PROPOSITION V. THEOREM 446. Of all polygons constructed with the same given sides, that which can be inscribed in a circle is the 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 ▲ C'D'F' equal to ▲ CDF, and join A'F'"'. But 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. B E Hyp. ABCDE is the maximum of all isoperimetric polygons of the same number of sides. 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 AB'CDE 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 area. B H E F Hyp. Regular pentagon ABCDE is isoperimetric with regular 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. PROPOSITION VIII. THEOREM 451. Of two equivalent regular polygons, that which has the greater number of sides has the smaller perim eter. A B Hyp. Square A≈ regular pentagon B. To prove perimeter of A> perimeter of B. Proof. Construct square C isoperimetric with B. Area C area B, с (449) or 452. COR. area Carea A. .. perimeter of C< perimeter of A. .. perimeter of B< perimeter of A. Q.E.D. 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 having one side equal to 3 times the radius, and the other to § 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 P, are respectively the perimeters of an inscribed and circumscribed regular polygon of n sides, and pen and P2n the perime-. ters of regular polygons of 2n sides, respectively inscribed and circumscribed about the same circle, prove that P2n is the harmonical mean between pn and P; i.e. = Pan2Pn Pn * Ex. 1031. Using the notations of Ex. 1030, prove that pan is the mean proportional between pn and P2n; i.e. p2n = √PnP2n 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. |