4PROBLEMS AND SOLUTIONS IN EUCLIDEAN GEOMETRY1.30. In an isosceles trapezoid, the base angles and the diagonals are equal toone another.INTRODUCTION TO CONCURRENCY1.31. The perpendicular bisectors of the sides of a triangle are concurrentin a point equidistant from the vertices of the triangle which is the center of thecircumscribed circle and called the circumcenter of the triangle.1.32. The bisectors of the angles of a triangle are concurrent in a pointequidistant from the sides of the triangle which is the center of the inscribedcircle and called the incenter of the triangle.
bisectors of the other, two exterior angles are concurrent in a point outside thetriangle which is equidistant from the sides (or produced) of the triangle andcalled an excenter of the triangle.
triangle: one inside the triangle, which is the incenter, and three outside it,which are the excenters.1.33. The altitudes of a triangle are concurrent in a point called the ortho-
1.34. The medians of a triangle are concurrent in a point I the distance fromeach vertex to the mid point of the opposite side. This point is called thecentroid of the triangle.
Solved Problems1.1. ABC is a triangle having BC = 2 AB. Bisect BC in D and BD in E.Prove that AD bisects LCAE.CONSTRUCTION: Draw DF 11 AC to meet AB in F (Fig. 1.)FIGURE IProof: '.' D is the mid-point of BC and DF 11 AC, .'. F is the mid-point of AB (Th. 1.26). Also, AB = BD = ,ABC..'. BF = BE.As ABE and DBF are congruent..'. LEAF = LEDF, but /BAD= /BDA (since BA = BD, Th. 1.12)..'. Subtraction gives LEAD