PROPOSITION XVII. PROBLEM. 394. To inscribe in a given circle a regular decagon. Let O be the centre of the given circle. Draw the radius 0 C, and divide it in extreme and mean ratio, so that 0 C shall be to O S as 0 S is to SC. § 311 Draw BC, BS, and B 0. Cons. Cons. Substitute for OS its equal B C, then OC : BC :: BC:SC. Moreover the LOCB=ZSC B, Iden. .. the A OC B and B C S are similar, $ 284 (having an Z of the one equal to an Zof the other, and the including sides proportional). § 160 (its sides 0 C and 0 B being radii of the same circle). .. the A BCS, which is similar to the AOC B, is isosceles, and and $ 114 Cons. .:.08= BS, Ax. 1 .. the A S O B is isosceles, and the 20=2 SBO, § 112 (being opposite equal sides). But the ZC SB=20+ Z SBO, § 105 (the exterior 2 of a A is equal to the sum of the two opposite interior &). ... the Z CSB = 2 2 0. ZSCB=2 C SB) = 2 2 0, $ 112 and ZOBC=2 SC B) = 2 20. § 112 is the sum of the 4s of the A OCB=5 20. .:15 20= 2 rt. 6, § 98 and 20= 1 of 2 rt. £, or 1 of 4 rt. E. .. the arc B C is to of the circumference, and .. the chord B C is a side of a regular inscribed decagon. Hence, to inscribe a regular decagon, divide the radius in extreme and mean ratio, and apply the greater segment ten times as a chord. Q. E. F. 395. COROLLARY 1. By joining the alternate vertices of a regular inscribed decagon, a regular pentagon may be inscribed. 396. Cor. 2. By bisecting the arcs BC, CF, etc., a regular polygon of 20 sides may be inscribed, and, by continuing the process, regular polygons of 40, 80, etc., sides may be inscribed. PROPOSITION XVIII. PROBLEM. 397. To inscribe in a given circle a regular pentedecagon, or polygon of fifteen sides. Let Q be the given circle. It is required to inscribe in Q a regular pentedecagon. Draw E H equal to a side of a regular inscribed hexagon, § 391 and E F equal to a side of a regular inscribed decagon. $ 394 Join FH. and the arc E F is Ib of the circumference; .. the chord FH is a side of a regular inscribed pentedecagon, and by applying FH fifteen times as a chord, we have the polygon required. Q.E.F. 398. COROLLARY. By bisecting the arcs FH, HA, etc., a regular polygon of 30 sides may be inscribed ; and by continuing the process, regular polygons of 60, 120, etc. sides may be inscribed. PROPOSITION XIX. PROBLEM. 399. To inscribe in a given circle a regular polygon similar to a given regular polygon. CDI D Let A B C D, etc., be the given regular polygon, and C'D' E' the given circle. It is required to inscribe in C'D' E' a regular polygon. similar to A B C D, etc. From O, the centre of the polygon A B C D, etc. draw 0 D and 0 C. draw O'C' and O'D', making the 20' = < 0. Draw C' D'. Then C' D' will be a side of the regular polygon required. For each polygon will have as many sides as the 20 (=20') is contained times in 4 rt. 4. .. the polygon C' D' E', etc. is similar to the polygon C D E, etc., § 372 (two regular polygons of the same number of sides are similar). Q. E. F. PROPOSITION XX. PROBLEM. 400. To circumscribe about a circle a regular polygon similar to a given inscribed regular polygon. BMC Let H MRS, etc., be a given inscribed regular polygon. It is required to circumscribe a regular polygon similar to HM RS, etc. At the vertices H, M, R, etc., draw tangents to the O, intersecting each other at A, B, C, etc. Then the polygon A B C D, etc. will be the regular polygon required. Since the polygon A B C D, etc. has the same number of sides as the polygon H MRS, etc., it is only necessary to prove that A B C D, etc. is a regular polygon. In the A BHM and CMR, § 372 § 363 HM= MR, |