to be inscribed in the triangle; and the triangle is circumscribed about the circle.

135. It will be readily seen that no circumference passing through the three points A, B, C, can also pass through the point E (fig. 73). A quadrilateral, there- Fig. 73. fore, cannot always be inscribed in a circle. It may be well to inquire under what circumstances a quadrilateral may be inscribed in a circle. If we consider the inscribed quadrilateral ABCD, we shall perceive that the angle ADC has for its measure half the arc ABC; if the angle ADC were less than it is, the others remaining the same, its vertex would be beyond the circumference; if it were greater, the vertex would be within the circumference; it must therefore be such as to be measured by the arc ABC. So also the three points A, D, C, remaining the same, the angle at B must be such as to have for its measure half the arc ADC. But these two angles have for their measure half of an entire circumference ; their sum, therefore, must be equal to two right angles. In the same manner we may show that the sum of the other two angles is equal to two right angles; whence we infer, that—A quadrilateral may be inscribed in a circle, when its opposite angles are supplements of each other.

A square, therefore, and any other rectangle, may be inscribed in a circle; but no other parallelogram.

136. In polygons, generally, of a greater number of sides, it would be more difficult to ascertain whether they could be inscribed.

Suppose, however, we take a certain number of equal lines, as AB, BC, CD and DE (fig. 75), and dispose Fig. 75. them as in the figure, so as to make equal angles whose vertices are at B, C, D. The circumference of only one circle can be drawn through the three points A, B, C; the relative position of these points, therefore, determines the curvature. But on account of the equality of the lines and angles, the relative position of any three contiguous vertices will be the same; the same curvature, therefore, would be required throughout, and the circular arc which passes through the three vertices A, B, C, if produced will pass through the other vertices. We say, therefore, that-Any polygon which has its sides equal and its angles equal, may be inscribed in a circle.

Remark. As the circumference of a circle may be divided into unequal parts, and chords may be drawn to

to the points of division, forming inscribed polygons; it will be perceived that there is an endless variety of polygons which may be inscribed in a circle.

137. If we construct the isosceles triangle ABC (fig. Fig. 76. 76), the angle ACB being any measure of four rightangles; the perpendicular C a will bisect the side AB and measure the distance of the point C from this side. If this triangle be made to revolve about the side BC, till the point A comes again into the same plane, it will occupy the place of the isosceles triangle BCA', equal to ABC; the perpendicular Ca will fall upon C a', which is therefore equal to C a; we have also the angle ABA' double the angle ABC, the angle at the base of the triangle. If we now revolve this triangle about the line CA till B meets the plane of the paper in B', we shall have C a falling upon C a" perpendicular to A' B' and therefore measuring the distance of the point C from this line; we have a polygonal angle BA'B' equal to twice the angle at the base of the isosceles triangle, and therefore equal to ABA'. If we continue this process, as the angle ACB is a measure of four right-angles, the triangle after a certain number of turns, will have occupied all the angular space about C, bringing one of its sides to the line AC, whence the triangle first departed. We shall therefore have the polygon ABA'B'A"B" of which all the sides are at the same distance from C; if, therefore, from the point C, with a radius equal to this distance Ca, we describe a circumference, it will touch each of the sides of the polygon; the polygon is therefore circumscribed about the circle. If we examine this polygon we shall see that the sides are all equal, and all the angles at the vertices are equal, being each equal to twice the angle at the base of the isosceles triangle used in generating this polygon. We say therefore-A polygon, whose sides are equal and whose angles are equal, may be circumscribed about a circle. Because CA, CB, CA', &c. are sides of the same isosceles triangle, they are all equal; and if from C as a centre with the radius CA, we describe a circumference, it will pass through the vertices of the polygon. The polygon will then be inscribed.

Which of the several classes of parallelograms (62) can be inscribed in a circle?

Which of the several kinds of parallelograms can be circumscribed about a circle ?

138. In the polygons whose sides are equal and angles equal, whichever side be taken for the base, the base will be of the same magnitude, the angles at the base will be the same, and the opposite vertices will be determined by equal triangles similarly situated. On account of this regularity of their parts, these figures are called regular polygons. We have seen (136, 137) that polygons whose sides are equal and angles equal, may be iuscribed in a circle, and also circumscribed about a circle. We say, therefore-Regular polygons may be inscribed in a circle, and also circumscribed about a circle.

139. In the regular hexagon ABA'B'A"B" (fig. 76), Fig. 76. the point C is called the centre of the polygon; the equal angles ACB, BCA', &c. are called angles at the centre. As regular polygons of the same number of sides must have equal angles at the centre (15), they will be composed of the same number of similar isosceles triangles (46), and will therefore be similar polygons (93), and as the sides of the isosceles triangles of which they are composed, are the radii of the circumscribed circles, and the heights of these triangles, the radii of the inscribed circles; it follows (92) that-In regular polygons of the same number of sides, the perimeters are to each other as the radii of the circumscribed circles, and also as the radii of the inscribed circles.

140. Each angle of an equilateral triangle is an angle. of 60° (40); the chord of 60° is equal to radius. Therefore (fig. 77)—To draw an equilateral triangle, Fig. 17. describe a circular arc with a radius equal to the given side, take a chord equal to this radius, and draw radii from the centre to the extremeiies of this chord.

141. It will be perceived that, as radius is equal to the chord of 60°, it may be applied six times to the circumference. This will give us a regular, inscribed hexagon, as ABDEFG. If we draw chords to the alternate angles of the hexagon, we shall have an inscribed equilateral triangle. And if we bisect each of the arcs subtended by the sides of the hexagon, and draw chords to the same arcs, we shall have a regular polygon of twelve sides, called a dodecagon. By continuing the process, we obtain regular inscribed polygons of 24, 48, 96, &c. sides. 142. If we draw two diameters perpendicular to each other (fig. 78) and draw the chords (AB, BC, &c.) to their extremities, we shall have an inscribed quadrilat

Fig. 78.

Fig. 78. eral, the sides of which are equal. We perceive that this figure is composed of four right-angled triangles, which are also isosceles; the acute angles therefore are each equal to half a right-angle; but each angle of the inscribed quadrilateral is composed of two of these equal angles, and is therefore a right-angle; consequently the figure is an inscribed square.

Fig. 79.

If we bisect the arcs AB, BC, &c. and draw the chords of these semi-arcs, we shall have an inscribed octagon. And by continuing the process of subdividing the arcs and drawing chords to them, we shall have inscribed figures of 16, 32, 64, &c. sides.

143. In the right-angled triangle AOB, we have (101) (AB) 2 (AO)2 + (OB)22, or (AB)2=2(AO)2, as AÓ is equal to OB; and if we extract the second root of both sides of this equation, we shall have AB= A0 × (2)3'; or considering the radius AO as unity or one, it gives AB (2). This is the side of the inscribed square. It is always the diagonal of a square whose side is AO, equal to unity. Now it is known that there is no numerical expression for this value (2) independent of the radical form; and as this quantity is incommensurable with unity, it follows that the diagonal is incommensurable with the side of a square (131). We see, however, that notwithstanding this incommensurableness, the geometrical process gives us their ratio exactly. The ratio of the diagonal to the side of the square, is (2)3.

1

2

2

2

The side of the inscribed equilateral triangle is obtained from figure 77, thus: (EB) (EA) (AB)2; but as AB is equal to radius or one, and EA equals twice radius or two, we have (EB)2 4-1=3; therefore EB = (3)

1

The side of the regular inscribed hexagon equals radius, or one. Having the side of an inscribed polygon, we have its perimeter, and radius entering into the expression, we have the ratio of the perimeter to radius.

144. If in the circle ABFGH, &c. (fig. 79) we divide the radius AC into extreme and mean ratio at E, EC being the larger part, take the chord AB equal to EC, and join EB and BC; the result of this division of AC EC ; but substituting for EC its

radius gives

EC

AE

equal AB, we have

AC AB

AB AE

We have, therefore,

in the two triangles ABC, ABE, the angle A common and the sides about this angle proportional; they are therefore similar (91); but as ABC is an isosceles triangle, ABE must be an isosceles triangle, and EB is equal to AB or to EC; the triangle EBC is therefore isosceles, and the angle ECB, which, on account of the similar triangles, is equal to ABE, is also equal to EBC; consequently each of the angles at the base of the isosceles triangle ABC is double the angle at the vertex; whence the angle ACB is of two right angles or of four right angles; the arc AB which subtends this angle, must therefore be of the circumference; and the chord AB is the side of a regular inscribed decagon.

1

ΤΟ

By joining the alternate vertices BG, GI, &c. of the regular inscribed decagon, we have the regular inscribed pentagon. And by bisecting the arcs and drawing chords, we obtain regular inscribed polygons of 20, 40, 80, &c. sides.

145. The arc subtended by the side of a regular inscribed decagon, is one tenth of the circumference; and the arc subtended by the side of the regular inscribed hexagon, is one sixth of the circumference. Let the arc BP be one sixth of the circumference; if we subtract from this the arc BA, which is, of the circumference, we shall have the arc AP: 1 of the circumference; and the chord AP will be the side of a regular inscribed polygon of 15 sides. By the process of bisecting the arcs, we obtain polygons of 30, 60, &c. sides.

6

10

1 10

15

146. If we could divide any arc into three equal parts, we could obtain successively the arcs subtended by the sides of regular inscribed polygons of 9, 18, 27, 36, 45, &c. sides; but this division refers itself to the famous problem of the trisection of the angle, whose solution was so much sought by the elder geometers; "being in point of difficulty, or rather perhaps of impossibility, on a footing with the other two celebrated problems, viz. the duplication of the cube, and the quadrature of the circle." Elementary geometry has not yet furnished us with the means of solving this problem.

147. Having a regular inscribed polygon of any number of sides, we may circumscribe about the circle a regular polygon of the same number of sides, by drawing tan