For, the angles AOB, BOC, &c. being equal, the chords AB, BC, &c. are also equal (Prop. III. Bk. III.); and the angles ABC, BCD, &c., being inscribed in semicircles, are right angles (Prop. XVIII. Cor. 2, Bk. III.). Hence

D

ABCD is a square, and A is inscribed in the circle ABCD.

354. Cor. Since the triangle AO B is right-angled and isosceles, we have (Prop. XI. Cor. 5, Bk. IV.),

AB: A0::2:1;

hence, the side of the inscribed square is to the radius as the square root of 2 is to unity.

PROPOSITION V. THEOREM.

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

Let ABCDEF be a regular

hexagon inscribed in a circle, the centre of which is O; then any side, as B C, will be equal to the radius O A.

Join BO; and the angle at the centre, A OB, is one sixth of four

F

E

D

A

B

C

right angles (Prop. II. Sch. 1), or one third of two right angles; therefore the two other angles, OA B, OBA, of the same triangle, are together equal to two thirds of two right angles (Prop. XXVIII. Bk. I.). But A O and BO being equal, the angles OA B, OBA are also equal (Prop. VII. Bk. I.); consequently, each is one third of two right angles. Hence the triangle A OB is equiangular; therefore AB, the side of the regular hexagon, is equal to A O, the radius of the circle (Prop. VIII. Cor. Bk. I.).

356. Cor. 1. To inscribe a regular hexagon in a given circle, apply the ra

E

F

D

dius, AO, of the circle six times, as a
chord to the circumference. Hence,
beginning at any point A, and applying
AO six times as a chord to the circum- A
ference, we are brought round to the
point of beginning, and the inscribed

figure ABCDEF, thus formed, is a regular hexagon.

357. Cor. 2. By joining the alternate angles of the inscribed regular hexagon by the straight lines A C, CE, EA, the figure AC E, thus inscribed in the circle, will be an equilateral triangle, since its sides subtend equal arcs, ABC, CDE, EFA, on the circumference (Prop. III. Bk. III.).

358. Cor. 3. Join O A, O C, and the figure ABCO is a rhombus, for each side is equal to the radius. Hence, the sum of the squares of the diagonals A C, O B is equivalent to the sum of the squares of the sides (Prop. XV. Bk. IV.); or to four times the square of the radius OB; that is, A C2+O B2 is equivalent to 4 A B2, or 4 O B2; and taking away O B2 from both, there remains A C2 equivalent to 3 0 B2; hence

A C2: 0 B2 : : 3:1, or AC: OB::√3:1; hence, the side of the inscribed equilateral triangle is to the radius as the square root of 3 is to unity.

PROPOSITION VI.-PROBLEM.

359. To inscribe a regular decagon in a given

circle.

Divide the radius, O A, of the given circle, in extreme and mean ratio, at the point M (Prob. XXVII. Bk. V.).

Take the chord AB equal to OM, and AB will be the side of a regular decagon inscribed in the circle. For we have by construction,

AO: 0M::OM: AM;

or, since A B is equal to O M,

AO:AB::AB: AM.

M

B

Draw MB and BO; and the triangles ABO, AMB have a common angle, A, included between proportional sides; hence the two triangles are similar (Prop. XXIV. Bk. IV.). Now, the triangle OAB being isosceles, AMB must also be isosceles, and A B is equal to B M; but AB is equal to O M, consequently M B is equal to MO; hence the triangle MBO is isosceles.

Again, the angle AM B, being exterior to the isosceles triangle BMO, is double the interior angle O (Prop. XXVII. Bk. I.). But the angle A M B is equal to the angle MA B; hence the triangle OA B is such, that each of the angles at the base, OAB, OBA, is double the angle O, at its vertex. Hence the three angles of the triangle. are together equal to five times the angle O, which consequently is a fifth part of two right angles, or the tenth part of four right angles; therefore the arc AB is the tenth part of the circumference, and the chord A B is the side of an inscribed regular decagon.

360. Cor. 1. By joining the vertices of the alternate angles A, C, &c. of the regular decagon, a regular pentagon may be inscribed. Hence, the chord AC is the side

of an inscribed regular pentagon.

361. Cor. 2. AB being the side of the inscribed regular decagon, let AL be the side of an inscribed regular hexagon (Prop. V. Cor. 1). Join BL; then BL will be the side of an inscribed regular pentedecagon, or regular polygon of fifteen sides. For A B cuts off an arc equal to a tenth part of the circumference; and A L subtends an

arc equal to a sixth of the circumference; therefore BL, the difference of these arcs, is a fifteenth part of the cir cumference; and since equal arcs are subtended by equal chords, it follows that the chord BL may be applied exactly fifteen times around the circumference, thus forming a regular pentedecagon.

362. Scholium. If the arcs subtended by the sides of any inscribed regular polygon be severally bisected, the chords of those semi-arcs will form another inscribed polygon of double the number of sides. Thus, from having an inscribed square, there may be inscribed in succession polygons of 8, 16, 32, 64, &c. sides; from the hexagon may be formed polygons of 12, 24, 48, 96, &c. sides; from the decagon, polygons of 20, 40, 80, &c. sides; and from the pentedecagon, polygons of 30, 60, 120, &c. sides.

NOTE. For a long time the polygons above noticed were supposed to include all that could be inscribed in a circle. In the year 1801, M. Gauss, of Göttingen, made known the curious discovery that the circumference of a circle could be divided into any number of equal parts capable of being expressed by the formula 2 + 1, provided it be a prime number. Now, the number 3 is the simplest of this kind, it being the value of the above formula when the exponent n is 1; the next prime number is 5, and this is contained in the formula. But the polygons of 3 and of 5 sides have already been inscribed. The next prime number expressed by the formula is 17, so that it is possible to inscribe a regular polygon of 17 sides in a circle. The investigations, however, which establish this geometrical fact involve considerations of a nature that do not enter into the elements of Geometry.

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

Let ABCDEF be a regular polygon inscribed in a circle whose centre is 0.

Through M, the middle point of the arc A B, draw the tangent, GH; also draw tangents at the middle points of

Since M is the middle point of the arc A B, and N the middle point of the equal arc BC, the arcs B M, B N are halves of equal arcs, and therefore are equal; that is, the vertex, B, of the inscribed polygon is at the middle point of the arc MN. Draw OH; the line OH will pass through the point B. For, the right-angled triangles OMH, ONH, having the common hypothenuse OH, and the side OM equal to ON, must be equal (Prop. XIX. Bk. I.), and consequently the angle MOH is equal to HION, wherefore the line OH passes through the middle point, B, of the arc MN. In like manner, it may be shown that the line OI passes through the middle point, C, of the arc NP; and so with the other vertices.

Since GI is parallel to A B, and HI to B C, the angle G III is equal to the angle A B C (Prop. XXVI. Bk. I.) ; in like manner, HIK is equal to B CD; and so with the other angles; hence, the angles of the circumscribed polygon are equal to those of the inscribed polygon. And, further, by reason of these same parallels, we have

:

GH AB:: OH: OB, and HI: BC:: OH: OB; therefore (Prop. X. Bk. II.),

GH:AB:: HI: BC.

But A B is equal to B C, therefore G H is equal to H I. For the same reason, HI is equal to IK, &c.; consequently, the sides of the circumscribed polygon are all equal; hence this polygon is regular, and similar to the inscribed one.