PROPOSITION XVII. PROBLEM 432 To construct a polygon similar to a given polygon and having a given ratio to it. DATA. P is a given polygon, m:n a given ratio. REQUIRED. To construct a polygon similar to P, and which shall be to P as m: n. SOLUTION Draw a line b, such that b2 : a2 = m : n. § 431 On the line b, homologous to a, construct the polygon Q similar to P. CONCLUSION. Q is the polygon required. § 390 Q. E. F. 962 A side of a given polygon is 6 in. Find the homologous side of a similar polygon with twice the area. 963 To construct a right triangle equivalent to a given triangle. 964 To bisect a triangle by drawing a line through the vertex. 965 To trisect a triangle by drawing lines through the vertex. PROPOSITION XVIII. PROBLEM 433 To construct a polygon similar to one of two given polygons and equivalent to the other.* DATA. P and Q are two given polygons, a being a side of P. REQUIRED. To construct a polygon similar to P and equivalent to Q. SOLUTION Find m and n, the sides of two squares respectively equiva lent to P and Q. Find b, the fourth proportional to m, n, and a. § 430 $385 On b, homologous to a, construct the polygon S similar to P. EXERCISE 966. On a given base to construct a triangle equivalent to a given triangle. * The solution of this problem is attributed to Pythagoras. MISCELLANEOUS EXERCISES PROBLEMS OF CONSTRUCTION 967 To divide a triangle into two parts in the ratio of 2 to 3. 968 On a given base to construct a right triangle equivalent to a given triangle. 969 On a given base to construct an isosceles triangle equivalent to a given triangle. 970 To construct an equilateral triangle equivalent to a given triangle. 971 To construct a triangle equivalent to the sum of two given triangles. 972 To construct a triangle equivalent to the difference of two given triangles. 973 To find a point within a triangle from which lines drawn to the vertices divide the triangle into three equivalent triangles. 974 To construct a square equivalent to three fifths of a given square. 975 On a given base to construct a parallelogram equivalent to a given triangle. 976 On a given base to construct a parallelogram equivalent to a given parallelogram. A 977 To construct an equilateral triangle equivalent to a given square. 978 To transform a trapezoid into an equivalent isosceles trapezoid. 979 To divide a triangle into two equivalent parts by drawing a line through a given point in a side. [Let P be the point in AB of the ▲ ABC. Draw the median AM. Join PM and draw AD || to PM. Then PD is the line required.] 980 To divide a triangle into two equivalent parts by drawing a line parallel to the base. [On the altitude AH as a diagonal construct a square. Take AE equal to a side of this square. Then DEG II to BC divides the ▲ ABC as required.] 981 To divide a triangle into two equivalent parts by drawing a line perpendicular to the base. P B M с 982 To bisect a quadrilateral by drawing a line through one of its vertices. 983 To bisect a quadrilateral by drawing a line through a given point in one of its sides. BOOK V REGULAR POLYGONS MEASUREMENT OF THE CIRCLE 434 DEFINITION. A regular polygon is a polygon which is equilateral and equiangular. The square and the equilateral triangle are regular polygons. PROPOSITION I. THEOREM 435 An equilateral polygon inscribed in a circle is a regular polygon. B A HYPOTHESIS. ABCDE is an equilateral polygon inscribed in a circle. That is, the polygon ABCDE is equiangular, and therefore regular. § 434 Q. E. D. PROPOSITION II. THEOREM 436 A circle may be circumscribed about, or inscribed in, any regular polygon. B HYPOTHESIS. ABCDE is a regular polygon. CONCLUSION. 1 A circle may be circumscribed about ABCDE. PROOF Describe a circumference through A, B, and C, and from its center O draw the radii OA, OB, OC. That is, the Oce described through A, B, and C passes through D. In the same manner it may be proved that the Oce through B, C, and D also passes through E, and so on for a regular polygon of any number of sides. .. A circle can be circumscribed about ABCDE. § 236 2 A circle may be inscribed in ABCDE. PROOF Draw OHL to AB. Then, since the sides of the inscribed polygon are equi |