AB;

therefore the sum of the straight lines AC, CD, and

DB are greater than AB.

Since, therefore, a line shorter than any of the curve or crooked lines that join A and B can be found, but none shorter than the straight line can be found, therefore it is the shortest

(Ax. 1).

COR. 1.-If two points A and B be joined by a straight line
AB, and a curve line ACB, or a

mixed line, which is concave towards
ACB is the least of all the

AB;

D

by joining two points in it by a straight line, a shorter line will be found than it, joining A and B, and fulfilling the other conditions; and this will be the case whether ADB be composed of straight lines, or be a mixed line, or a curve line. And as this can be proved of every line fulfilling the required conditions, except of ACB, therefore (Ax. 2) ACB is the least.

COR. 2.-If. two intersecting curves, whose curvatures lie in one and the same direction, be cut by a straight line, inclined to the interior at an angle not greater than a right angle, the arc which it intercepts on this line from the point of intersection will be less than the corresponding arc of the other line.

Let AB and AC intersect in A, and be cut by CD, so that the angle at B is not greater than a right angle, then AC AB.

For if CE, BE be curves equal to AC, AB in every respect, on the other side of CD, but in a reverse position, and having respectively the same inclinations to CD; then A since the angles at B do not exceed two right angles, (last Cor.) ACE > ABE,

C

ABE is concave towards D,
and therefore AC

B

E

D

and hence

AB.

COR. 3.-Hence the circumference of a circle is greater than the perimeter of any inscribed polygon, and less than that of any one circumscribing it.

PROPOSITION III. THEOREM.

If from the greater of two unequal magnitudes there be taken away its half, and from the remainder its half, and so on, there shall 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.

and let DE be

equal to C.

and from the

K

A

D

(Const.) For C may be multiplied so as at length to become greater than AB. Let it be so multiplied, and let DE its multiple be greater than AB, divided into DF, FG, and GE, each From AB take BH equal to its half, 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, and HB; and the divisions in ED be DF, FG, and GE. (Dem.) And because DE is greater than AB, and that 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

H

B C E

GD is greater than HA, and that GF is not
greater than the half of GD,
the half of HA;
the remainder AK;
greater than AK;

but HK is equal to
remainder FD is greater than

therefore the
and FD is equal to C;

that is, AK is less than C.

therefore C is

PROPOSITION IV. THEOREM.

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

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

(Const.) 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.

(Dem.) Because the chords AB, BC, CD, DE, EF, and FA are all equal, the arcs AB, BC, CD, DE, EF, and FA are also

equal (III. 28);

for the same reason, the arcs GH, HI, IK, KL, LM, and MG are all equal, and they are equal in nuni ber to the others; therefore whatever part the arc AB is of the whole circumference ABD, the same part 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 (VI. 33, Cor. 2); and the angle GOH is the same part of four right angles that the arc GH is of the circumference GHK; therefore the angles ANB and GOH are each of them

the same part of four right angles, and therefore they are equal to one another.

and

The isosceles triangles ANB and GOH are therefore equiangular (VI. 6), the angle ABN equal to the angle GHO; in the same manner, by joining NC, OI,

it may be proved that the angles NBC and OHI are equal to one another, and to the angle ABN; therefore the whole angle ABC is equal to the whole angle GHI; and the same may be proved of the angles BCD and 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; and because similar polygons are as the squares on their homologous sides (VI. 20), the polygon ABCDEF is to the polygon GHIKLM as the square on AB to the square on GH; triangles ANB and GOH are equiangular, is to the square on GH as the square on AN to the square on GO (VI. 4, and 22, Cor.), or as four times the square on AN to four times the square on GO; that is, as the square on AD to the square on GK; therefore also the polygon ABCDEF is to the polygon GHIKLM as the square on AD to the square on GK;

but because the the square on AB

and they have also been shewn to be similar.

COR.-Every equilateral polygon inscribed in a circle is also equiangular. 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, which are the doubles of these angles, are therefore also equal.

PROPOSITION V. PROBLEM.

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.

Given ABCDEF 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.

(Const.) 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 meet

ing GA and GB produced in K and L; KL is the side of the polygon

in H,

(Dem.) Because the arc AB is bisected

the angle AGH is equal to the

angle BGH, and because KL touches the circle in H, the angles LHG and KHG are right angles; 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 and GL is equal to GK. Again, in the triangles KGL and 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 and the angle GKL to GKN; but because the triangle KGN is isosceles, the angle GKN is equal to the angle GNK, and the angles GMK and GMN are both right angles by construction; wherefore the triangles GMK and GMN have two angles of the one equal to two angles of the other, and they have also the side GM common;8

therefore they are equal, 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 and GKM, the two sides GK and KH are equal to the two sides GK and KM, each to each; and the angles GKH, GKM are also equal, therefore GM is equal to GH; wherefore the point M is in the circumference 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 described about the circle, sides of which will each be equal to KL, and will be equal in number to the sides of the inscribed polygon; therefore KL is the side of an equilateral polygon described about the circle of the same number of sides with the inscribed polygon ABCDEF, which was to be found.

the

COR.-Because GL, GK, and GN, and the other straight lines drawn from the centre G to the angular points of the polygon

if a circle

the

and therefore

described about the circle ABD, are all equal;
be described from the centre G, with the distance GK,
polygon will be inscribed in that circle;

it is similar to the polygon ABCDEF (4).

PROPOSITION VI. THEOREM.

A circle being given, two similar polygons may be found, the one described about the circle, and the other inscribed in it, which shall differ from one another by a space less than any given space.

Let ABC be the given circle, and the square on D any given space; to prove that a polygon may be inscribed in the circle ABC, and a similar polygon described about it, that the difference between them shall be less than the square on D.

(Const.) In the circle ABC apply the straight line AE equal to D, and let AB be a fourth part of the circumference of the circle. From the circumference AB take away its half, and from the remainder its half, and so on, till the circumference AF is found less than the circumference AE. Find the centre draw the diameter AC, as also the straight lines AF and FG; and having bisected the circumference AF in K, join KG, and draw HL, touching the circle in K, and meeting GA and GF produced in H and L; join CF.

(Dem.) Because the isosceles triangles HGL and AGF have the common angle AGF, they are equiangular (VI. 6),

B

and the

angles GHK and GAF are therefore equal to one another; but the angles GKH and CFA are also equal, for they are right angles; therefore the triangles HGK and ACF are likewise equiangular.

And because the arc AF was found by taking from the arc AB its half, and from that remainder its half, and so on,

AF will be contained a certain number of times exactly in the arc AB,

and therefore it will also be contained a certain number of times exactly in the whole circumference

H

LF

KM

D

ABC; and the straight line AF is therefore the side of an equilateral polygon inscribed in the circle ABC; wherefore also HL is the side of an equilateral polygon of the same number of sides described about ABC (5). about the circle be called M, called N; then because these

Let the polygon described and the polygon inscribed be polygons are similar (5, Cor.),