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shall be equal to the sum of the rect- SECTION 4.-Regular polygons, and

angles under its opposite sides.

Let ACBD be

a quadrilateral in

scribed in the circle ABC; and let AB, CD be its diagonals: the rectangle under AB, CD shall be equal to the sum of the rectangles under AD, BC, and A C, BD.

B

At the point A make the angle D AF equal to BAC, and let A F meet CD in F.

Then, because the angle ABC is equal to AD F in the same segment (15.), and that B A C was made equal to DAF, the triangles A B C, ADF are equiangular: therefore, (II. 31.) AB: BCAD: D F, and (II. 38.) the rectangle under AB, D F is equal to the rectangle under A D, B C.

Again, because the angles BA C, DAF are equal to one another, let the angle BAF be added to each; therefore the whole angle FAC is equal to the whole angle D A B; and the angle FCA is equal to the angle DBA in the same segment (15.); therefore, the triangles AFC, AD B are equiangular. There fore (II. 31.) ABBD::AC: CF, and (II. 38.) the rectangle under AB, CF is equal to the rectangle A C, B D. Therefore, the sum of the rectangles under AB, DF and AB, CF, that is, (I. 30. Cor.) the rectangle under AB, CD, is equal to the sum of the rectangles under A D, B C, and A C, B D.

Therefore, &c.

Cor. Hence, a quadrilateral may be constructed, which shall have its sides equal to four given straight lines, in a given order, each to each, and its angular points lying in the circumference of a circle. For, by the 24th proposition, the ratio of the diagonals, and by that which has been just demonstrated, their rectangle is given: therefore, (II. 63.) the diagonals may be found, and (I. 50.) the quadrilateral constructed.

It is only essential to the possibility of the construction that of the four given straight lines, every three be greater than the fourth (I. 10. Cor. 2.). It is remarkable that, although the diagonals will be different in different orders of the given sides, the circumscribing circle has the same magnitude whatsoever be their order. (See Sect. 5. Prop. 41.

Scholium.)

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approximation to the area of the circle.

Def. 11. A regular polygon is that which has all its sides equal, and likewise all its angles equal.

A figure of five sides is called a pentagon; a figure of six sides a hexagon; of ten sides a decagon; and of fifteen sides a pente-decagon. There is so seldom any occasion, however, to specify the number of sides of an irregular figure, as distinct from a multilateral figure in general, that it has become common to appropriate these names with others of similar derivation (as by way of preeminence) to the regular figures-“ a hexagon," for instance, is understood to mean a regular figure of six sides, and so of the rest.

It is evident, that regular polygons, which have the same number of sides, are similar figures; for their angles are equal, each to each, because they are contained the same number of times in the same number of right angles (I. 20.); and their sides about the equal angies are to another in the same ratio, viz. the ratio of equality.

12. The centre of a regular polygon is the same with the common centre of the inscribed and circumscribed circles (see Prop. 26.): and the perpendicular which is drawn from the centre to any one of the sides is called the apothem,

which subtend equal angles at the centre. Similar sectors and segments are those which are bounded by similar arcs.

13. Similar arcs of circles are those

PROP. 26. (Euc. iv. 13 and 14.).

If any two adjoining angles of a regular polygon be bisected, the intersection of the bisecting lines shall be the common centre of two circles, the one circumscribing, the other inscribed in, the polygon.

Let ABCDEF be any regular polygon, and let the angles at A and B be bisected by the straight lines AO, BO; which meet in some point O,

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(I. 15. Cor. 4.) because each of the angles FA B, CBA is less than two right angles, and therefore each of their halves OAB, OBA less than a right angle, and the two together less than two right angles The point O shall be the centre

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of two circles, one passing through all the points A, B, C, D, E, F, and the other in contact with all the sides A B, BC, CD, DE, E F.

Join O C, O D, O E, OF, and draw the perpendiculars Oa, Ob, Oc, Od, Oe, Of Then, because the triangles OB C, OBA have two sides of the one equal to two sides of the other, each to each, and the included angles OB C, OBA equal to one another, (I. 4.) the base O C is equal to the base O A, and the angle OCB to the angle OA B. But OAB is the half of FAB, and FAB is equal to DCB: therefore OCB is the half of DCB, and the latter angle is bisected by the line O C. By a similar demonstration, therefore, it may be shown that OD is equal to O B, OE to OC, and OF to OD. And, because the angles O AB, OBA, being halves of equal angles, are equal to one another, OB is equal to OA (J. 6.). Therefore the straight lines drawn from O to the angular points of the figure are equal to one another, and O is the centre of a circle passing through those points. And because A B, BC, &c. are equal chords of the same circle, they are at equal distances from the centre O (4. Cor.): that is, the perpendiculars O a, Ob, &c. are equal to one another, and O is likewise the centre of a circle described with the apothem Oa or Ob for its radius, and (2.) touching the sides in their middle points (3.), a, b, c, d, e, f. Therefore, &c.

PROP. 27.

If the circumference of a circle be divided into any number of equal parts, the chords joining the points of division shall include a regular polygon inscribed in the circle; and the tangents drawn through those points shall include a regular polygon of the same number of sides circumscribed about the circle. Let the circumference of the circle ACF be divided into any number of equal parts in the points A, B, C, D, E, F. The figure which is included by the straight lines joining those points shall be a regular polygon.

For its sides being

the chords of equal

arcs of the same cir

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B

B

E

and its angles standing upon equal arcs, viz. the differences between the whole circumference, and two of the former, are likewise equal (14. Cor. 2.).

Next, let abcdef be the figure which is included by tangents drawn through the points ABCDEF: this shall bikewise be a regular polygon.

Let O be the centre of the circle, and join OA, OB, O a, Ob, OC. Then because a A, a B, are tangents drawn from the same point, they are equal to one And because the another (2. Cor. 3.). triangles A O a, BO a have the three sides of the one equal to the three sides of the other, each to each, the angle BO a is equal to AO a, that is, to the it may half of A OB. In like manner, be shown that the angle BOb is equal to the half of BOC; and AOB is equal to BOC, because the arc A B is equal to the arc BC (12.); therefore the angle BO a is equal to B Ob. Therefore BOa, BOb are triangles which have two angles of the one equal to two angles of the other, each to each, and the interjacent side OB common to both: consequently, (I. 5.) they are equal in every respect, and B a is equal to Bb; therefore ab is bisected in B. In the same manner it may be shown that af is bisected in A; and it has been shown that a B, a A, are equal to one another; therefore ab is equal to af. And by a like demonstration it may be shown that the other sides of the figure are each of them equal to ab or a f. Therefore, the figure abcdef has all its sides equal to one another. And because its angles, as a, b, are supplements (I. 20.Cor.) of equal angles, as AOB, BOC, they are likewise all equal to one another. Therefore it is a regular polygon.

Therefore, &c.

Cor. 1. (Euc. iv. 12.). If any regular polygon be inscribed in a circle, a similar polygon may be circumscribed about the circle by drawing tangents through the angular points of the former; and conversely.

Cor. 2. If, any regular polygon being inscribed in a circle, a tangent be drawn parallel to one of its sides, and be terminated both ways by radii passing through the extremities of that side, such terminated tangent shall be a side of a regular polygon of the same number of sides circumscribed about the circle. For, since the radius drawn to the point of contact bisects the side

cle, are (12.Cor. 1.) equal to one another; (3. Cor. 1.) at right angles, and there,

fore also bisects the angle formed by the radii passing through its extremities (I. 6. Cor. 3.) it is obvious from I. 5. that the parts of the tangent in question are equal to one another, and to the halves of any side of the regular circumscribed polygon of the same number of sides.

PROP. 28. (EUc. iv. 10. and 15. Cor.)

The side of a regular hexagon is equal to the radius of the circle in which it is inscribed; the side of a regular decagon is equal to the greater segment of the radius divided medially; and the side-square of a regular pentagon is greater than the square of the radius by the side-square of a regular decagon

inscribed in the same circle.

First, let AB be the side of a hexagon inscribed in the circle ADB, the centre of which is C. Join CA, CB, and let AC produced meet the circum

ference in D. Then, because the arc A B is contained in the whole circumference six times, it is contained three times in the semi-circumference ABD, and twice in the arc B D. But the angle BAC is measured by half of the are BD, (14. Cor. 1.) and the angle ACB by the arc A B. Therefore the angle BAC is equal to A CB, and the side A B is equal to BC; that is, the side of the regular hexagon is equal to the radius of the circle.

Secondly, let AB be the side of a regular decagon inscribed in the circle ADB the centre of which is C. Join CA, CB, and let AC produced meet the circumfer

B

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ence in D. Then, because the arc A B is contained ten times in the whole circumference, it is contained five times in the semi-circumference A B D, and four times in the arc BD. But the angle BAC is measured by half of the arc BD (14. Cor. 1.) and the angle A CB is measured by the arc A B. Therefore the angle BAC is equal to twice the angle ACB. Let the angle BAC be bisected by the line A E, and let A E meet CB in E. Then, because the angle EAC is equal to ACE, the side EA is equal to EC; and because A E B is equal to the two E AC, ACE, that is

double of AC E, and that ABE being equal (I. 6.) to B A C is likewise double of ACE, A E B is equal to A B E, and AB is equal to AE or EC (I. 6.). But, because the straight line AE bisects the angle BAC, CA: AB::CE:EB (II. 50.). Therefore CB CE::CE: EB, that is the radius C B is medially divided in E; and A B, the side of the decagon, is equal to the greater segment CE.

Lastly, let AB be the side of a regular pentagon incribed in the circle AD B, the

centre of which is C.

Bisect the angle
ACB by the radius
CE (I. Post. 4.):

E

B

then the arcs AE, E B measuring equal angles, are equal to one another (12.). Join A E; take CF equal to A E, and join A F. Then, because the arc A E is the half of A B, it is contained ten times in the whole circumference, and the chord AE is the side of a regular decagon inscribed in the circle A D B. Again, the angle E AC being measured by half the arc E B D, that is, by twofifths of the semi-circumference AED, is equal to the angle FCA, which is measured by the are AB, that is likewise by two-fifths of the semi-circumference AED; therefore, because the triangles E A C, FCA, have two sides of the one equal to two sides of the other, each to each, and the included angles equal to one another, the base AF (I.4.) is equal to CE or AC. And, because from the vertex A of the isosceles triangle ACF, the straight line AB is drawn to meet the base produced in B, the square of A B is greater than the square of AC (I. 39.) by the rectan gle CB, BF. But CF being equal to AE, the side of a regular decagon, the radius CB is medially divided in F, as shown in the second part of the proposition; and therefore the rectangle CB, BF is equal to the square of CF or AE (II.38.Cor.1.). Therefore the square of AB is greater than the square of AC by the square of AE; that is, the side-square of a regular pentagon is greater than the square of the radius by the side-square of a regular decagon inscribed in the same circle.

Therefore, &c.

PROP. 29.

The area of any regular polygon vớ

F

equal to half the rectangle under its equal to one another (II. ax. 2.), as perimeter and apothem.

E

Let ABCDE F be any regular polygon, O its centre, and OK, which is drawn perpendicular to A B, its apothem: the area of the polygon shall be equal to half the rectangle under the 2pothem OK and the perimeter ABC DEF.

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Take A L equal to the sum of the sides or perimeter A B C D E F, and join OL, OB. Then because the sides are equal to one another, the base AL contains the base AB, and therefore the triangle OA L contains the triangle OAB, (I. 27.) as many times as the polygon has sides. But, because the triangles OAB, OBC, &c. having equal bases and altitudes (I. 27.), are equal to one another, the polygon likewise contains the triangle OAB as many times as it has sides. Therefore the polygon is equal to the triangle OA L, (II. ax. 1.) that is, (I. 26. Cor.) to half the rectangle under the perimeter AL and apothem OK, Therefore, &c

PROP. 30. (Euc. xii. 1.)

The perimeters of similar regular polygons are as the radii of the inscribed or circumscribed circles, and their areas are as the squares of the radii.

Let O, o be the centres of two regular polygons having the same number of sides, A B, a b, any two sides, and OK, ok lines drawn from the centres

u k

perpendicular to them respectively:

then (26.) O A, o a are the radii of the circumscribed circles, and OK, ok the radii of the inscribed circles. The perimeters of the polygons shall be to one another as OA to o a, or OK to ok and their areas as OAo to o a2, or OK to o ko. Because the polygons have the same number of sides, the angles A OB, ao b are contained the same number of times in four right angles, and are therefore

also their halves (I. 6. Cor. 3.) the angles A OK, a ok. Therefore the triangles AO B, a ob (II. 32.) as also AOK, a ok, (II. 31. Cor. 1.) are similar, and AB is to ab as OA to oa, or as OK to ok (II. 31.). But, because the polygons are similar, their perimeters are to one another as AB to ab, and their areas as AB to ab (II. 43.). Therefore their perimeters are to one another as OA to oa, or as OK to ok (II. 12.); and their areas as OA2 to o a2, or as OK to o k2. (II. 37. Cor. 4.) Therefore, &c.

PROP. 31.

Any circle being given, similar regular polygons may be, the one inscribed, and the other circumscribed, such that the difference of their perimeters or areas shall be less than any given difference.

Let C be the centre and C A the radius of any given circle: and let K be any given straight line; there may be found two regular polygons of the A same number of sides, the one inscribed in the cir cle, the other circumscribed about

RD

M

B

the circle, which shall have the difference of their perimeters less than the straight line K, or the difference of their areas less than the square of K.

And first of the perimeters. Let L be a straight line equal to the perimeter of some circumscribed polygon; and let the radius AC be divided in the point D in 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 are 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 PQ parallel to E F and cutting the

radius C A in R. Then, because the double of the arc AP is contained a

The necessity of having the lines and letters in the neighbourhood of A clear and distinct has led the engraver to tax the reader's imagination somewhat more than was absolutely necessary in the figures of this proposition. He is requested, therefore, to suppose that the point M bisects the arc AB, the point N the are A M, and the point P the are AN. regard to the operation of bisecting the arc, we should remark that it may be effected by bisecting the angle at the centre (12.).

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certain number of times in the quadrant A B, it is contained four times as often in the whole circumference. But because the radius CA cuts the chord PQ at right angles, the arc PAQ is equal to the double of AP (3. and I. 6. Cor. 3.). Therefore the arc which it subtends being contained exactly a certain number of times in the whole circumference, the chord PQ is the side of an inscribed polygon (27.). And because the perimeter of this inscribed polygon is to the perimeter of the similar circumscribed polygon as CR to CA (30.), the difference of their perimeters (invertendo and dividendo) is to the perimeter of the inscribed polygon as AR to RC, that is, in a less ratio than that of A D to DC, or of K to L. But even a magnitude which should have been to the perimeter of the inscribed polygon in the same ratio as that of K to L would have been less than K (II. 18.), because the perimeter of the inscribed polygon is less than L (I. 10. Cor. 3.). Much more, then, is the difference of the perimeters of the inscribed and circumscribed polygons less than K, that is, less than the given difference.

In the next place, of the areas. Let Mbe 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 ratio of K-square to M-square, and let CD be taken (II. 51.) a mean proportional between C A and C d. Then, as before, there may be found an inscribed polygon whose apothem CR is greater than CD. Take Ĉr (II. 52.) a third proportional to CA and CR; and, because CD2: CRo :: CA × C d: CAx Cr (II. 38. Cor. 1.), in which proportion the first term is less than the second, the third is also less than the fourth (II. 14.), and therefore Cd is less than Cr. And because the area of the inscribed polygon is to the area of the similar polygon eircumscribed as CR2 to CA2 (30.), that is (II. 35.), as Cr to CA (invertendo and dividendo), the difference of their areas is to the area of the inscribed polygon as Ar to Cr, or in a less ratio than that of Ad to Cd, or of K2 to M2. But even a magnitude which should have

been to the inscribed polygon as K to Me would have been less than K3, because the inscribed polygon is less than M. Much more, then, is the difference between the inscribed and circumscribed polygons less than K square, that is, less than the given difference. Therefore, &c.

Cor. 1. Any circle being given, a regular polygon may be inscribed (or circumscribed) which shall differ from the circle, in perimeter or in area, by less than any given difference. For the difference between the circle and either of the polygons is less than the difference of the two polygons.

Cor. 2. Any two circles being given, similar regular polygons may be inscribed (or circumscribed), which shall differ from the circles, in perimeter or in area, by less than any the same given difference.

PROP. 32.

the rectangle under the radius and cirThe area of a circle is equal to half cumference.

Let C be the centre, and CA the radius of any circle: from the point A let there be drawn AB perpendicular to

B

CA, and suppose any line AB equal to the circumference of the circle, and join CB: the circle shall be equal to the triangle CAB.

For, if not equal, it must be either greater or less than the triangle. First, let it be supposed greater, and therefore equal to some triangle CAD, the base AD of which is greater than AB. Then, because (31.) there may be circumscribed about the circle a regular polygon, the perimeter of which approaches more nearly to that of the circle (A B) than by any given difference, as BD, a polygon may be circumscribed, the perimeter of which shall be less than AD. But the area of any regular polygon is equal to half the rectangle under its perimeter and apothem (29.). Therefore the area of such circumscribed polygon will be less than the triangle CAD, less that is, than the supposed area of the circle; which is absurd.

Neither can the area of the circle be

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