ELEMENTS

OF

GEOMETRY.

SUPPLEMENT.

BOOK I.

OF THE QUADRATURE OF THE CIRCLE.

LEMMA

Any curve line, or any polygonal line, which envelopes a convex line from one end to the other, is longer than the enveloped line.

Let AMB be the enveloped line; then will it be less than the line APDB which envelopes it.

We have already said that by the term convex line we understand a line, polygonal, or curve, or partly curve and partly polygonal, such that a straight line cannot cut it in more than two points. If in the line AMB there were any sinuosities or re-entrant portions, it would cease to be convex, because a straight line might cut it in more than

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M

B

two points. The arcs of a circle are essentially convex; but the present proposition extends to any line which fulfils the required conditions.

This being premised, if the line AMB is not shorter than any of those which envelope it, there will be found among the latter, a line shorter than all the rest, which is shorter than AMB, or, at most, equal to it. Let ACDEB be this enveloping line: any where between those two lines, draw the straight line PQ, not meeting, or at least only touching, the line AMB. The straight line PQ is shorter than PCDEQ; hence, if instead of the part PCDEQ, we substitute the straight line PQ, the enveloping line APQB will be shorter than APDQB. But, by hypothesis, this latter was shorter than any other; hence that hypothesis was false; hence all of the enveloping lines are longer than AMB

COR. 1. Hence the perimeter of any polygon inscribed in a circle is less than the circumference of the circle.

COR. 2. If from a point two straight lines be drawn, touching a circle, these two lines are together greater than the arc intercepted between them; and hence the perimeter of any polygon described about a circle is greater than the circumference of the circle.

PROP. I. THEOR.

If from the greater of two unequal magnitudes there be taken away its half, and from the remainder its half; and so on; There will at length remain a magnitude less than the least of the proposed magnitudes.

Let AB and C be two unequal magnitudes, of which AB is the greater. If from AB there be taken away its half, and from the remainder its half, and so on; there shall at length remain a magnitude less than C.

D

A

F

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H

For C may be multiplied so as, at length, to become greater than AB. Let DE, therefore, be a multiple of C, which is greater than AB, and let it contain the parts DF, FG, GE, each equal to C. From AB take BH equal to its half; and from the remainder AH, take HK equal to its half, and so on, until there be as many divisions in AB as there are in DE; And let the divisions in AB be AK, KH, HB. And because DE is greater than AB, and EG taken from DE is not greater than its half, but BH taken from AB is equal to its half; therefore the remainder GD is greater than the remainder HA. Again, because GD is greater than HA, and GF is not greater than the half of GD, but HK is equal to the half of HA; there fore the remainder FD is greater than the remainder AK. And FD is equal to C, therefore C is greater than AK; that is, AK is less than C.

PROP. II. THEOR.

B CE

Equilateral polygons, of the same number of sides, inscribed in circles, are similar, and are to one another as the squares of the diameters of the circles.

Let ABCDEF and GHIKLM be two equilateral polygons of the same number of sides inscribed in the circles ABD and GHK; ABCDEF and GHIKLM are similar, and are to one another as the squares of the diameters of the circles ABD, GHK.

Find N and O the centres of the circles, join AN and BN, as also GO and HO, and produce AN and GO till they meet the circumferences in D and K.

Because the straight lines AB, BC, CD, DE, EF, FA, are all equal the arcs AB, BC, CD, DE, EF, FA are also equal (28. 3.). For the same reason, the arcs GH, HI, IK, KL, LM, MG are all equal, and they

are equal in number to the others; therefore, whatever part the arc AB 18 of the whole circumference ABD, the same is the arc GH of the circumference GHK. But the angle ANB is the same part of four right angles, that the arc AB is of the circumference ABD (33. 6.); and the angle GOH is the same part of four right angles, that the arc GH is of the circumference GHK (33. 6.), therefore the angles ANB, GOH are each of them the same part of four right angles, and therefore they are equal to one another. The isosceles triangles ANB, GOH are therefore equiangular, and the angle ABN equal to the angle GHO; in the same manner, by joining NC, OI, it may be proved that the angles NBC, OHI are equal to one another, and to the angle ABN. Therefore the whole angle ABC

is equal to the whole GHI; and the same may be proved of the angles BCD, HIK, and of the rest. Therefore, the polygons ABCDEF and GHIKLM are equiangular to one another; and since they are equilateral, the sides about the equal angles are proportionals; the polygon ABCDEF is therefore similar to the polygon GHIKLM (def. 1.6.). And because similar polygons are as the squares of their homologous sides (20. 6.), the polygon ABCDEF is to the polygon GHIKLM as the square of AB to the square of GH; but because the triangles ANB, GOH are equiangular, the square of AB is to the square of GH as the square of AN to the square of GO (4. 6.), or as four times the square of AN to four times the square (15. 5.) of GO, that is, as the square of AD to the square of GK, (2. Cor. 8.2.). Therefore also, the polygon ABCDEF is to the polygon GHIKLM

as the square of AD to the square of GK; and they have also been shewn to be similar.

COR. Every equilateral polygon inscribed in a circle is also equiangu lar: For the isosceles triangles, which have their common vertex in the centre, are all equal and similar; therefore, the angles at their bases are all equal, and the angles of the polygon are therefore also equal.

PROP. III. PROB.

The side of any equilateral polygon inscribed in a circle being given, to find the side of a polygon of the same number of sides described about the circle.

Let ABCDEF be an equilateral polygon inscribed in the circle ABD ; it is required to find the side of an equilateral polygon of the same number of sides described about the circle.

Find G the centre of the circle; join GA, GB, bisect the arc AB in H; and through H draw KHL touching the circle in H, and meeting GA and GB produced in K and L; KL is the side of the polygon required.

Produce GF to N, so that GN may be equal to GL; join KN, and from G draw GM at right angles to KN, join also HG.

K

H

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L

E

D

Because the arc AB is bisected in H, the angle AGH is equal to the angle BGH (27. 3.); and because KL touches the circle in H, the angles LHG, KHG are right angles (18. 3.); therefore, there are two angles of the triangle HGK, equal to two angles of the triangle HGL, each to each. But the side GH is common to these triangles; therefore they are equal (26.1.), and GL is equal to GK. Again, in the triangles KGL, KGN, because GN is equal to GL; and GK common, and also the angle LGK equal to the angle KGN; therefore the base KL is equal to the base KN (4. 1.). But because the triangle KGN is isosceles, the angle GKN 18 equal to the angle GNK, and the angles GMK, GMN are both right angles by construction; wherefore, the triangles GMK, GMN have two an gles of the one equal to two angles of the other, and they have also the side GM common, therefore they are equal (26.1.), and the side KM is equal to the side MN, so that KN is bisected in M. But KN is equal to KL, and therefore their halves KM and KH are also equal. Wherefore, in the triangles GKH, GKM, the two sides GK and KH are equal to the two GK and KM, each to each; and the angles GKH, GKM, are also equal, therefore GM is equal to GH (4. 1.); wherefore, the point M is in the cir cumference of the circle; and because KMG is a right angle, KM touches the circle. And in the same manner, by joining the centre and the other angular points of the inscribed polygon, an equilateral polygon may be

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