CONSTRUCTION. With center A and radius AB, describe the circle BCD. With center B and radius BA, describe O ACF (using for "circle "). (102. A circle may be described with any center and radius.) Join a point C, at which the circles cut one another, to the points A and B. (100. A line may be drawn from any one point to any other point.) Then will ABC be an equilateral triangle. PROOF. Because A is the center of O BCD, .. AB = AC, being radii of the same circle ; and because B is the center of the O ACF, BA = BC, being radii of the same circle. Therefore AC = AB BC, (87. Things equal to the same thing are equal to one another.) and an equilateral triangle has been described on AB. PROBLEM II. 133. On a given line, to mark off a sect equal to a given sect. GIVEN, AB, a line; a, a sect. REQUIRED, to mark off a sect on AB equal to a. CONSTRUCTION. From any point O as a center, on AB, describe the arc of a circle with a radius equal to a. If D be the point in which the arc intersects AB, then OD will be the required sect. PROOF. All the radii of the circle around O are, by construction, equal to a. OD is one of these radii, therefore it is equal to a. CONSTRUCTION. From O as a center, with any radius, OA, describe the arc of a circle, cutting the arms of the angle in the points A and B. Join AB. On AB, on the side remote from O, describe, by 132, an equilateral triangle, ABC. Join OC. The line OC will bisect the given angle AOD. PROOF. In the triangles CAO and CBO we have OA = OB, CA = CB by construction, and the side CO common. ▲ CAO A CBO, (129. Triangles with the three sides respectively equal are congruent.) EXERCISES. I. Describe an isosceles triangle having each of the equal sides double the base. 2. Having given the hypothenuse and one of the sides of a right-angled triangle, construct the triangle. PROBLEM IV. 135. Through a given point on a given line, to draw a perpendicular to this line. GIVEN, the line AB, and the point C in it. EXERCISES. 5. Solve 134 without an equilateral triangle. 6. Divide a given angle into four equal parts. 3. Find a point in a line at a given distance from a given. point. When is the problem impossible? 4. Find a point in a line equally distant from two given points without the line. 5. From a given point without a given line, draw a linemaking with the given line an angle equal to one-half a right angle. 6. A is a point without a given line, BC. Construct an isosceles triangle, having A as vertex, whose base shall lie along BC, and be equal to a given sect. II. If a right-angle triangle have one acute angle double the other, the hypothenuse is double one side. 132. GIVEN, the sect AB. REQUIRED, to bisect it. CONSTRUCTION. On AB describe an equilateral triangle ABC, by By 134, bisect the angle ACB by the line meeting AB in D. Then AB shall be bisected in D. PROOF. In the triangles ACD and BCD BD, and AC = BC by construction, and CD is common. AD = .. (124. Triangles are congruent if two sides and the included angle are equal in each.) 137. COROLLARY I. The line drawn to bisect the angle at the vertex of an isosceles triangle, also bisects the base, and is perpendicular to it. 138. COROLLARY II. The line drawn from the vertex of an isosceles triangle to bisect the base, is perpendicular to it, and also bisects the vertical angle. EXERCISES. 12. Construct a right-angled isosceles triangle on a given sect as hypothenuse. PROBLEM VI. 139. From a point without a given line, to drop a perpendicu lar upon the line. GIVEN, the line AB, and the point C without it. REQUIRED, to drop from C a perpendicular upon AB. CONSTRUCTION. Take any point, D, on the other side of AB from C, and, by 102, from the center C, with radius CD, describe the arc FDG, meeting AB at F and G. By 136, bisect FG at H. Join CH. CH shall be 1 AB (using for the words "perpendicular to "). PROOF. Because in the triangles CHF and CHG, by construction, HG, and CH is common, CF CG, and HF = (129. Triangles with three sides respectively equal are congruent.) FHC, being half the st. FHG, is a rt. ; and CH is perpendicular to AB. 140. COROLLARY. The line drawn from the vertex of an isosceles triangle perpendicular to the base, bisects it, and also bisects the vertical angle. EXERCISES. 13. Instead of bisecting FG, would it do to bisect the angle FCG? |