B В 10. The two sides BF, FK, are equal to the two sides CF, FK, each to each; 11. And the base BK was shown to be equal to the base CK; 12. Therefore the angle BFK is equal to the angle CFK (I. 8); and the angle BKF to the angle CKF; (I. 4.) 13. Therefore the angle BFC is double of the angle CFK, and the angle BKC is double of the angle CKF. 14. For the same reason, the angle CFD is double of the angle CFL, and the angle CLD is double of the angle CLF. 15. And because the arc BC is equal to the arc CD, the angle BFC is equal to the angle CFD; (III. 27.) 16. And the angle BFC is double of H the angle CFK, and the angle CFD is double of the angle CFL; 17. Therefore the angle CFK is equal to the angle CFL. (ax. 7.) 18. And the right angle FCK is equal to the right angle FCL. 19. Therefore in the two triangles FCK, FCL, there are two angles of the one equal to two angles of the other, each to each; 20. And the side FC, which is adjacent to the equal angles in each, is common to both; 21. Therefore the other sides are equal, each to each, and the third angle of the one equal to the third angle of the other; 22. Therefore the straight line CK is equal to the straight line CL, and the angle FKC to the angle FLC. (I. 26.) 23. And because CK is equal to CL, LK is double of CK. 24. In the same manner it may be shown that HK is double of BK. 25. And because BK is equal to CK, as was shown, and that HK is double of BK, and LK double of CK, therefore HK is equal to LK. (ax. 6.) 26. In the same manner it may be shown that GH, GM, ML, are each of them equal to HK or LK; 27. Therefore the pentagon GHKLM is equilateral. 28. It is also equiangular. 29. For, since the angle FKC is equal to the angle FLC, and that the angle HKL is double of the angle FKC, and the angle KLM double of the angle FLC, as was shown, 30. Therefore the angle HKL is equal to the angle KLM. (ax. 6.) 31. In the same manner it may be shown that each of the angles KHG, HGM, GML, is equal to the angle HKL or KLM; 32. Therefore the pentagon ĜHKLM is equiangular. 33. And it has been shown to be equilateral. Conclusion.—Therefore, about the given circle a regular pentagon has been described. Q. E. F. PROPOSITION 13.-PROBLEM. M B To inscribe a circle in a given regular pentagon. (References-Prop. I. 4, 9, 12, 26; III. 16.) Given. Let ABCDE be the given regular pentagon. Sought. It is required to inscribe a circle in it. Construction.-1. Bisect the angles BCD, CDE by the straight lines CF, DF. (I. 9.) 2. From the point F, in which CF and DF meet, draw the straight lines FB, FA, FE. 3. And from the point F draw FG, FH, FK, FL, FM, perpendicular to the sides of the pentagon. (I. 12.) 4. Also from the centre F, at the distance FG, FH, FK, FL, or FM, describe the circle GHKLM. Then GHKLM shall be inscribed as required in the pentagon ABCDE. Proof.-1. Because BC is equal to DC (hyp.), and CF is common to the two triangles BCF, DCF; 2. The two sides BC, CF, are equal to the two sides DC, CF, each to each; 3. And the angle BCF is equal to the angle DCF; (const.) 4. Therefore the base BF is equal to the base DF, and the other angles to the other angles to which the equal sides are opposite; (I. 4.) 5. Therefore the angle CBF is equal to the angle CDF. 6. And because the angle CDE is double of the angle CDF, and that the angle CDE is equal to the angle CBA, and the angle CDF is equal to the angle CBF, 7. Therefore the angle CBA is double of the angle CBF; 8. Therefore the angle ABF is equal to the angle CBF; 9. Therefore the angle ABC is bisected by the straight line BF. 10. In the sam manner it may be shown that the angles BAE, AED, are bisected by the straight lines AF, EF. 11. And because the angle FCH is equal to the angle FCK, and the right angle FHC equal to the right angle FKC; 12. Therefore in the two triangles FHC, FKC, there are two angles of the one equal to two angles of the other, each to each; 13. And the side FC, which is opposite to one of the equal angles in each, is common to both; 14. Therefore the other sides are equal, each to each, and therefore the perpendicular FH is equal to the perpendicular FK. (I. 26.) 15. In the same manner it may be demonstrated that FL, FM, FG, are each of them equal to FH or FK. 16. Therefore the five straight lines FG, FH, FK, FL, FM, are equal to one another, and the circle described from the centre F, at the distance of one of them will pass through the extremities of the other four. 17. And because the angles at the points G, H, K, L, M, are right angles (const.), and that a straight line drawn from the extremity of a diameter, at right angles to it, touches the circle; (III. 16.) 18. Therefore each of the straight lines AB, BC, CD, DE, EA, touches the circle. Conclusion.-Therefore a circle has been described in the given regular pentagon ABCDE. Q. E. F. PROPOSITION 14.-PROBLEM. (References-Prop. I. 6, 9.) Construction.—1. Bisect the angles BCD, CDE, by the straight lines CF, DF. (I. 9.) 2. From the point F, in which CF and DF meet, draw the straight lines FB, FA, FE, to the points B, A, E. 3. And from the centre F, at the distance FA,FB, FC, FD, or FE, describe the circle ABCDE. Then ABCDE shall be described as required about the given pentagon. Proof.—1. It may be demonstrated, as in the preceding proposition, that the .angles CBA, BAE, AED, are bisected by the straight lines BF, AF, EF. 2. And because the angle BCD is equal to the angle CDE, and that the angle FCD is half of the angle BCD, and the angle FDC is half of the angle CDE, В. 3. Therefore the angle FCD is equal to the angle FDC; (ax. 7.) 4. Therefore the side FC is equal to the side FD. (I. 6.) 5. In the same manner it may be shown that FB, FA, FE, are each of them equal to FC or FD;, 6. Therefore the five straight lines FA, FB, FC, FD, FE, are equal to one another, and the circle described from the centre F, at the distance of any one of them, will pass through the extremities of the other four. Conclusion.—Therefore, the circle ABCDE is described about the regular pentagon ABCDE. Q. E. F. PROPOSITION 15.–PROBLEM. To inscribe a regular hexagon in a given circle. (References-Prop. I. 5 cor., 13, 15, 32; III. 1, 26, 27, 29.) Given. Let ABCDEF be the given circle. Sought.-It is required to inscribe a regular hexagon in it. Construction.-1. Find the centre G of the circle ABCDEF (III. 1), and draw the diameter AGD. 2. From the centre D, at the distance DG, describe the circle EGCH. 3. Join EG, CG, and produce them to the points B, F; and join AB, BC, CD, DE, EF, FA. Then the figure ABCDEF shall be the hecagon required. Proof.-1. Because G is the centre of the circle ABCDEF, GE is equal to GD; 2. And because D is the centre of the circle EGCH, DE is equal to DG; 3. Therefore GE is equal to DE (ax. 1), and the triangle EGD is equilateral; 4. Therefore the three angles, EGD, GDE, DEG, are equal to one another. (I. 5 cor.) 5. But the three angles of a triangle are together equal to two right angles; (I. 32.) 6. Therefore the angle EGD is the third part of two right angles. 7. In the same manner it may be shown that the angle DGC is the third part of two right angles. 8. And because the straight line GC makes with the straight IC line EB the adjacent angles EGC, CGB, together equal to two right angles, (I. 13.) 9. Therefore the remaining angle CGB is the third part of two right angles; 10. Therefore the angles EGD, DGC, CGB, are equal to one another. 11. And to these are equal the vertical opposite angles BGA, AGF, FGE; (I. 15.) 12. Therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE, are equal to one another. 13. But equal angles stand on equal arcs; (III. 26.) 14. Therefore the six arcs AB, BC, CD, DE, EF, FA, are equal to one another. 15. And equal arcs are subtended by equal chords; (III. 29.) 16. Therefore the six straight lines are equal to one another, and the hexagon ABCDEF is equilateral. 17. It is also equiangular. 18. For, since the arc AF is equal to the arc ED; 19. To each of these add the arc ABCD; 20. Therefore the whole arc FABCD is equal to the whole arc ABCDE; 21. And the angle FED stands on the arc FABCD, and the angle AFE stands on the arc ABCDE; 22. Therefore the angle FED is equal to the angle AFE. (III. 27.) 23. In the same manner it may be shown that the other angles of the hexagon ABCDEF are each of them equal to the angle AFE or FED; 24. Therefore the hexagon is equiangular; and it was shown to be equilateral. Conclusion.—Therefore, a regular hexagon has been inscribed in the circle ABCDEF. Q. E. F. Corollary.-1. From this it is manifest that the side of the hexagon is equal to the radius of the circle. 2. And if through the points A, B, C, D, E, F, there be drawn straight lines touching the circle, a regular hexagon will be described about the circle, as may be shown from what was said of the pentagon; and a circle may be inscribed in a given regular hexagon, and circumscribed about it, by a method like that used for the pentagon. |