or (B. II., P. XII., C. 2),

AC : OA :: √3: 1;

that is, the side of an inscribed equilateral triangle is to the radius, as the square root of 3 is to 1.

PROPOSITION VI. THEOREM.

If the radius of a circle be divided in extreme and mean ratio, the greater segment will be equal to one side of a regular inscribed decagon.

Let ACG be a circle, OA its radius, and AB, equal to OM, the greater segment of OA when divided in extreme and mean ratio: then will AB be equal to the side of a regular inscribed decagon.

Draw OB and BM. We

have, by hypothesis,

AO OM OM: AM; ::

or, since AB is equal to OM, we have,

AO AB :: AB : AM;

M

A

B

hence, the triangles OAB and BAM have the sides about their common angle BAM, proportional; they are, therefore, similar (B. IV., P. XX.). But, the triangle OAB is isosceles; hence, BAM is also isosceles, and consequently, the side BM is equal to AB. But, AB is equal to OM, by hypothesis: hence, BM is equal to OM, and consequently, the angles MOB

and MBO are equal.

The angle AMB being an exterior angle of the triangle OMB, is equal to the sum of the angles MOB and MBO,

or

to twice the angle MOB; and because AMB is equal to OAB, and also to OBA, the sum of the angles OAB and OBA is equal to four times the angle AOB: hence, AOB is equal to one-fifth of two right angles, or to one-tenth of four right angles; and conse

M

quently, the arc AB is equal

B

to one-tenth of the circumfer

ence : hence, the chord AB is equal to the side of regular inscribed decagon; which was to be proved.

Cor. 1. If AB be applied ten times as a chord, the resulting polygon will be a regular inscribed decagon.

Cor. 2. If the vertices A, C, E, G, and I, of the alternate angles of the decagon be joined by straight lines, the resulting figure will be a regular inscribed pentagon.

Scholium 1. If the arcs subtended by the sides of any regular inscribed polygon be bisected, and chords of the semiarcs be drawn, the resulting figure will be a regular inscribed polygon of double the number of sides.

Scholium 2. The area of any regular inscribed polygon. is less than that of a regular inscribed pclygon of double the number of sides, because a part is less than the whole.

Tu circumscribe a polygon about a circle which shall be similar to a given regular inscribed polygon.

Let TNQ be a circle, O its centre, and ABCDEF, a regular inscribed polygon.

HI to BC, the angle H is equal to the angle B. In like manner, it may be shown that any other angle of the circumscribed polygon is equal to the corresponding angle of the inscribed polygon: hence, the circumscribed polygon is equiangular.

2o. Draw the lines OG, OT, OH, ON, and OI. Then, because the lines HT and HN are tangent to the circle, OH will bisect the angle NHT, and also the angle NOT (B. III., Prob. XIV., S.); consequently, it will pass through the middle point B of the arc NBT In like manner, it

may be shown that the line drawn from the centre to the vertex of any other angle of the circumscribed polygon, will pass through the corresponding vertex of the inscribed polygon.

The triangles OHG and OHI have the angles OHG

and OHI equal, from what has just been shown; the angles GOH and HOI equal, because they are measured by the equal arcs AB and

gon being both equiangular and equilateral, is regular; and since it has the same number of sides as the inscribed polygon, it is similar to it.

Cor. 1. If lines be drawn from the centre of a regular circumscribed polygon to its vertices, and the consecutive points in which they intersect the circumference be joined by chords, the resulting figure will be a regular inscribed polygon similar to the given polygon.

Cor. 2. The sum of the lines HT and HN is equal to the sum of HT and TG, or to HG; that is, to one of the sides of the circumscribed polygon.

Cor. 3. If at the vertices A, B, C, &c., of the inscribed polygon, tangents be drawn to the circle and prolonged till they meet the sides of the circumscribed polygon, the resulting figure will be a circumscribed polygon of double the number of sides.

Cor. 4. The area of any regular circumscribed polygon

is greater than that of a regular circumscribed polygon of double the number of sides, because the whole is greater than any of its parts.

Scholium. By means of a circumscribed and inscribed square, we may construct, in succession, regular circumscribed and inscribed polygons of 8, 16, 32, &c., sides. By means of the regular hexagon, we may, in like manner, construct regular polygons of 12, 24, 48, &c., sides. By means of the decagon, we may construct regular polygons of 20, 40, 80, &c., sides.

PROPOSITION VIII. THEOREM.

The area of a regular polygon is equal to half the product of its perimeter and apothem.

Let GHIK be a regular polygon, O its centre, and OT its apothem, or the radius of the inscribed circle: then will the area of the polygon be equal to half the product of the perimeter and the apothem.

For, draw lines from the centre to the vertices of the polygon. These lines will divide the polygon into triangles whose bases will be the sides of the polygon, and whose altitudes will be equal to the apothem. Now, the area of any triangle, as OHG, is equal to half the product of the side HG

and the apothem: hence, the area.

H T

K

of the polygon is equal to half the product of the perimeter and the apothem; which was to be proved.