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Book IV.

. 9. 1.

PROP. XIII. PROB. 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.

Bisecta 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 64. 1. BF is equal b to the base FD, and the other angles to the

other angles, to which the equal sides are opposite; there-
fore the angle CBF is equal to the angle CDF: And be-
cause the angle CDE is double of CDF, and that CDE is
equal to CBA, and CDF to
CBF; CBA is also double of

A

tis
the angle CBF; therefore the
angle ABF is equal to the an-

I see
gle CBF; wherefore the an- :
gle ABC is bisected by the BF
straight line BF: In the same
manner it may be demon- .
strated, that the angles BAE, 1
AE!, are bisected by the

straight lines 'AF, FE: From c12. 1. the point F, drawc FG, FH

KDY 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, and the side FC,

which is opposite to one of the equal angles in each, is * 26. 1. common to both ; therefore the other sides shall be equal,

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

M

of these five, shall pass through the extremities of the other Book IV. 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 e thee 16. 3. circle: Therefore 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.

PROP. XIV. PROB.

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.

Bisecta the angles BCD, CDE by the straight lines CF, . 9. 1. FD, and from the point F, in which they meet, draw the straight lines FB, FA, FE, to the points B, A, E. It may be demonstrated, in the same manner, as in the preceding propo. E șition, that 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 CBE; the angle FCD is equal to FDC; wherefore the side CF is equalb to the side FD: In like manner . 6. 2. it may be demonstrated 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 equiangular pentagon ABCDE. Which was to be done.

Book IV.

PROP. XV. PROB.

See n. To inscribe an equilateral and equiangular hexagon

in a given circle.

Let ABCDEF be the given circle; it is required to irscribe 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 equal to GD: 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, because the angles at the base of an isosceles * 5. 1. triangle are equala; and the three angles of a triangle 32. 1. are equalb to two right angles; therefore the angle EGD

is the third part of two right angles: In the same man. ner it may be demonstrated, that the angle DGC is also the third part of two right angles: And because the straight line GC makes with EB

the adjacent angles EGC, CGB € 13. 1. equalc to two right angles: the F

remaining angle CGB is the third part of two right angles: there'fore the angles EGD, DGC, CGB

are equal to one another: And to 14 15. 1. these are equald the vertical oppo

site angles BGA, AGF, FGE:
Therefore the six angles EGD,
DGC, CGB, BGA, AGF, FGE,

are equal to one another: But * 26. 3. equal angles stand upon equal e circumferences; therefore the six ore the six

H circumferences AB, BC, CD, DE, EF, FA are equal - to one another: And equal circumferences are subf 29. 3. tended 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 cir

cumference AF is equal to ED, to each of these add the Book IV. circumference ABCD; therefore the whole circumference FABCD shall be equal to the whole EDCBA: And the angle FED stands upon the circumference FABCD, and the angle AFE upon EDCBA; 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 semidiameter 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 equiangular hexagon shail be described about it, which may be demonstrated from what has 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 mei thod like to that used for the pentagon.

PROP. XVI. PROB. To inscribe an equilateral and equiangular quin- See N. decagon 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 inscribed a • 2. 4. in the circle, and AB the side of an equilateral and equiangular pentagon inscribedb in the same; therefore, of such - 11. 4. equal parts as the whole circumference ABCDF contains fifteen, the circumference ABC, being the third part of the whole, contains five; and the circumference AB, which is the fifth part of at the whole, contains three; therefore BC their difference contains El tivo of the same parts : Bisectc BC

€30. 3. in £; therefore BE, EC are, each of them, the fifteenth part of the whole circumference ABCD: .

Book IV. Therefore, if the straight lines. BE, EC he drawn, and

straight lines equal to them be placed round d in the whole circle, an equilateral and equiangular quindecagon shall be inscribed in it, which was to be done.

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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