Trusses are networks of triangles. They are used in the construction of roofs and bridges. Since a triangle is a rigid figure, a network of two or more triangles is also rigid. 124. Bisecting an angle. Draw ABC (Fig. 233) to represent any given angle. With B as center and a convenient radius, draw arcs cutting AB and CB in points D and E. With D and E as centers and a radius longer than one half the distance DE, draw arcs intersecting at F. Draw BF. Line BF divides angle ABC into two equal parts. To test FIG. 233 the accuracy of your construction, measure 4ABF and CBF with the protractor. It is easier to prove the equality of these angles than to show it by measuring. For AFDB and FEB have the corresponding sides equal by construction. They are therefore congruent (§ 123). It follows that the corresponding angles FBD and FBE are equal. When a line drawn through the vertex of an angle divides the angle into two equal parts, it bisects the angle. The dividing line is the bisector of the angle. 125. Constructing an angle of 60°. We have seen that the angles of an equilateral triangle are equal to each other, and each equal to 60°. Hence, to construct 126. Construct a regular hexagon. Divide a circle (Fig. 242) into six equal arcs (Exercise 2, §125). Join the points of division, forming the hexagon ABCDEF. This is the required hexagon. The correctness of the construction may be proved as follows: Proof: Draw radii OA, OB, OC, Show that the six triangles formed are congruent and equilateral. It follows that the sides AB, BC, CD, etc., are equal to each other. Prove that FAB=120°. E FIG. 242 Similarly, 4ABC, BCD, CDE, DEF, and EF A are each 120°. Therefore ABC= LBCD= 4 CDE etc. Why? It has been shown that (1) the hexagon ABCDEF is equilateral and (2) equiangular. A polygon which is equilateral and equiangular is called a regular polygon. Hence ABCDEF is a regular hexagon. A polygon all of whose vertices lie on a circle is an inscribed polygon. EXERCISES 1. Draw an equilateral inscribed triangle. Suggestion: Divide a circle into 6 equal parts and draw 3 line segments joining alternate points of division. 2. Construct an angle of 120 degrees. Suggestion: Draw two 60-degree angles on opposite sides of the same segment. 3. Draw a regular 12-sided polygon (dodecagon). Suggestion: Draw a regular inscribed hexagon. Bisect the arcs by drawing the bisectors of the angles at the center. 127. Bisect a line segment. Let AB (Fig. 243) be the segment to be bisected. Using A as center and a convenient radius, draw arcs above and below AB. With the same radius and center at B, draw arcs cutting the first two arcs at C and E FIG. 243 D. Draw CD intersecting AB at E. i.e., AE= EB. We may prove that AE= EB as follows: ACAD≥^CBD, because the sides of one triangle are respectively equal to the sides of the other. It follows that x=y. Why? Then AACE ABCE, because two sides and the included angle of one are equal to the corresponding parts of the other. ..AE= EB Furthermore, CD1AB, because the two adjacent angles at E are equal. Since CD is perpendicular to AB and bisects AB, it is the perpendicular bisector of AB. EXERCISES Construct the following figures, using only straightedge and compass: 1. Draw an equilateral triangle (Fig. 244). Bisect the sides and complete the figure as shown. 2. Draw a triangle and bisect each side. Join each of the points of bisection to the opposite vertex. 3. Draw Figs. 245 to 247. B FIG. 244 FIG. 245 FIG. 246 FIG. 247 4. Draw a triangle ABC. Construct the perpendicular bisectors of the sides. If your construction is made carefully, the three bisectors will meet in a point. Using the point as a center, draw a circle passing through the vertices of AABC. 5. Make the construction of Exercise 4 for a triangle having an obtuse angle; for a triangle having a right angle. |