PROP. XVII. PROBLEM 302. To construct a triangle equivalent to a given polygon. Construction. 1. Take three consecutive vertices, A, B, C; draw diagonal AC, and line BF AC, meeting DC prolonged at F, also line AF. 2. ▲ ACF ▲ ABC; then, polygon AFDE ≈ ABCDE (having common part ACDE), and has a number of sides less by 1. 3. Draw diagonal AD, and line EG | AD, meeting CD prolonged at G, also line AG; then, ▲ AED≈ ▲ AGD, whence ▲ AFG polygon AFDE, or to ABCDE. Proof. 4. A ACF and ABC have same base and altitude. Note. By aid of §§ 302 and 298, a square may be constructed equivalent to a given polygon. Ex. 49. The bases of a trapezoid are 8 and 10, respectively, the altitude 6. Construct an equivalent equilateral triangle. [An equilateral triangle can be constructed when its altitude is known.] Ex. 50. The bases of a trapezoid are 6 and 8, respectively; the area is 35. Construct the trapezoid and an equivalent equilateral triangle. Ex. 51. Given A ABC. Draw a line from A to BC which shall divide the triangle into two triangles the ratio of whose areas is 1 to 2. PROP. XVIII. PROBLEM 303. To construct a polygon similar to a given polygon, and having a given ratio to it. Required to construct a polygon similar to AC, and having to it the ratio n m Construction. 1. Let A'B' be side of square having to the square described on AB the required ratio. (§ 301) 2. Polygon A'C', similar to AC, will have required ratio to it. 304. To construct a polygon similar to one of two given polygons, and equivalent to the other. Required to construct a polygon similar to M, and ≈ N. Construction. 1. Find m and n sides of squares M and N. (Note, § 302) 2. Polygon P, similar to M, with A'B' a fourth proportional to m, n, and AB, will be ≈ N. Ex. 52. To inscribe in a given triangle a parallelogram whose area is one-half the area of the triangle. Can a parallelogram be inscribed whose area is still greater? Ex. 53. Through a given point P, either within or without a given angle, to draw a line which shall form with the sides of the angle a triangle of given area. Ex. 54. To construct a triangle whose angles shall be equal, respectively, to the angles of a given triangle, and whose area shall be four times the area of the given triangle. BOOK V REGULAR POLYGONS.-MEASUREMENT OF THE CIRCLE.-LOCI 305. Def. A regular polygon is a polygon which is both equilateral and equiangular. PROP. I. THEOREM 306. A circle can be circumscribed about any regular polygon. Given regular polygon ABCDE. To Prove that a O can be circumscribed about it. Proof. 1. Draw circumference through A, B, C'; also, radii OA, OB, OC, and line OD. 2. In OAB, OCD, OB = OC, AB = CD. 3. LOBA = 4. Since ABC (?) ABC-ZOBC, OCD =/ BCD- OCB. = 5. Then A OAB = 6. Then circumference through A, B, C passes through D; and similarly through E. 307. Since AB, BC, CD, etc., are equal chords of the circumscribed O, they are equally distant from 0. (§ 164) Then, a drawn with O as a centre, and a line OF to any side AB as a radius, will be inscribed in ABCDE. Hence, a circle can be inscribed in any regular polygon. 308. Defs. The centre of a regular polygon is the common centre of the circumscribed and inscribed circles. The angle at the centre is the angle between the radii drawn to the extremities of any side; as AOB. The radius is the radius of the circumscribed circle, 04. 309. From equal ▲ OAB, OBC, etc., we have Whence, the angle at the centre of a regular polygon is equal to four right angles divided by the number of sides. PROP. II. THEOREM 310. If a circumference be divided into equal arcs, their chords form a regular inscribed polygon. Given circumference ACD divided into five equal arcs, AB, BC, CD, etc., and chords AB, BC, etc. To Prove polygon ABCDE regular. Proof. 1. Sides of polygon are equal by § 159. 2. Each equal arcs. is measured by one-half the sum of three of the (§ 192) |