shall be equal to the sum of the rect- Section 4.- Regular polygons, and angles under its opposite sides. approximation to the area of the Let ACBD be circle. à quadrilateral inscribed in the circle Def. 11. A regular polygon is that ABC; and let AB, which has all its sides equal, and likewise CD be its diagonals: all its angles equal. the rectangle under A figure of five sides is called a pentaAB, CD shall be gon; a figure of six sides a hexagon; of equal to the sum of ten sides a decagon; and of fifteen sides the rectangles under a pente-decagon. There is so seldom AD, BC, and AC, any occasion, huwever, to specify the BD. number of sides of an irregular figure, At the point A make the angle DAF as distinct from a multilateral figure in equal to B AC, and let AF meet CD general, that it has become common to in F. appropriate these names with others of Then, because the angle ABC is similar derivation (as by way of preequal to AD F in the same segment eminence) to the regular figures—"a (15.), and that B A C was made equal hexagon," for instance, is understood to to DAF, the triangles ABC, ADF mean a regular figure of six sides, and are equiangular: therefore, (11.31.) AB: so of the rest. BC :: AD: DF, and (II. 38.) the rect It is evident, that regular polygons, angle under A B, DF is equal to the which have the same number or sides, rectangle under AD, B C. are similar figures; for their angles are Again, because the angles B AC, equal, each to each, because they are DAF are equal to one another, let the contained the same number of times in angle B AF be added to each ; therefore the same number of right angles (I. 20.); the whole angle FAC is equal to the whole and their sides about the equal angies angle D AB; and the angle FCA is are to another in the same ratio, viz. the equal to the angle D BA in the same ratio of equality. segment (15.); therefore, the triangles 12. The centre of a regular polygon AFC, AD B are equiangular. There is the same with the common centre of fore (II. 31.) AB:BD::AC:CF, the inscribed and circumscribed circles and (II. 38.) the rectangle under AB, (see Prop. 26.): and the perpendicular CF is equal to the rectangle A C, B D. which is drawn from the centre to any Therefore, the sum of the rectangles one of the sides is called the apothem. under A B, D F and A B, C F, that is, which subtend equal angles at the centre. 13. Similar arcs of circles are those (I. 30. Cor.) the rectangle under A B, CD, is equal to the sum of the rectangles Similar sectors and segments are those under AD, BC, and A C, B D. which are bounded by similar arcs. Therefore, &c. Cor. Hence, a quadrilateral may be PROP. 26. (Euc. iv. 13 and 14.). constructed, which shall have its sides If any two adjoining angles of a equal to four given straight lines, in a regular polygon be bisected, the intergiven order, each to each, and its angular section of the bisecting lines shall be the points lying in the circumference of a common centre of two circles, the one circle. For, by the 24th proposition, the circumscribing, the other inscribed in, ratio of the diagonals, and by that which the polygon. has been just demonstrated, their rect Let ABCDEF angle is given: therefore, (I1.63.) the be any regular podiagonals may be found, and (1.50.) the lygon, and let the quadrilateral constructed. angles at A and B It is only essential to the possibility be bisected by the of the construction that of the four given straight lines AO, straight lines, every three be greater B0; which meet than the fourth (I. 10. Cor. 2.). It is in some point 0, remarkable that, although the diagonals (I. 15. Cor. 4.) because each of the anwill be different in different orders of the gles FAB, CBA is less than two right given sides, the circumscribing circle angles, and therefore each of their halves has the same magnitude whatsoever be OAB, OBA less than a right angle, and their order. (See Sect. 5. Prop. 41. the two together less than two right an cuolaira gles The point I shall be the centre C e F A. B Scholium.) Dis 2 ܝ of two circles, one passing through all and its angles standing upon equal arcs, the points A, B, C, D, E, F, and the viz. the differences between the whole other in contact with all the sides AB, circumference, and two of the former, BC, CD, DE, E F. are likewise equal (14. Cor. 2.). Join O C, OD, O E, O F, and draw Next, let abcdef be the figure which the perpendiculars Oa, Ob, O c, Od, is included by tangents drawn through Oe, Of. Then, because the triangles the points ABCDEF: this shall likeOB C, OB A have two sides of the one wise be a regular polygon. equal to two sides of the other, each to Let O be the centre of the circle, and each, and the included angles O BC, join OA, OB, O a, Ob, OC. Then beOB A equal to one another, (I. 4.) the cause a A, a B, are tangents drawn from base 0 C is equal to the base o A, and the same point, they are equal to one the angle O CB to the angle O AB. another (2. Cor. 3.). And because the But OAB is the half of FAB, and FAB triangles A O a, B O a have the three is equal to DCB: therefore 0 CB is sides of the one equal to the three sides the half of DCB, and the latter angle is of the other, each to each, the angle bisected by the line 0 C. By a similar B Oa is equal to AO a, that is, to the demonstration, therefore, it may be half of A OB. In like manner, it may shown that OD is equal to 0 B, oE to be shown that the angle Bob is equal OC, and OF to OD. And, because the to the half of BOC; and A O B is angles O A B, O BA, being halves of equal to BOC, because the arc AB equal angles, are equal to one another, is equal to the arc BC (12.); thereOB is equal to OA (I. 6.). Therefore the fore the angle B O a is equal to B Ob. straight lines drawn from 0 to the angu- Therefore B O a, Bob are triangles lar points of the figure are equal to one which have two angles of the one equal another, and O is the centre of a circle to two angles of the other, each to each, passing through those points. And be- and the interjacent side O B common cause AB, BC, &c. are equal chords to both : consequently, (I. 5.) they are of the same circle, they are at equal equal in every respect, and B a is equal distances from the centre O (4. Cor.): to Bb; therefore ab is bisected in B. that is, the perpendiculars O a, Ob, &c. In the same manner it may be shown are equal to one another, and O is that a f is bisected in A; and it has likewise the centre of a circle described been shown that a B, a A, are equal to with the apothem o or 0 b for its ra- one another; therefore ab is equal to dius, and (2.) touching the sides in their af. And by a like demonstration it middle points (3.), a, i, c, d, e, f. may be shown that the other sides of Therefore, &c. the figure are each of them equal to ab PROP. 27. or a f. Therefore, the figure abcdef has all its sides equal to one another. If the circumference of a circle be di- And because its angles, as a, b, are supvided into any number of equal parts, plements (I. 20.Cor.) of equal angles, as the chords joining the points of division AOB, BOC, they are likewise all equal shall includea regular polygon inscribed to one another. Therefore it is a regular in the circle; and the tangents drưwn polygon. through those points shall include a re- Therefore, &c. gular polygon of the same number of Cor. 1. (Euc. iv. 12.). If any regular sides circumscribed about the circle. polygon be inscribed in a circle, a similar Let the circumfe. polygon may be circumscribed about the rence of the circle circle by drawing tangents through the ACF be divided angular points of the former; and coninto any number of versely. equal parts in the Cor. 2. If, any regular polygon being points A, B, C, D, E, inscribed in a circle, a tangent be drawn F. The figure which parallel to one of its sides, and be teris included by the minated both ways by radii passing straight lines joining through the extremities of that side, those points shall be be such terminated tangent shall be a side a regular polygon. of a regular polygon of the same numFor its sides being ber of sides circumscribed about the the chords of equal circle. For, since the radius drawn to arcs of the same cir the point of contact bisects the side cle, are (12.Cor. 1.) equal to one another; (3. Cor. 1.) at right angles, and there: B D B fore also bisects the angle formed by the double of AC E, and that A B E being radii passing through its extremities equal (I. 6.) to B AC is likewise double (I. 6. Cor. 3.) it is obvious from 1. 5. of ACE, A E B is equal to A B E, and that the parts of the tangent in question AB is equal to AE or E C (I. 6.). But, are equal to one another, and to the because the straight line AE bisects the halves of any side of the regular circum- angle B AC, CA : AB::CE: E B scribed polygon of the same number of (II. 50.). Therefore CB : CE:: CE : sides. EB, that is the radius C B is medially divided in E; and A B, the side of the Prop. 28. (Euc. iv. 10. and 15. Cor.) decagon, is equal to the greater seg. The side of a regular hexagon is ment CE. equal to the radius of the circle in which it is inscribed ; the side of a regular the side of a regular Lastly, let A B be decagon is equal to the greater segment pentagon incribed in of the radius divided media’ly ; aid the the circle AD B, the side-square of a regular pentagon is centre of which is C. greater than the square of the radius by Bisect the angle the side-square of a regular decugon ACB by the radius inscribed in the same circle. CE (1. Post. 4.): First, let A B be then the arcs AE, E B measuring equal the side of a hexa angles, are equal to one another (12.). gon inscribed in the Join A E; take C F equal to A E, and circle ADB, the cen join A F. Then, because the arc A E tre of which is C. is the half of A B, it is contained ten Join CA, CB, and times in the whole circumference, and let AC produced the chord AE is the side of a regular meet the circum decagon inscribed in the circle A D B. ference in D. Then, because the arc Again, the angle E A C being measured A B is contained in the whole circum- by half the arc E B D, that is, by twoference six times, it is contained three fifths of the semi-circumference AED, is times in the semi-circumference A BD, equal to the angle FC A, which is meaand twice in the arc B D. But the an- sured by the arc AB, that is likewise gle BAC is measured by half of the arc by two-fifths of the semi-circumference BD, (14. Cor. 1.) and the angle ACB AED; therefore, because the trianby the arc A B. Therefore the angle gles EA C, FCA, have two sides BAC is equal to ACB, and the side of the one equal to two sides of the A B is equal to BC; that is, the side of other, each to each, and the includec. the regular hexagon is equal to the ra- angles equal to one another, the base dius of the circle. AF (1.4.) is equal to CE or AC. And, Secondly, let AB because from the vertex A of the isosbe the side of a regu celes triangle ACF, the straight line lar decagon inscribed AB is drawn to meet the base produced in the circle ADB in B, the square of A B is greater than the centre of which the square of AC (1.39.) by the rectanis C. Join CA, CB, gle CB, BF. But CF being equal to AE, and let AC produced the side of a regular decagon, the radius meet the circumfer C B is medially divided in F, as shown in ence in D. Then, because the arc A B the second part of the proposition; and is contained ten times in the whole cir- therefore the rectangle CB, BF is equal cumference, it is contained five times in to the square of C For AE (II.38.Cor.1.). the semi-circumference A B D, and four Therefore the square of A B is greater times in the arc B D. But the angle than the square of AC by the square of BAC is measured by half of the arc AE; that is, the side-square of a reguBD (14. Cor. 1.) and the angle A C B is lar pentagon is greater than the square measured by the arc A B. Therefore of the radius by the side square of a the angle BAC is equal to twice the regular decagon inscribed in the same angle ACB. Let the angle B A C be circle. bisected by the line A E, and let A E Therefore, &c. meet C B in E. Then, because the angle EAC is equal to ACE, the side EA PROP. 29. is equal to EC; and because A E B is equal to the two E AC, ACE, that is The area of any regular polygon 18 B D B M P equal to half the rectangle under its equal to one another (II. ax. 2.), as perimeter and apothem. also their halves (1.6. Cor. 3.) the angles Let ABCDEF be any regular po A OK, ao k. Therefore the triangles lygon, O its centre, and 0 K, which is AO B, a ob (II. 32.) as also AOK, a ok, drawn perpendicular to A B, its apo- (11. 31. Cor. 1.) are similar, and AB is to them: the area of the polygon shall be ab as OA to o 0, or as OK to ok (II.31.). equal to half the rectangle under the But, because the polygons are similar, apothem 0 K and the perimeter ABC their perimeters are to one another as DEF. AB to ab, and their areas as AB* to a he (II. 43.). Therefore their perimeters are to one another as QA to oa, or as OK to ok II. 12.); and their areas as 0 Ao to o a’, or as OK to o ko. (11. 37. Cor. 4.) Therefore, &c. A K B PROP. 31. Take A L equal to the sum of the Any circle being given, similar resides or perimeter A B C D E F, and join gular polygons may be, the one in() L, O B. Then because the sides are scribed, and the other circumscribed, equal to one another, the base AL con such that the difference of their peritains the base AB, and therefore the tri- meters or areas shall be less than any angle O A L contains the triangle OAB, given difference. (I. 27.) as many times as the polygon Let C be the has sides. But, because the triangles centre and C A the OAB, OBC, &c. having equal bases and radius of any given altitudes (I. 27.), are equal to one ano circle: and let Kbe ther, the polygon likewise contains the any given straight triangle_OAB as many times as it has line; there may be sides. Therefore the polygon is equal found two regular to the triangle O AL, (II. ax. 1.) That polygons of the is, (I. 26. Cor.) to half the rectangle un same number of der the perimeter AL and apothem OK. sides, the one inTherefore, &c scribed in the cir. cle, the other cirProp. 30. (Euc. xii. 1.) cumscribed about The perimeters of similar regular the circle, which shall have the difference polygons are as the radii of the in- of their perimeters less than the straight scribed or circumscribed circles, and line K, or the difference of their areas their areas are as the squares of the less than the square of K. radii. And first of the perimeters. Let L be Let o, o be the centres of two re a straight line equal to the perimeter of gular polygons having the same number some circumscribed polygon; and let the of sides, A B, a b, any two sides, and radius AC be divided in the point D in OK, o k lines drawn from the centres the ratio of K to L (II. 55.). Through D draw the chord E F perpendicular to CA, and draw the radius C B likewise perpendicular to CA: bisect the arc AB in M,* the arc A M in N, the arc AN in P, and so on till the point of bisection fall between A and E; let P be the first point so falling ; draw the chord perpendicular to them respectively : P Q parallel to E F and cutting the radius CA in R. Then, because the then (26.) O A, 0 a are the radii of the double of the arc AP is contained a circumscribed circles, and OK, o k the radii of the inscribed circles. The peri The necessity of having the lines and letters in meters of the polygons shall be to one the neighbourhood of A clear and distinct has led the another as QA to o a, or OK to ok and engraver to tax the reader's imagination somewhat their areas as OA' to o a', or OK to ok. more than was absolutely necessary in the figures of this proposition. He is requested, therefore, to supBecause the polygons have the same pose that the point M bisects the arc AB, the point number of sides, the angles A O B, ao b N the arc AM, and the point P the arc AN. With are contained the same number of times regard to the operation of bisecting the are, we should remark that it may be effected by bisecting the angle in four right angles, and are therefort at the centre (12.). certain number of times in the quadrant been to the inscribed polygon as Ko to A B, it is contained four times as often MR would have been less than Ks, bein the whole circumference. But be cause the inscribed polygon is less than cause the radius C A cuts the chord M. Much more, then, is the difference PQ at right angles, the arc PAQ is between the inscribed and circumscribed equal to the double of AP (3. and 1. 6. polygons less than K square, that is, Cor. 3.). Therefore the arc which it less than the given difference. subtends being contained exactly a cer Therefore, &c. tain numberof times in the whole circum Cor. 1. Any circle being given, a reference, the chord P Q is the side of an gular polygon may be inscribed (or cirinscribed polygon (27.). And because the cumscribed) which shall differ from the perimeter of this inscribed polygon is to circle, in perimeter or in area, by less ihe perimeter of the similar circum- than any given difference. For the scribed polygon as CR to CA (30.), the difference between the circle and either difference of their perimeters (invertendo of the polygons is less than the difference and dividendo) is to the perimeter of of the two polygons. the inscribed polygon as A R to RC, Cor. 2. Any two circles being given, that is, in a less ratio than that of A D similar regular polygons may be in to DC, or of K to L. But even a mag- scribed (or circumscribed), which shall . nitude which should have been to the differ from the circles, in perimeter or perimeter of the inscribed polygon in the in area, by less than any the same given same ratio as that of K to L would have difference. been less than K (II. 18.), because the perimeter of the inscribed polygon is PROP. 32. less than L (I. 10. Cor. 3.). Much more, The area of a circle is equal to half then, is the difference of the perimeters the rectangle under the radius and cirof the inscribed and circumscribed po- umference. lygons less than K, that is, less than the Let C be the centre, and CA the ragiven difference. dius of any circle: from the point A let In the next there be drawn AB perpendicular to place, of the areas. Let M be a straight line, such that the square of M is equal to the area of some circumscribed polygon let AC (11.55.) be divided in d in the CA, and suppose any line AB equal to ratio of K-square the circumference of the circle, and join to M-square, and CB: the circle shall be equal to the trilet CD be taken angle CAB. (II. 51.) a mean For, if not equal, it must be either proportional between CA and C d. greater or less than the triangle. First, Then, as before, there may be found an let it be supposed greater, and therefore inscribed polygon whose apothem CR equal to some triangle CAD, the base is greater than CD. Take Ĉr (II. 52.) a AD of which is greater than AB. Then, third proportional to CA and CR; and, because (31.) there may be circumbecause CD : CR:: CAXCd:CA“ scribed about the circle a regular polyCr (II. 38. Cor. 1.), in which proportion gon, the perimeter of which approaches the first term is less than the second, the more nearly to that of the circle (AB) third is also less than the fourth (II. 14.), than by any given difference, as B D, a and therefore C dis less than Cr. And polygon may be circumscribed, the pebecause the area of the inscribed poly- rimeter of which shall be less than AD. gon is to the area of the similar polygon But the area of any regular polygon is circumscribed as CRP to CA2 (30.), that equal to half the rectangle under its periis (II. 35.), as Cr to C A (invertendo meter and apothem ( 29.). Therefore the and dividendo), the difference of their area of such circumscribed polygon will areas is to the area of the inscribed po- be less than the triangle C A D, less lygon as Ar to Cr, or in a less ratio that is, than the supposed area of the than that of Ad to Cd, or of Ko to M$. circle ; which is absurd. But even a magnitude which should have Neither can the area of the circle be |