#9. 1. Bisect the angles BCD, CDE by the straight lines CF, DF, and with the point F, in which they meet, as centre, and FC or FD as distance, if a circle be described, it will be that required. 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 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; there +7 Ax. *6. 1. B fore the angle FCD is equal to FDC; 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. To inscribe an equilateral and equiangular hexagon in a given circle. +1. 3. and 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, draw the diameter AGD; and from D, as a centre, at the distance DG, describe the circle EGCH, draw EG, CG, and prolong them to the points B, F; and draw 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 +1 Ax. + Cor. 5.1. *32. 1. 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: but the three angles of a triangle are equal to two right angles; therefore the angle EGD is the third part of two right angles: in a similar manner 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 *31. 1. equal to two right angles; the remaining angle CGB is the third part of two right angles: therefore the angles EGD, DGC, CGB are equal to one another: and to these are equal the vertical or opposite angles BGA, AGF, FGE: therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE, are equal to #15. 1. #26. 2. #29.3. F EX H one another: but equal angles stand upon equal* arcs; therefore the six arcs AB, BC, CD, DE, EF, FA are equal to one another: and equal arcs are subtended by equal* straight lines: therefore the six straight lines are equal to one another, and the bexagon ABCDEF is equilateral. It is also equiangular: for, since the arc AF is equal to ED, to each of these add the arc ABCD; therefore the whole arc FABCD is equal to the whole EDCBA: and the angle FED stands upon the arc FABCD, and the angle AFE upon EDCBA; therefore the angle AFE is equal to FED: in a similar 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. 127.3. 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. N 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, as 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 method like to that used for the pentagon. PROP. XVI. PROB. 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. *2.4. Let AC be the side of an equilateral triangle inscribed in the circle, and AB the side of an equilateral and equiangular pen #11.4. tagon inscribed in the same: therefore, of such equal parts as the whole circumference ABCDF contains fifteen, the arc ABC, being the E #30. 3. B F 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 quindecagon will be inscribed in it; and that it is equiangular is plain, because each angle stands upon an arc which is equal to the entire circumference #1. 4. diminished by the two arcs, which its sides subtend, and these arcs are, by construction, all equal. Therefore an equilateral and equiangular quindecagon is inscribed in the circle. 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. THE ELEMENTS OF EUCLID. wwwwww BOOK V. DEFINITIONS. I. Of two unequal magnitudes, the greater is said to contain the less as many times as there are parts in the greater, equal to the less. II. If the greater of two magnitudes contain the less a certain number of times without leaving a remainder, it is called a multiple of the less; and the less is in this case called a submultiple of the greater, or a measure of the greater. III. Magnitudes which have a common measure, that is, which are multiples of any other quantity, are said to be commensurable. But if it be impossible that any such common measure can exist, then the magnitudes are called incommensurable. It is very important that the student have a clear conception of the difference between commensurable and incommensurable magni |