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• 26. i.
KCF, and the right angle FHC equal to the right angle FKC; therefore in the triangles FHC, FŘC there are two angles of the 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 are 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 of these five, will 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 * the 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.
Bisect * the angles BCD, CDE by the straight lines • 9. 1. CF, 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 de
E monstrated, in the same manner as in the preceding proposition, 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 CDE; therefore the angle FCD is equal+ to FDC; + 7 Ax. wherefore the side CF is equal* to the side FD: in like manner 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, will pass through the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE. Which was to be done.
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
inscribe an equilateral and equiangular hexagon in it. + 1.3.
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 shall be 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 + Cor. 5. 1. equal+ to one another: but the three angles of a triangle * 32. 1.
are equal* to two right angles; therefore the angle EGD
* 13. 1.
* 15. 1.
* 26. 3.
* 29. 3.
cumference FABCD is 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 equalt to FED: in the same + 27. 3. 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 shewn; 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 will be described about it, which
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 method like to that used for the pentagon.
PROP. XVI. PROB.
* 11. 4.
To inscribe an equilateral and equiangular quindecagon
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 * in the circle, and AB the side of an equilateral * 2. 4. and equiangular pentagon inscribed* in the same: therefore, of such equal parts, as the whole circumference ABCDF contains fifteen, the circumference ABC,
El being the third part of the whole, contains five; and the circumference AB, which is the fifth part of the whole, contains three; therefore BC, their difference, contains two of the same parts : bisect * BC in E; therefore BE, EC are each of them the fifteenth part of the whole circumference ABCD: therefore if the straight lines BE, EC be drawn, and straight lines equal to them be placed round* in the whole circle, an equilateral and * 1. 4.
* 30. 3.
equiangular quindecagon will 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 will 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.
ELEMENTS OF EUCLID.
A Less magnitude is said to be a part of a greater
magnitude when the less measures the greater ; that is, when the less is contained a certain number of
times exactly in the greater.'
A greater magnitude is said to be a multiple of a less, when the greater is measured by the less, that is,
when the greater contains the less a certain number 6 of times exactly.'
III. “ Ratio is a mutual relation of two magnitudes of the See N.
“ same kind to one another, in respect of quantity.”
Magnitudes are said to have a ratio to one another,
when the less can be multiplied so as to exceed the other.
The first of four magnitudes is said to have the same
ratio to the second, which the third has to the fourth,