47. What is the magnitude of an angle of a regular octagon?
48. Through two given points draw two straight lines forming with a straight line given in position an equilateral triangle.
49. If the straight lines bisecting the angles at the base of an isosceles triangle be produced to meet, they will contain an angle equal to an exterior angle of the triangle.
50. A is the vertex of an isosceles triangle ABC, and BA is produced to D, so that AD is equal to BA; and DC is drawn: shew that BCD is a right angle.
51. ABC is a triangle, and the exterior angles at B and C are bisected by the straight lines BD, CD respectively, meeting at D: shew that the angle BDC together with half the angle BAC make up a right angle.
52. Shew that any angle of a triangle is obtuse, right, or acute, according as it is greater than, equal to, or less than the other two angles of the triangle taken together.
53. Construct an isosceles triangle having the vertical angle four times each of the angles at the base.
54. In the triangle ABC the side BC is bisected at E and AB at G; AE is produced to F so that EF is equal to AE, and CG is produced to H so that GH is equal to CG: shew that FB and HB are in one straight line.
55. Construct an isosceles triangle which shall have one-third of each angle at the base equal to half the vertical angle.
56. AB, AC are two straight lines given in position: it is required to find in them two points P and Q, such that, PQ being joined, AP and PQ may together bo equal to a given straight line, and may contain an angle equal to a given angle.
57. Straight lines are drawn through the extremities of the base of an isosceles triangle, making angles with it on the side remote from the vertex, each equal to one-third of one of the equal angles of the triangle and meeting the sides produced: shew that three of the triangles thus formed are isosceles.
58. AEB, CED are two straight lines intersecting at E; straight lines AC, DB are drawn forming two triangles ACE, BED; the angles ACE, DBF are bisected by the straight lines CF, BF, meeting at F. Shew that the angle CFB is equal to half the sum of the angles EAC, EDB.
