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 e the circle : therefore each e 16. 3. 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. B 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, a 9.1 and from the point F, in which they meet, draw the straight lines FB, FA, FE to the points B, А 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 F 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 D equal to FDC; wherefore the side 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 equiangular pentagon ABCDE. Which was to be done. Book IV. PROP. XV. PROB. Sce 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. 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 a 5. 1. the base of an isosceles triangle are equal a ; and the three angles b 32. 1. of a triangle are equal b to two right angles; therefore the angle EGD is the third part of two right angles: in the same А B G c 13. 1. adjacent angles EGC, CGB equal c to two right angles; the remaining EGD, DGC, CGB are equal to one d 15. 1. another: and to these are equal d the vertical opposite angles BGA, AGF, D equal to one another : but equal e 26. 3. angles stand upon equal e circumfe rences; therefore the six circumfe rences AB, BC, CD, DE, EF, FA are equal to one another : f 29. 3. and equal circumferences are subtended by equal f struight 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; с and the angle FED stands upon the circumference FABCD, and Book IV. 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 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 circumscribed about it, by a method like to that used for the pentagon. a 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 in a 2. 4. the circle, and AB the side of an equilateral and equiangular pentagon inscribed b in the same; therefore, of such equal parts b 11.4. as the whole circumference ABCDF contains fifteen, the circumference ABC, being the third A part of the whole, contains five; and the circumference AB, which is the fifth part of the whole, contains three; therefore BC their difference B F contains two of the same parts: bi E c 30. 3. sect BC in E; therefore BE, EC are, each of them, the fifteenth part с of the whole circumference ABCD: therefore, if the straight lines BE, EC be drawn, and straight lines equal to them be placed d around in the whole circle, an equila-d 1. 4. teral and equiangular quindecagon shall be inscribed in it. Which was to be done. 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. THE ELEMENTS OF EUCLID. BOOK V. DEFINITIONS. e I. A LESS magnitude is said to be a part of a greater magnitude, Book V. when the less measures the greater, that is, when the less is contained a certain number of times exactly in the greater.' II. greater is measured by the less, that is, ' when the greater III. • Ratio is a mutual relation of two magnitudes of the same kind See N. ' to one another, in respect of quantity.' IV. V. second, which the third has to the fourth, when any equimul- |