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Book IV.

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

Bisect a the angles BCD, CDE by the straight lines CF, DF, and from the point F, in which they meet, draw the straight lines FB, FA, FE ; therefore, since BC is equal to CD, and CF common to the triangles BCF,DCF, the two sides BC,CF are equal to the two DC, CF; and the angle BCF is equal to the angle DCF; therefore the base BF is equalb to the base FD, and the other angles to the other arglęs, to which the equal sides are opposite; therefore 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

А
CBF;CBA is also double of the
angle CBF; therefore the angle G

M
ABF; is equal to the angle CBF;
wherefore the angle ABC is bi-

B
sected by the straight line BF:

E in the same manner it may be demonstrated, that the angles BAE, AED are bisected by the H straight lines AF, FE: from the point F drawc FG, FH, FK, FL, FM perpendiculars to the

C K D straight lines AB, BC, 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, and the side FC, which is opposite to one of the equal angles in each, is common to both; therefore the other sides shall be equals, each to each; wherefore the perpendicular FH is equal to the perpendicular FK: in the same manner it may be demonstrated that FL, FM, FG are each of them equal to FH, or FK; therefore the five straight lines FG, FH, FK, FL, 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

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C 12. 1.

d 26.1.

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touch the straight lines AB, BC, CD, DE, EA, because the Book IV. 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 the circle: therefore e 16.2 each of the straight lines AB, BC, CD, DE, EA touches the circle; wherefore it is inscribed in the pentagon ABCDE. Which was to be done,

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TO describe a circle about a given equilateral and cquiangular pentagon.

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

Bisect a the angles BCD, CDE by the straight lines CF, FD, . 9. 1. and from the point F, in which they meet, draw the straight lines FB, FA, FE, to the points B,

A A, E. It may be demonstrated, in the same manner as in the preceding proposition, that the angles CBA,

B BAE, AED are bisected by the

T

E 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; the angle FCD is с

D CF is equal b to the side FD :' in like manner it may be demon.b'6. 1, strated that FB, FA, FE are each of them equal to FC or FD :: 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 one of them, shall pass through the extremities of the other four, and be described about the equilateral and cquiangular pentagon ABCDE. Which was to be done.

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Book IV

PROP. XV. PROB.

See Note TO inscribe an equilateral and equiangular hexa

gon

in a given circle.

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a 5. 1. b32. 1.

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

„Find 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, FA: the hexagon ABCDEF is equilateral and equiangular.

Because G is the centre of the circle ABCDEF, GE is cqual to ĜD: and because D is the centre of the circle EGCH, DE is equal to DG; wherefore GE is equal to ED, and the triangle EGD is equilateral; and therefore its three angles EGD, GDE, DEG are equal to one another,

c. Because the angles at the base of an'isosteles triangle are equala ; and the three angles of a triangle are equalb to two right angles; therefore the angle EGD is the third part of two right angles: in the same manner it may be demon

Α.
strated, that the angle DGO is also the
third part of two right angles : and F
because the straight line GC makes

B
with EB the adjacent angles EGC,
CG3 equal to two right angles; the
remaining angle CGB is the third
part of two right angles; therefore
the angles EDG, DGC, CGB are
equal to one another; and to these
( are equals the vertical opposite an-

D gles BGA, AGF, FGE: therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE are equal to one another: but equal angles stand upon H equale circumferences; therefore the. six circumferences AB, BC, CD, DE, EF, FA are equal to. one another: and equal circumferences are subtended by equalf straight lines; therefore the six straight lines are equal to one another, and 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 the whole circumference FABCD shall be equal to the whole EDCBA: and the angle FED stands upon

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1 the circumference FABCD, and the angle AFE upon Book iv. EDCBÀ; therefore the angle AFE is equal to FED: in the same manner it may be demonstrated that the other angles of the hexagon ABCDEF are each of them equal to the angle AFE or FED; therefore the hexagon is equiangular ; and it is equilateral, as was shown; and it is inscribed in the given circle ABCDEF. Which was to be done.

Cor. From this it is manifest, that the side of the hexagon is equal to the straight line from the centre, that is, to the semi-diameter of the circle.

And if through the points A, B, C, D, E, F there be drawn straight lines touching the circle, an equilateral and equiangúlar hexagon shall be described about it, which may be de.. monstrated from what Kás been said of the pentagon; and likewise a circle may be inscribed in a given equilateral and equiangular hexagon, and circumscribed about it, by a mer thod like to that used for the pentagon.

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PROP. XVI. PROB.

To inscribe an ecrilateral and equiangular quinde. See Note. cagon in a given circle.

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

Let AC be the side of an equilateral triangle instribed in a 2. 4. the circle, and AB the side of an equilateral and equiangular pentagon inscribedb in the same; therefore, if such equal parts b 11. 4. as the whole circumference ABCDF contains fifteen, the circumference ABC, being the third

A parts of the whole, contains five and the circunference AB, which is the fifth part of the whole, contains three;

F therefore BC their difference con-B tains two of the same parts : bisect

< 11. BC in E; therefore BE, EC are, each of them, the fifteenth part of the whole circumference ABCD: there. fore, if the straight lines BEZEC be? drawn, and straight lines 'equal to them be placed d around in the whole circle, an equilateral d 1.4. and equiangular quindecagon shall be inscribed in it. Which was to be done,

E

Book IV.

And in the same manner as was done in the pentagon, if through the points of division made by inscribing the quindecagon, straight lines be drawn touching the circle, an equilateral and equiangular quindecagon shall be described about it: and likewise as in the pentagon, a circle may be inscribed in a given equilateral and equiangular quindecagon, and circumscribed about it.

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