PROPOSITION V. THEOREM 446. Of all polygons constructed with the same given sides, that which can be inscribed in a circle is the тахітит. Hyp. Polygon ABCDE, which is inscribed in a circle, is mutually equilateral with polygon A'B'C'D'E', which cannot be inscribed in a circle. To prove area ABCDE > area A'B'C'D'E'. Proof. From A draw diameter AF, and join F to the two nearest vertices C and D. On C'D', construct A C'D'F' equal to A CDF, and join A'F'. Then area ABCDE > area A'B'C'D'E'. (Ax. 5.) Q.E.D. PROPOSITION VI. THEOREM 447. Of all isoperimetric polygons of the same number of sides, the equilateral is the maximum. Hyp. ABCDE is the maximum of all isoperimetric polygons of the same number of sides. To prove AB= BC= CD= DE= EA. Proof. The polygon must be equilateral, for suppose any two sides AB and BC were not equal. Then construct on the diagonal AC the isosceles triangle AB'C, isoperimetric with ABC. Area AB'C would be greater than area ABC. (443) :: polygon ABCDE would be greater than the isoperimetric polygon ABCDE, which would contradict the hypothesis. Hence AB= BC. Q.E.D. 448. Cor. Of all isoperimetric polygons of the same number of sides, the maximum is regular. Ex. 1018. In a given segment to inscribe an angle so that the sum of its arms is a maximum. Ex. 1019. In a given semicircle to inscribe a trapezoid whose area is a maximum. PROPOSITION VII. THEOREM 449. Of two isoperimetric regular polygons, that which has the greater number of sides has the greater Hyp. Regular pentagon ABCDE is isoperimetric with reg. ular hexagon H. To prove area H > area ABCDE. Proof. Let F be any point in CD. ABCFDE may be considered a hexagon, having one of its angles equal to a straight angle. ... area H > area ABCDE. (Why?) Q.E.D. 450. COR. The area of a circle is greater than the area of any polygon whose perimeter equals the circumference of the circle. Ex. 1020. In a given circle to inscribe a triangle, having a maximum perimeter. Ex. 1021. The circumscribed regular polygon has a smaller area than any other polygon circumscribed about the same circle. Ex. 1022. Of all equivalent parallelograms on the same base, which has the minimum perimeter ? PROPOSITION VIII. THEOREM 451. Of two equivalent regular polygons, that which has the greater number of sides has the smaller perimeter. To prove Hyp. Square A = regular pentagon B. perimeter of A > perimeter of B. Proof. Construct square C isoperimetric with B. Area C < area B, area C < area A. (449) or Q.E.D. 452. Cor. The circumference of a circle is less than the perimeter of any equivalent polygon. MISCELLANEOUS EXERCISES Ex. 1023. If two equal lines are divided externally so that the product of the segments of one is equal to the product of the segments of the other, the segments are equal respectively. *Ex. 1024. Two triangles are equal if the base, the opposite angle, and its bisector of one are respectively equal to the base, the opposite angle, and its bisector of the other. Hint. — Circumscribe circles, produce bisectors until they meet the circumference, and join the point of intersection to one end of the base. (Ex. 733 and Ex. 1023.) Ex. 1025. If two bisectors of a triangle are equal, the corresponding sides are equal, and the triangle is isosceles. (Ex. 1024.) * Ex. 1026. The square of the side of a regular pentagon increased by the square of one of its diagonals is equal to five times the square of the radius. * Ex. 1027. A regular pentagon is equivalent to a rectangle hạving one side equal to 3 times the radius, and the other to ã of a diagonal of the pentagon. * Ex. 1028. The product of the diagonals of an inscribed quadrilateral is equal to the sum of the products of the opposite sides. (Ptolemy's Theorem.) * Ex. 1029. To construct a triangle having given the three feet of the altitudes. *Ex. 1030. If Pn and Pi, are respectively the perimeters of an inscribed and circumscribed regular polygon of n sides, and pen and P2n the perimeters of regular polygons of 2 n sides, respectively inscribed and circumscribed about the same circle, prove that P2n is the harinonical mean between Pn and Pr; i.e. 2 Pn Pn Pn+ Pn * Ex. 1031. Using the notations of Ex. 1030, prove that P2n is the mean proportional between Pn and P2n; i.e. P2n VPnP2n. Ex. 1032. If circles be circumscribed about the four triangles into which a quadrilateral is divided by its diagonals, their centers form the vertices of a parallelogram. P2n |