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PRO P. XIII. PRO B.
To infcribe a circle in a given equilateral and equi
Let ABCDE be the given equilateral and equiangular penta
gon; it is required to inscribe a circle in the pentagon ABCDE. a. y. I. Bisect the angles BCD, CDE by the straight lines CF, DF,
and from the point F in which they meet draw the straight lines
two DC,CF; and the angle BCFis equal to the angle DCF; thereb. 4. Si fore the base BF is equal o to the base FD, and the other angles
to the other angles, to which the equal fides are opposite; there
posite to one of the equal angles in each, is common to both; d. 26.1. therefore the other side shall be equal d, each to each ; wherefore
the perpendicular FH is equal to the perpendicular FK. in the
f. 12. I.
lines AB, BC, CD, DE, EA, because the angles at the points G, Book IV. 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 each of the straight lines AB, BC,c. 16. 3. CD, DE, EA touches the circle ; wherefore it is inscribed in the pentagon ABCDE. Which was to be done.
PRO P. XIV.
To describe a circle about a given equilateral and
Let ABCDE be the given equilateral and equiangular pentagon; it is required to describe a circle about it.
Bifect the angles BCD, CDE by the straight lines CF, FD, a. 9. 1. 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 fame man
A ner as in the preceding proposition, that the angles CBA, BAE, AED are bisected by the straight lines FB, FA,B
E 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
D is equal to FDC; wherefore the side CF is equal to the side FD. in like manner it may be demon-b. 6. t. 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 center F, at the distance of one of them, shall pass thro' the extremities of the other four, and be described about the equilateral and equiangular pen. tagon ABCDE. Which was to be done.
A. $. I.
PRO P. XV. PRO B. "O 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 the center G of the circle ABCDEF, and draw the diameter AGD; and from D as a center, 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 center of the circle ABCDEF, GE is equal to GD. and because D is the center 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 ifofceles
triangle are equal ?. and the three angles of a triangle are equal bto b. 32. 1.
two right angles; therefore the angle EGD is the third part of two
with EB the adjacent angles LGC,CGB C. 13. 1. equal to two right angles; the re
maining angle CGB is the third part of,
EGD, DGC, CGB are equal to one d. 15.1. another. and to these are equal - the
H 6. 26. 2
stand upon equal circumferences; therefore the fix circumferences
AB,BC, CD, DE, EF, FA are equal to one another. and equal f. 29. 3.
circumferences are fubtended by equal f straight lines; therefore the fix 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
the circumference FABCD, and the angle AFE upon EDCBA ; Book IV.
Cor. From this it is manifest, that the side of the hexagon is
And if thro' the points A, B, C, D, E, F there be drawn straight lines touching the circle, an equilateral and equiangular hexagon shall 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 circumfcribed about it, by a method like to that used for the pentagon.
PROP. XVI. PROB.
Tinscribe an equilateral and equiangular quinde- see N.
cagon in a given circle.
Let ABCD be the given circle; it is required to infcribe an equilateral and equiangular quindecagon in the circle ABCD.
Let AC be the fide of an equilateral triangle inscribed in the a. 2.6
c. 30. 3.
Book IV. And in the same manner as was done in the pentagon, if thro'
the points of division made by infcribing the quindecagon, straighe