PROPOSITION 12. PROBLEM. To circumscribe a regular pentagon about a given circle. 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. Proof. Because and FK is common; III. 1. and KB KC, being tangents to the circle from the same point K; III. 17, Cor. I. 8. I. 8, Cor. ..the BFC = the CFD; and the halves of these angles are equal, that is, the CFK the CFL. IV. 11. III. 27. Hence KL is double of KC; similarly HK is double of KB. And since KC = KB, III. 17, Cor. L Again, it has been proved that the FKC = the FLC, and that the HKL, KLM are respectively double of these angles : 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 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. Let ABCDE be the given regular pentagon. It is required to inscribe a circle within the figure ABCDE. Construction. Bisect two consecutive CF and DF which intersect at F. Proof. But the .. also the Join FB; and draw FH, FK perp. to BC, CD. In the ▲ BCF, DCF, BC = DC, BCD, CDE by I. 9. I. 12. Hyp. Because and CF is common to both; .. the CBF the CDF. = CDF is half an angle of the regular pentagon : CBF is half an angle of the regular pentagon : that is, FB bisects the ABC. So it may be shewn that if FA, FE were joined, these lines would bisect the 4 at A and E. Again, in the a FCH, FCK, the Because and the FCH the FCK, = FHC = the Constr. FKC, being rt. angles ; I. 26. 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. 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 2 at H, K, L, M, G are rt. 2s. Constr. .. the 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. PROPOSITION 14. PROBLEM. 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 BCD, CDE by CF, DF, intersecting at F. I. 9. Proof. Join FB, FA, FE. .. the CBF the Because and CF is common to both; But the .. also the Hyp. CDF. I. 4. CDF is half an angle of the regular pentagon : So it may be shewn that FA, FE bisect the ▲3 at A and E. Now the given regular pentagon; .. the FCD = the 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. H.S.E. T |