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is drawn from the centre F, FC is perpendicular (III, 18.) to KL, therefore

1. Each of the angles at C is a right angle: for the same reason,

2. The angles at the points B, D, are right angles : and because FCK is a right angle, (I. 47.)

3. The square of FK is equal to the squares of FC, CK; for the same reason,

4. The square of FK is equal to the squares of FB, BK: therefore (Ax. 1.)

5. The squares of FC, CK, are equal to the squares of FB, BK, of which the square of FC is equal to the square of FB; therefore (Ax. 3.)

6. The remaining square of CK is equal to the remaining square of BK, and the straight line CK equal to BK: and because EB is equal to FC, and FK common to the triangles BFK, CFK, the two BF, FK, are equal to the two CF, FK; and the base BK was proved equal to the base KC; therefore (I. 8.)

7. The angle BFK is equal to the angle KFC, and (I. 4.) the angle BKF to FKC: wherefore

8. The angle BFC is double of the angle KFC, and BKC double of FKC: for the same reason,

9. The angle CFD is double of the angle CFL, and CLD double of CLF: and because the circumference BC is equal to the circumference CD, (III. 27.)

10. The angle BFC is equal to the angle CFD; and BFC is double of the angle KFC, and CFD donble of CFL; therefore (Ax. 7.)

11. The angle KFC is equal to the angle CFL; and the right angle FCK is equal to the right angle FCL: therefore, in the two triangles FRC, FLC, there are two angles of one equal to two angles of the other, each to each, and the side FC, which is adjacent to the equal angles in each, is common to both; therefore (I. 26.) the other sides shall be equal to the other sides, and the third angle to the third angle; therefore

12. The straight line KC is equal to CL, and the angle FKC to the angle FLC:

and because KC is equal to CL,

13. KL is double of KC. In the same manner it may be shown, that

14. HK is double of BK: and because BK is equal to KC, as was demonstrated, and that KL is double of KC, and HR double of BK, (Ax. 6.)

15. HK is equal to KL. In like manner it may be shown, that

16. GH, GM, ML, are each of them equal to HK or KL: therefore

17. The pentagon GHKLM is equilateral. It is also equiangular; for, since the angle FKC is equal to the angle FLC, and the angle HKL double of the angle FKC, and KLM double of FLC, as was before demonstrated, (Ax. 6.)

1. The angle HKL is equal to KLM. And in like manner it

may be shown, that 2. Each of the angles KHG, HGM, GML, is equal to the angle HKL or KLM; therefore, the five angles GHK, HKL, KLM, LMG, MGH, being equal to one another,

3. The pentagon GHKLM is equiangular : and it is equilateral, as was demonstrated; and it is described about the circle ABCDE. Q.E.F.

PROP. XIII.- PROBLEM.
To inscribe a circle in a given equilateral and equiangular pentagon.

Let ABCDE be the given equilateral and equiangular pentagon; it is required to inscribe a circle in the pentagon ABCDE.

Bisect (I. 9.) the angles BCD, CDĚ, by the straight lines CF, DF, and from the point F, in which they meet, draw the straight lines FB, FA, FE.

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Therefore, since BC is equal (Hyp.) to CD, and CF common to the triangles BCF, DCF, the two sides BC, CF, are equal to the two DC, CF, each to each ; and the angle BCF is equal (Constr.) to the angle DCF; therefore (I. 4.)

1. The base BF is equal to the base FD,

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and the other angles to the other angles, to which the equal sides are opposite; therefore

2. The angle CBF is equal to the angle CDF: and because the angle CDE is double of CDF, and that CDE is equal to CBA, and CDF to CBF; also

3. CBA is double of the angle CBF; therefore

4. The angle ABF is equal to the angle CBF; wherefore

5. The angle ABC is bisected by the straight line BF. In the same manner it may be demonstrated that

6. The angles BAE, AED, are bisected by the straight lines AF, FE. From the point F draw (1. 12.) FG, FH, FK, FL, FM, perpendiculars to the straight lines AB, BC, CD, DE, EA; and because the angle HCF is equal to KCF, and the right angle FHC equal to the right angle FKC; in the triangles FHC, FKC, there are two angles of one equal to two angles of the other, each to each; and the side FC, which is opposite to one of the equal angles in each, is common to both; therefore the other sides shall (I. 26.) be equal, each to each; wherefore

7. The perpendicular FH is equal to the perpendicular FK. In the same manner it may be demonstrated that

8. FL, FM, FG, are each of them equal to FH or FK: therefore

9. The five straight lines FG, FH, FK, FI, FM, are equal to one another; wherefore the circle described from the centre F, at the distance of one of these five, shall pass through the extremities of the other four, and touch the straight lines AB, BC, CD, DE, EA, because the angles at the points G, H, K, L, M, are right angles, and that a straight line drawn from the extremity of the diameter of a circle at right angles to it, touches (III. 16.) the circle; therefore

10.

Each of the straight lines AB, BC, CD, DE, EA, touches the circle; wherefore it is inscribed in the pentagon ABCDE.

Q.E.F.

PROP. XIV. PROBLEM. To describe a circle about a given equilateral and equiangular pentagon.

Let ABCDE be the given equilateral and equiangular pentagon; it is required to describe a circle about it.

Bisect (I. 9.) the angles BCD, CDE, by the straight lines CF, FD, and from the point F, in which they meet, draw the straight lines FB, FA, FE, to the points B, 4, E.

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It may be demonstrated, in the same manner as in the preceding proposition, that

1.

The angles CBA, BAE, AED, are bisected by the straight lines FB, FA, FE; and because the angle BCD is equal to the angle CDE, and that FCD is the half of the angle BCD, and CDF the half of CDE; (Ax. 7.)

2. The angle FCD is equal to FDC; wherefore (I. 6.)

3. The side CF is equal to the side FD. In like manner it may be demonstrated that

4. FB, FA, FE, are each of them equal to FC or FD; therefore

5. 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 one of them, shall pass through the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE. Q.E.F.

PROP. XV.- PROBLEM.
To inscribe an equilateral and equiangular hexagon in a given circle.

Let ABCDEF be the given circle; it is required to inscribe an equilateral and equiangular hexagon in it.

Find (I. 3.) the centre G of the circle ABCDEF, and draw the diameter AGD; and from D as a centre, at the distance DG, describe the circle EGCH; join EG, CG, and produce them to the points B, F; and join AB, BC, CD, DE, EF, ; the hexagon ABCDEF is equilateral and equiangular.

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Because G is the centre of the circle ABCDEF,

1. GE is equal to GD; and because D is the centre of the circle EGCH,

2. DE is equal to DG; wherefore (Ax. 1.)

3. GE is equal to ED, and the triangle EGD is equilateral ; and therefore (I. 5. Cor.)

4. The three angles EGD, GDE, DEG, are equal to one another, but the three angles of a triangle are equal (I. 32.) to two right angles : therefore

5. The angle EGD is the third part of two right angles ; in the same manner it may be demonstrated that also

6. The angle DGC is the third part of two right angles : and because the straight line GC makes with EB the adjacent angles EGC, CGB, equal (I. 13.) to two right angles,

7. The remaining angle CGB is the third part of two right angles ; therefore

8. The angles EGD, DGC, CGB, are equal to one another ; and (I. 15.)

9. To EGD, DGC, CGB, are equal the vertical opposite angles BGA, AGF, FGE; therefore

10. The six angles EGD, DGC, CGB, BGA, AGF, FGE, are equal to one another : but equal angles stand (III. 26.) upon equal circumferences; therefore

11. The six circumferences AB, BC, CD, DE, EF, FA, are equal to one another; and equal circumferences are subtended (III. 29.) by equal straight lines ; therefore the six straight lines are equal to one another, and

12. The hexagon ABCDEF is equilateral. It is also equiangular; for, since the circumference AF is equal to ED, to each of these add the circumference ABCD; therefore

1. The whole circumference FABCD is equal to the whole EDCBA: and the angle FED stands upon the circumference FABCD, and the angle AFE upon EDCBA; therefore (III. 27.)

2. The angle AFE is equal to FED: in the same manner it may be demonstrated that the other angles of the

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