1 of the angle CBP, therefore the angle ABF is equal to the angle CBF; wherefore the angle ABC is G M may E strated that the angles BAE, AED, BG are bisected by the straight lines AF, EF : from the point F draw L (12. 1.) FG, FH, FK, FL, FM per- H pendiculars to the straight lines AB, BC, CD, DE, EA: and because the angle HCF is equal to KCF, and the D 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 common to both; therefore, the other sides shall be equal (26. 1.), 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 that 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 (Cor. 16. 3.) the circle : therefore each of the straight lines AB, BC, CD, DE, EA touches the circle ; wherefore the circle is inscribed in the pentagon ABCDE. Which was to be done. : PROP. XIV. PROB. To describe a circle about a given equilateral and equiangular penta gon. Let ABCDE be the given equilateral and equiangular pentagon ; it is required to describe a circle about it. Bisect (9. 1.) the angles BCD, CDE by the straight lines CF, FD, and from the point F, in which they A meet, draw the straight lines FB, FA, FE to the points B, A, E. It may be demonstrated, in the same manner as in E the preceding proposition, that the an- В. gles CBA, BAE, AED are bisected by the straight lines FB, FA, FE : and be. cause that 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 equal D to FDC; wherefore the side CF is equal (6. 1.) 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. : Let ABCDEF be the given circle ; it is required to inscribe ap 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 (Cor. 5. 1.); and the three angles of a triangle are equal (32. 1.) to two right angles; therefore the angle EGD is the third part of two right angles : In the same manner it may be demonstrated that A the angle DGC is also the third part of two right angles : and because the straight T B line GC makes with EB the adjacent angles EGC, CGB equal (13. 1.) to two right angles : the remaining angle CGB is the third part of two'right angles : there. fore the angles EGD, DGC, CGB, are EY equal to one another: and also the angles vertical to them, BGA, AGF, FGE (15.1.); therefore the six angles EGD, DGC,CGB, D BGA, AGF, FGE are equal to one another. But equal angles at the centre stand upon equal (26. 3.) arches; there- H fore the six arches AB, BC, CD, DE, EF, FA are equal to one another : and equal arches are subtended by equal (29. 3.) 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 arch AF is equal to ED, to each of these add the arch ABCD ; therefore the whole arch FABCD shall be equal to the whole EDCBA: and the angle FED stands upon the arch FABCD, and the angle AFE upon EDCBA ; therefore the angle IG : 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; it is also 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 radius of the circle. And if through the points A, B, C, D, E, F, there be drawn straight lines toucbing the circle, an equilateral and equiangular bexagon 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. 70 inscribe an equilateral and equiangular quindecagon in a given circle. Let ABCD be the giren circle; it is required to inscribe an equi. lateral and equiangular quiodecagoo in the circle ABCD. Let AC be the side of an equilateral triangle inscribed (2. 4.) in the circle, and AB the side of an equilateraland equiangular pentagon inscribed (11. 4.) in the same; therefore, of such equal parts as the whole circumference ABCDF contains fifteen, the B arch ABC, being the third part of the whole contains five ; and the arch AB, E which is the fifth part of the whole, contains three ; therefore BC their difference contains two of the same parts : bisect (30. 3.) 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 (1. 4.) around in the whole circle, an equilateral and 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 may 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. OF GEOMETRY. BOOK V. IN N the demonstrations of this book there are certain " signs or characters which it has been found convenient to employ. “ 1. The letters A, B, C, &c are used to denote magnitudes of any « kind. The letters m, n, p, q, are used to denote numbers only. “ 2. The sign + (plus), written between two letters, that denote “ magnitudes or numbers, signifies the sum of those magnitudes or “ numbers. Thus, A+B is the sum of the two magnitudes denoted s7 by the letters A and B ; mtn is the sum of the numbers denoted ; by m and n. « 3. The sign (minus), written between two letters, signifies the * excess of the magnitude denoted by the first of these letters « which is supposed the greatest, above that which is denoted by the « other. Thus, A-B signifies the excess of the magnitude A above “ the magnitude B. . " 4. When a number, or a letter denoting a number, is written " close to another letter denoting a magnitude of any kind, it sig" nifies that the magnitude is multiplied by the number. Thus, “ 3A signifies three times A ; mB, m times B, or a multiple of B by 6m. When the number is intended to multiply two or more mag“ nitudes that follow, it is written thus, m (A+B), which signifies " the sum of A and B taken m times ; m (A - B) is m times the ex"cess of A above B. “ Also, when two letters that denote numbers are written close to “ one another, they denote the product of those numbers, when “ multiplied into one another. Thus, mn is the product of m into n; and mnA is A multiplied by the product of m into n. “5. The sign signifies the equality of the magnitudes denoted . by the letters that stand on the opposite sides of it; A=B signifies " that A is equal to B A+B=C-D signifies that the sum of A and “ B is equal to the excess of C above Ď. “6. The sign 7 is used to signify that the magnitudes between “ which it is placed are unequal, and that the magnitude to which the opening of the lines is turned is greater than the other. Thus A “7 B signifies that A is greater than B ; and A B signifies that A es is less than B.” DEFINITIONS. a 1. 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 II. er is measured by the less, that is, when the greater contains the III. IV. tiplied so as to exceed the greater; and it is only such magnitudes V. taken of the first and third, and any equimultiples whatsoever of VI. ratio to the second that the third has to the fourth ; and the third to whatever be their number. “ When four magoitudes, A, B, C, D are proportionals, it is usual to “ say that A is to B as C to D, and to write them thus, A:B:: “ C:D, or thus, A : B=C: D.” |