Scholium. The radius is to the side of the inscribed square as 1 is to √2.

If a regular hexagon be inscribed in a circle, any side will be equal to the radius of the circle.

Let ABD be a circle, and ABCDEH a regular inscribed hexagon: then will any side, as AB, be equal to the radius of the circle.

Draw the radii OA and OB. Then will the angle AOB be equal to one-sixth of four right angles, or to two-thirds of one right angle, because it is an gle at the centre (P. II., D. 2). The sum of the two angles OAB

an

and OBA is, consequently, equal

H

E

A

to four-thirds of a right angle (B. I., P. XXV., C. 1); but, the angles OAB and OBA are equal, because the opposite sides OB and OA are equal: hence, each is equal

two-thirds of a right angle. The three angles of the triangle AOB are therefore, equal, and consequently, the triangle is equilateral hence, AB is equal to OA; which was to be proved.

PROPOSITION V. PROBLEM.

To inscribe a regular hexagon in a given circle.

Let ABE be a circle, and its centre.

Beginning at any point of

the circumference, as A, ap

ply the radius OA six times

as a chord ;

then will

triangle ACE will be equilateral (P. II., S.).

B

E

Cor. 2. If we draw the radii OA and OC, the figure AOCB will be a rhombus, because its sides are equal : hence (B. IV., P. IV., C.), we have,

AB2 + BC2 + OA2 + OC2 = AC2 + OB2;

or, taking away from the first member the quantity ÕÃ3, and from the second its equal OB, and reducing, we have,

30A2 =
2

AC2;

whence (B. II., P II.),

AC : 04

2 : 3 : 1;

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,

M

B

AO : AB :: AB : AM; 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 MOR

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 consequently, the arc AB is equal to one-tenth of the circumfer

M

A

B

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

Cor. 1. If AB 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.

PROPOSITION VII. PROBLEM.

Tɩ 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,

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 •quiangular.

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