Problem IV. In a given circle to construct a Triangle.
Fig. 4
Take the radius of the circle, and with it mark off six points on the circumference. Take two of these lengths of the radius and join their extreme points A and B, which will be the base. Now take this base as a radius and describe alternately two arcs cutting each other at C. Join A, C, and B, C, and a triangle is formed, whose sides being equal is termed an equilateral triangle.
In order to ensure its being upright, erect a perpendicular at the centre, and let the two sides A, C, and B, C, meet that perpendicular where it intersects the circumferences. Or, begin the triangle at this point, and mark off two lengths of the radius, joining the extreme points as before; and do this at each side of the perpendicular; finally connecting the distant extremities of the two sides for a base.
Problem V. To construct an upright square in a given circle.
Let fall a perpendicular, I, E, from the centre to the circumference, and with that as a radius and E as a centre, cut the circumference at A, B, C, and D, and join the points. The four-sided figure called a square is thus formed.
Fig. 5
