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

Join OC, OD, O E, OF, and draw the perpendiculars Oa, Ob, O c, 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 O B C, OBA equal to one another, (I. 4.) the base OC 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, O BA, 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 O a 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.

.A.

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.

6

B

BA

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 likewise 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 another (2, Cor. 3.). And because the triangles AOa, 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 half of A OB. In like manner, it may be shown that the angle BOb is equal to the half of BOČ; and AOB is equal to BOC, because the arc AB is equal to the arc BC (12.); therefore the angle BO a is equal to B O b. 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 a b 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.

For its sides being the chords of equal arcs of the same cir

C

E

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 A B 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 A B D, and twice in the arc B D. But the angle BAC is measured by half of the arc 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 AD B the centre of which is C. Join CA, CB, and let AC produced meet the circumfer

B

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 B D. 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 AB. 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 C B 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 EAC, ACE, that is

double of ACE, 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 E C (I. 6.). But, because the straight line AE bisects the angle BA C, 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 A D 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 AE; take CF equal to A E, and join A F. Then, because the arc AE is the half of A B, it is contained ten times in the whole circumference, and the chord A E is the side of a regular decagon inscribed in the circle AD B. Again, the angle E A C being measured by half the arc E B D, that is, by twofifths of the semi-circumference AED, is equal to the angle FC A, which is measured by the arc A B, that is likewise by two-fifths of the semi-circumference AED; therefore, because the triangles EA C, F CA, 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 rectangle 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 is

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 O AL, (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, o k lines drawn from the centres

A K B

a k o

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 OA2 to o a2, or OK2 to o k2.

Because the polygons have the same number of sides, the angles AO B, ao b are contained the same number of times in four right angles, and are therefore

equal to one another (II. ax. 2.), as also their halves (I. 6. Cor. 3.) the angles AOK, ao k. Therefore the triangles AO B, a o b (II. 32.) as also AOK, a ok, (II.31. Cor. 1.) are similar, and AB is to a b as OA to o a, 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 a h (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 CA 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 circle, the other circumscribed about

P

R

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 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 PQ parallel to E F and cutting the radius C A in R. Then, because the double of the arc A P 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 arc A M, and the point P the are A N. regard to the operation of bisecting the are, we should remark that it may be effected by bisecting the angle at the centre (12.).

With

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 circum scribed 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.

A

D

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: R let AC (II. 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 CA and C d. Then, as before, there may be found an inscribed polygon whose apothem CR is greater than CD. Take Cr (II.52.) a third proportional to CA and CR; and, because CD2: CR2 :: CA× Cd: 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 C dis less than Cr. And because the area of the inscribed polygon is to the area of the similar polygon circumscribed as CR2 to CA2 (30.), that is (II. 35.), as C r to C A (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 K2 to Me would have been less than K2, because the inscribed polygon is less than M2. 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 cir cumscribed) 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 C AD, the base AD of which is greater than AB. Then, because (31.) there may be circumscribed about the circle a regular poly. gon, the perimeter of which approaches more nearly to that of the circle (AB) 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