle, is constant.
63. The rectangle contained by the perpendiculars from the extremities of the base of a triangle, on the internal bisector of the vertical angle, is equal to the rectangle contained by the external bisector and the perpendicular from the middle of the base on the internal bisector.
64. State and prove the corresponding theorem for perpendiculars on the external bisector.
65. If R, R′ denote the radii of the circles inscribed in the triangles into which a right-angled triangle is divided by the perpendicular from the right angle on the hypotenuse; then, if c be the hypotenuse, and s the semiperimeter, R2 + R′2 = (s − c)2.