339. 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. Let AB be any side of M. Construct m, the side of a square M, and n, the side of a square N. (§ 337) Construct A'B', a fourth proportional to m, n, and AB. Upon side A'B', homologous to AB, construct polygon P similar to M. (§ 295) 61. To construct a triangle equivalent to a given square, having given its base and the median drawn from the vertex to the base. (Draw a || to the base at a distance equal to the altitude of the A.) What restriction is there on the values of the given lines? 62. To construct a rhombus equivalent to a given parallelogram, having one of its diagonals coincident with a diagonal of the parallelogram, (Ex. 60.) 63. To draw through a given point within a parallelogram a straight line dividing it into two equivalent parts. (Ex. 49, p. 178.) 64. To construct a parallelogram equivalent to a given trapezoid, having a side and two adjacent angles coincident with one of the nonparallel sides and the adjacent angles, respectively, of the trapezoid. (Ex. 23, p. 176.) 65. To construct a triangle equivalent to a given triangle, having given two of its sides. (Ex. 57.) (Let m and n be the given sides, and take m as the base.) Discuss the solution when the altitude is < n. = n. > n. 66. To construct a right triangle equivalent to a given square, having given its hypotenuse. (Ex. 96, p. 119.) (Find the altitude as in Ex. 56.) What restriction is there on the values of the given parts? 67. To construct a right triangle equivalent to a given triangle, having given its hypotenuse. (Ex. 61.) What restriction is there on the values of the given parts? 68. To construct an isosceles triangle equivalent to a given triangle, having given one of its equal sides equal to m. (Draw a || to the given side at a distance equal to the altitude.) Discuss the solution when the altitude is <m. 69. To draw a line parallel to the base of a triangle dividing it into two equivalent parts. (§ 319.) = m. > m. B (▲ ABC and AB'C' are similar.) B C 70. To draw through a given point in a side of a parallelogram a straight line dividing it into two equivalent parts. 71. To draw a straight line perpendicular to the bases of a trapezoid, dividing the trapezoid into two equivalent parts. (A str. line connecting the middle points of the bases divides the trapezoid into two equivalent parts.) 72. To draw through a given point in one of the bases of a trapezoid a straight line dividing the trapezoid into two equivalent parts. (A str. line connecting the middle points of the bases divides the trapezoid into two equivalent parts.) 73. To construct a triangle similar to two given similar triangles, and equivalent to their sum. (Construct squares equivalent to the ▲.) 74. To construct a triangle similar to two given similar triangles, and equivalent to their difference. BOOK V. REGULAR POLYGONS. - MEASUREMENT OF THE CIRCLE. 340. Def. A regular polygon is a polygon which is both equilateral and equiangular. 341. A circle can be circumscribed about, or inscribed in, any regular polygon. D Given regular polygon ABCDE. To Prove that a O can be circumscribed about, or inscribed in, ABCDE. Proof. Let O be the centre of the circumference described through vertices A, B, and C (§ 223). Draw radii OA, OB, OC, and OD. In ▲ OAB and OCD, OB: = OC. And since, by def., polygon ABCDE is equilateral, (?) AB = CD. Again, since, by def., polygon ABCDE is equiangular, ZABC BCD. = And since A OBC is isosceles, ▲ OBC= ≤ OCB. (?) Or, .. ZABC – Z OBC = ≤ BCD – ≤ OCB. Then, the circumference which passes through A, B, and C also passes through D. In like manner, it may be proved that the circumference which passes through B, C, and D also passes through E. Hence, a can be circumscribed about ABCDE. Again, since AB, BC, CD, etc., are equal chords of the circumscribed O, they are equally distant from O. (§ 164) Hence, a O described with O as a centre, and with a line OF to any side AB as a radius, will be inscribed in ABCDE. 342. Def. 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; as OA. The apothem is the radius of the inscribed circle; as OF. 343. Cor. From the equal ▲ OAB, OBC, etc., we have LAOB = L BOC = ≤ COD, etc. (?) Then, each of these is equal to four rt. divided by the number of sides of the polygon. ($ 35) That is, the angle at the centre of a regular polygon is equal to four right angles divided by the number of sides. EXERCISES. Find the angle, and the angle at the centre, 1. Of a regular pentagon. 2. Of a regular dodecagon. 3. Of a regular polygon of 32 sides. 4. Of a regular polygon of 25 sides. PROP. II. THEOREM. 344. If the circumference of a circle be divided into any number of equal arcs, I. Their chords form a regular inscribed polygon. II. Tangents at the points of division form a regular circumscribed polygon. Given circumference ACD divided into five equal arcs, AB, BC, CD, etc., and chords AB, BC, etc. Also, lines LF, FG, etc., tangent to O ACD at A, B, etc., respectively, forming polygon FGHKL. To Prove polygons ABCDE and FGHKL regular. Proof. Chord AB chord BC chord CD, etc. (§ 158) = Again, arc BCDE =arc CDEA = = arc DEAB, etc., for each is the sum of three of the equal arcs AB, BC, etc. Therefore, polygon ABCDE is regular. Again, in A ABF, BCG, CDH, etc., we have (§ 193) (§ 340) |