PROPOSITION IV. THEOREM. 385. A regular polygon being inscribed in a circle, a similar polygon may be circumscribed about the circle. To Prove: A similar polygon can be circumscribed about ACE. Through each vertex A, B, C, etc., draw tangents NG, GH, (being formed by tangents and chords of equal arcs.) Since each chord is less than a diameter, each of these equal is less than a rt. Z; (267) .. the tangents through adjacent vertices will meet. (114) Let them meet in G, H, K, L, etc. Since isos. A GAB isos. ▲ HBC = = isos. A KCD, etc., (63) N is equilateral, with the same no. of sides as P; N is a regular polygon similar to P. Q.E.D. (370) 386. COR. If a circumference be divided into n equal arcs (n being >2), the tangents drawn at the points of division will form a regular circumscribed polygon of n sides. PROPOSITION V. THEOREM. 387. Any regular polygon is equivalent to the rectangle of its apothem and half its perimeter. A B A Given: A straight line AB, and another line AHG To Prove: P is equivalent to rectangle n AB · OR. Draw AO, BO. Since OAB≈rect. AB · OR, (331) and PnA OAB, (371) Q.E.D. Pn rect. AB · OR≈rect. 1⁄2n AB · OR. SCHOLIUM. The theorem may also be stated thus: The area of a regular polygon is measured by one half the product of its apothem and perimeter. EXERCISE 557. In the diagram for Prop. IV., if GR be the perpendicular from G to AB, the number of sides being n, show that GR2 is 4n times the difference of the sums of the squares of the sides of the circumscribed and inscribed polygons. 558. In the same diagram, n being the number of sides, each angle formed like GAB by an interior and an exterior side, is 1 of a straight angle. n 559. In the diagram for Prop. V., if a parallel to AB be drawn through the mid point of BC, what lines will it bisect ? 560. In the same diagram, if the mid points of the radii be joined, what figure will be formed, and what ratio will its area have to that of P? PROPOSITION VI. THEOREM. 388. As the number of sides of a regular inscribed polygon indefinitely increases, the apothem increases towards the radius as its limit. Given: AB, a side, and OP, the apothem, of a regular polygon of n sides inscribed in a circle whose radius is OA; To Prove: As n increases, OP increases towards OA as limit. Since OP is to AB, OP2 ≈ 0A2 — AP2. (377) (347) Now as n increases, AB, and therefore AP, decreases (181), and when ʼn becomes indefinitely great, AP becomes indefinitely small, while 04, the radius, is constant; .. OP2 has for limit 0Ã2; .. OP has for limit 04. (235) Q.E.D. EXERCISE 561. In the diagram for Prop. VI., if AO be produced, show that it will pass through a vertex of the polygon. 562. In any polygon of an even number of sides, the lines joining opposite vertices are diameters of the circumscribed circle. 563. In the diagram for Prop. VI., if the inscribed polygon is a regular hexagon, what is the ratio of OP to AP? 564. In the same diagram, if OB be joined, and PC, PD, be drawn to the mid points of OA, OB, resp., OCPD will be a rhombus, unless ▲ AOB is a right angle. In what case will ZAOB be a right angle? 565. The area of the regular inscribed hexagon is half the area of the circumscribed equilateral triangle. 389. A straight line is the least of all the lines that terminate in two given points. Given: A straight line AB, and another line AHG nating in A, B; To Prove: AB is less than AHG ... B. D B Join alternate points AG, GE, EC. Since AG AH + HG, GE < GF + FE, EC<ED+ DC, (88) AGECBAHGFEDCB. (Ax. 4) By continuing the process, as the number of parts is always decreasing, we shall evidently obtain at last a broken line of two parts, which is less than any of the preceding broken lines, and greater than AB. Hence AB is less than any broken line terminating in A, B. As this reasoning holds true, no matter how numerous or how small the parts of the broken line may be, it holds true in regard to curves also, which are the limits towards which tend the broken lines formed by joining points of the curve taken indefinitely near to each other. Hence AB is less than any other line terminating in 4 and B. Q.E.D. 390. COR. 1. An arc ACB of a circle is less than any enveloping line ADB that terminates in the same points, A and B. about a circle is Join AB. Of all the lines enveloping the area ACB, there must be a least one. Now on are joined by ADB cannot be the least, for drawing EF r hexagon. tangent to ACB at C, we have AEFB<ADB, ? since EFEDF. In the same way it can be shown that no other line than ACB can be the least line enveloping the area ACB. 391. COR. 2. A circumference is greater than the perimeter of any inscribed polygon, and less than that of any circumscribed polygon. PROPOSITION VIII. THEOREM. 392. As the number of their sides indefinitely increases, the perimeters of regular inscribed and circumscribed polygons tend towards the circumference as their common limit, and their areas towards the circle as their common limit. Given: The perimeters p, p', of two regular polygons P, P', of n sides, circumscribed about, and inscribed in, a circle whose circumference is C and whose area is S; 1°. and p' each tend towards C as limit. To Prove: {10. p tend as limit. Let R be the apothem of P and radius of P', and r the Now when the number of sides becomes indefinitely regat approaches its limit R (388), so that Rr becomes 564. In the samill. Hence regular a R - 1° and its equal p-p' R p become to the mid points Call; that is, p-p', the difference of the perimZAOB is a right olygons, becomes less than any assignable 565. The area (. ing not much greater than C. Now as C the circumscribed' |