82

TRIANGLE EQUIVALENT TO POLYGON. [BOOK VI.

point A, draw the diagonal AE, and from F, draw FG

For the triangles AEF, AEG, have the same base AE, and the same altitude, since their vertices F, G, are in the same line FG parallel to the base; therefore these triangles are equivalent (B. V. Prop. 3, Cor. 1). Now if to the triangle AEF we add the figure ABCDE, there will result the polygon ABCDEF; and if to the triangle AEG we add the same figure, there will result the polygon GBCDE, hence these polygons are equivalent (B. I. Ax. 4).

Now, having reduced our original polygon of six sides, to its equivalent one GBCDE, which has but five sides, proceed as before; that is, draw GD, and from E draw EI parallel to GD, cutting the base produced in I; thén join ID, and because the triangles GDE, GDI,are equivalent, if we add each of them to GBCD, we shall find IBCD equivalent to GBCDE, therefore to ABCDEF. Proceed in like manner with the angle C and we shall find the triangle IHD equivalent to the polygon ABCDEF.

Scholium. We have already seen that every triangle may be changed into an equivalent square (Prob. 6.), and as we can find a triangle equivalent to any polygon, we can always find a square equivalent to a given polygon; this is called squaring the polygon, or finding the quadrature of it.

BOOK VII.

DEFINITIONS.

1. POLYGONS of five sides are called pentagons; those of six sides, hexagons; of seven, heptagons; of eight, octagons, &c.

2. Polygons which are at once equilateral and equiangular, are called regular polygons.

PROP. I. THEOREM.

Two regular polygons of the same number of sides, are similar.

Let the figures in the margin be two regular polygons, having the same number of sides.

The sum of all the

each figure, the sum of all the angles in each must be the same. Now since all the angles are equal (Def. 2), any one angle is found by dividing the sum of all the angles by the number of angles; hence it follows that the angles of the two polygons are equal, each to each.

Again, since the sides in each polygon are all equal, it follows that AB: BC:: ab: bc; and so for all the sides; hence these polygons, having the angles of the

one respectively equal to the angles of the other, and the sides including the equal angles proportional, are similar (B. VI. Def. 1).

PROP. II. THEOREM.

The side of a regular hexagon, inscribed in a circle, is equal to the radius of that circle.

Let ABCDEF be a regular hexagon. Now since the

Now the inscribed angle ADE, standing on the arc AFE, is equal to BED, standing on the equal arc BCD; and both are equal to the angle at the centre EOD, standing on the arc ED, half of AFE, or BCD (B. II. Prop. 2, and Cor. 2); hence the triangle EOD, having all its angles equal, is equilateral (B. I. Prop. 7, Cor.), and ED is equal to EO.

Scholium 1. Hence in order to inscribe a regular hexagon in a circle, apply the radius six times round the circumference. By joining the alternate points by the lines AC, CE, EA, an inscribed equilateral triangle will be formed.

Scholium 2. If we bisect the arcs AB, BC, &c., and from the points A, B, C, &c., draw lines to the points of bisection, it is evident that a regular inscribed figure will be formed, having twelve sides; and that the surface of this figure will be more nearly equal to that of the circle, than is the surface of the hexagon.

BOOK VII.] INSCRIBED HEXAGON AND SQuare. 85

Again, if we bisect the arcs of our twelve-sided figure, one of twenty-four sides will be formed, which will approach still nearer to the circle than the last; therefore we may conclude generally, that the greater number of sides, which a regular inscribed figure has, the more nearly will its surface approach to that of the circle.

PROP. III. PROBLEM.

To inscribe a square in a given circle.

Draw the diameters AC, BD cutting each other at

right angles; join their extremities A, B, C, D ; the inscribed figure formed will be a square. For the angles AOD, DOC, AOB, BOC, being equal, the chords AD, DC, CB, BA, are equal (B. II. Prop. 3, Cor. 1);

Fig. 90.

D

B

Q

and the angles ADC, DCB, CBA, BAD, being each subtended by a semi-circumference, are right angles (B. II. Prop. 2, Cor. 3).

PROP. IV. PROBLEM.

A regular inscribed polygon being given, to circumscribe a similar one about the same circle.

Let ABCDEF be the regular inscribed polygon. Bisect the arcs AB, BC, &c., in the points S, R, &c., and at these points draw GH, HI, &c., tangents to the circle; these lines will be

S

GKA

Fig. 91.

H

R I

K

respectively parallel to AB, BC, &c. (because GH, AB, &c. are perpendicular to OS, &c.)

Now these tangents, by their intersections, will form a circumscribed polygon GHIKLP, which we have to prove will be similar to the inscribed one.

Draw from H, the point where the tangents HG, HI, intersect, the line HO; we shall show that it will pass through B. The arcs BRC, BSA, being equal, their halves BR, BS, must be equal, and consequently the

Fig. 91.

H R

I

P

L

K

point B is the middle of the arc SR; now the right-angled triangles SOH, HOR, having the sides OS, OR, equal, and OH common, are equal (B. I. Prop. 20), and consequently the angle SOH=HOR; hence the line from H to O bisects the arc SBR, and must therefore pass through the point B.

Since the lines GH, HI, are respectively parallel to AB, BC, the angles GHI, ABC,are equal (B. I. Prop. 8); in the same manner all the angles of the circumscribing polygon are equal to the angles of the inscribed one, and since these latter are all equal to each other, the former are so also.

The triangles GOH, AOB,being equiangular, we have (B. V. Prop. 15)

OH OB:: HG: BA;

and the triangles OHI, OBC, being equiangular,

hence (B. IV. Prop. 4, Cor. 1)

HG BA: HI: BC;

but BA is equal to BC, hence (B. IV. Prop. 8, Cor.)