PROPOSITION 12. PROBLEM. III. 1. K с L Let ABCD be the given circle. It is required to circumscribe a regular pentagon about the O ABCD. Construction. Inscribe a regular pentagon in the O ABCD, IV. 11. and let A, B, C, D, E be its angular points. At the points A, B, C, D, E draw GH, HK, KL, LM, MG, tangents to the circle. III. 17. Then shall GHKLM be the required regular pentagon. Find F the centre of the O ABCD; and join FB, FK, FC, FL, FD. In the AS BFK, CFK, and FK is common ; Because and KB=KC, being tangents to the circle from the same point K; III. 17, Cor. .:. the L BFK=the L CFK, I. 8. also the BKF=the < CKF. I. 8, Cor. Hence the BFC = twice the L CFK, and the _ BKC=twice the _ CKF. Similarly it may be shewn that the L CFD=twice the į CFL, and that the L CLD=twice the L CLF. But since the arc BC = the arc CD, IV. 11. .:. the L BFC=the L CFD; III. 27. and the halves of these angles are equal, that is, the - CFK= the L CFL. Then in the A$ CFK, CFL, the _ CFK=the L CFL, Proved. I. 26. IlI. 17, Cor. ... KL= HK. In the same way it may be shewn that every two consecutive sides are equal; .. the pentagon GHKLM is equilateral. Again, it has been proved that the 2 FKC=the 2 FLC, and that the * HKL, KLM are respectively double of these angles : ... the L HKL=the < KLM. In the same way it may be shewn that every two consecutive angles of the figure are equal ; .. the pentagon GHKLM is equiangular. .. the pentagon is regular, and it is circumscribed about the O ABCD. Q.E.F. COROLLARY. Similarly it may be proved that if tangents are drawn at the vertices of any regular polygon inscribed in a circle, they will form another regular polygon of the same species circumscribed about the circle. [For Exercises see p. 293.] PROPOSITION 13. PROBLEM. To inscribe a circle in a given regular pentagon. M E H I. 12. 1. 4. с K D Bisect two consecutive — BCD, CDE by CF and DF which intersect at F. 1. 9. In the A8 BCF, DCF, Нур. and the - BCF=the 2 DCF; Constr. the 2 CBF=the L CDF. But the 2 CDF is half an angle of the regular pentagon : ... also the į CBF is half an angle of the regular pentagon : that is, FB bisects the 2 ABC. So it may be shewn that if FA, FE were joined, these lines would bisect the < at A and E. Again, in the A' FCH, FCK, Constr. also FC is common ; .:: FH= FK. Similarly if FG, FM, FL be drawn perp. to BA, AE, ED, it may be shewn that the five perpendiculars drawn from F to the sides of the pentagon are all equal. 1. 26. a With centre F, and radius FH, describe a circle ; this circle must pass through the points H, K, L, M, G; and it will be touched at these points by the sides of the pentagon, for the 28 at H, K, L, M, G are rt. 28. Constr. .:. the O HKLMG is inscribed in the given pentagon. Q.E.F. COROLLARY. The bisectors of the angles of a regular pentagon meet at a point. NOTE. In the same way it may be shewn that the bisectors of the angles of any regular polygon meet at a point. [See Ex. 1, p. 294.] [For Exercises on Regular Polygons see p. 293.] MISCELLANEOUS EXERCISES. 1. Two tangents AB, AC are drawn from an external point A to a given circle : describe a circle to touch AB, AC and the convex arc intercepted by them on the given circle. 2. ABC is an isosceles triangle, and from the vertex A a straight line is drawn to meet the base at D and the circumference of the circumscribed circle at E: shew that AB is a tangent to the circle circumscribed about the triangle BDE. 3. An equilateral triangle is inscribed in a given circle: shew that twice the square on one of its sides is equal to three times the area of the square inscribed in the same circle. 4. ABC is an isosceles triangle in which each of the angles at B and C is double of the angle at A; shew that the square on AB is equal to the rectangle AB, BC with the square on BC. To circumscribe a circle about a given regular pentagon, Let ABCDE be the given regular pentagon. It is required to circumscribe a circle about the figure ABCDE. Construction. Bisect the 4 BCD, CDE by CF, DF, intersecting at F. I. 9. Join FB, FA, FE. I. 4. Proof. In the A' BCF, DCF, Нур. and the BCF = the DCF; Constr. .:: the L CBF = the L CDF. But the 2 CDF is half an angle of the regular pentagon : ... also the 2 CBF is half an angle of the regular pentagon: that is, FB bisects the L ABC. So it may be shewn that FA, FE bisect the 28 at A and E. Now the < FCD, FDC are each half an angle of the given regular pentagon ; .. the 2 FCD=the 2 FDC, IV. Def. 2. .:. FC= FD. I. 6. Similarly it may be shewn that FA, FB, FC, FD, FE are all equal. With centre F, and radius FA, describe a circle : this circle must pass through the points A, B, C, D, E, and therefore is circumscribed about the pentagon. Q.E.F. : NOTE. In the same way a circle may be circumscribed about any regular polygon. |