COROLLARY 2. In any triangle, a mid-line between two sides and themedian to the third side bisect each other.1.27. In a right triangle, the median from the right vertex to the hypotenuseis equal to half the hypotenuse.1.28. If one angle of a right triangle is 30°, the side opposite this angle isequal to half the hypotenuse.COROLLARY. Conversely, f one side of a right triangle is half the hypote-nuse, the angle opposite to it is 30°.1.29. The median of a trapezoid is parallel to the parallel bases and is equalto half their sum.COROLLARY. The line joining the mid-points of the diagonals of a trape-zoid is parallel to the parallel bases and is equal to half their difference.

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

TRIANGLES AND POLYGONS5= /FDA. But /FDA = /DAC (since, DF 11 AC).LEAD= LDAC.1.2. ABC is a triangle. D and E are any two points on AB and AC. Thebisectors of the angles ABE and ACD meet in F. Show that LBDC +/BEC = 2 /BFC.CONSTRUCTION: Join AF and produce it to meet BC in G (Fig. 2).FIGURE 2Proof: LBDC is exterior to AADC..'. /BDC = LA + /ACD(Th. 1.8). Also, /BEC is exterior to AAEB.LBEC = LA+ /ABE; hence /BDC + /BEC = 2 /A + LA CD + LABE (1).Similarly, /s BFG, CFG are exterior to As AFB, AFC. .'. /BFG+ /CFG = /BFC = LA + } LABE + 4 LACD(2).