BOOK VI.

REGULAR POLYGONS.-MEASUREMENT OF THE

CIRCLE.

DEFINITIONS.

1. A regular polygon is one which is equilateral and equiangular.

2. A polygon of five sides is called a pentagon; of six sides, a hexagon; of eight sides, an octagon; of ten sides, a decagon; of fifteen sides, a pentedecagon, etc., etc.

3. The apothem is the line drawn from the centre of a regular polygon at right angles to one of its sides.

Proposition 1.

Problem.—To inscribe a square in a given circle.

Let ABCD be the given circle; it is required to inscribe a square in ABCD.

Draw the diameters AC, BD at right angles to each other, and join AB, BC, CD, DA; because in the triangles BEA, DEA, BE, EA and the angle BEA are equal to DE, EA and the angle DEA, each to each; therefore (I. 4) BA is equal to AD; and, for the same reason, BC, CD are each of them equal to BA or AD; therefore the quadrilateral figure

132

E

ABCD is equilateral. It is also rectangular, for the straight line BD being a diameter of the circle ABCD, BAD is a semicircle; wherefore the angle BAD is a right angle (III. 21); for the same reason each of the angles ABC, BCD, CDA is a right angle; therefore the quadrilateral figure ABCD is rectangular; therefore it is a square, and it is inscribed in the circle ABCD.

Scholium. Since the triangle AED is right-angled and isosceles, we have (I. 42, Cor. 2)

AD: AE :: V2 : 1, or AD=AE√2.

Proposition 2.

Problem. To describe a square about a given circle.

Let ABCD be the given circle; it is required to describe a square about it.

B

G

A

F

D

Draw two diameters, AC, BD, of the circle at right angles to each other, and through the points A, B, C, D draw (III. 15) FG, GH, HK, KF, touching the circle. Hence the angles at A, B, C, D are right angles (III. 14, Cor. 2); and because the angle A EB is a right angle, as likewise is EBG, GH is parallel (I. 26) to AC; for the same reason AC is parallel to FK, and, in like manner, GF, HK may be shown to be parallel to BED; therefore the figures GK, GC, AK, FB, BK are parallelograms, and GF is therefore equal (I. 32) to HK, and GH to FK; and because AC is equal to BD, and also to each of the two GH, FK, and BD to each of the two GF, HK, therefore GH, FK are each of them equal to GF or HK; therefore the quadrilateral figure FGHK is equilateral. It is also rectangular; for GBEA being a parallelogram, and AEB a right angle, AGB (I. 32)

H

C

K

is likewise a right angle; in the same manner, it may be shown that the angles at H, K, F are right angles; therefore the quadrilateral figure FGHK is rectangular, and therefore a square.

Proposition 3.

Problem.-To inscribe a circle in a given regular polygon.

Let ABCD be a regular polygon; it is required to inscribe a circle in it.

A

F

B

Ꭰ

K

Bisect any two adjacent angles, A and B, of the polygon by the lines AE and BE, cutting each other in E. From E draw EF, EG at right angles to AB, BC; join EC. Then, because in the two triangles FBE, GBE, EB is common, the angle EBF equal to the angle EBG, and the angles at F and G right angles, therefore EF is equal (I. 24) to EG. Also in the triangles EBA, EBC, EB, BA and the included angle EBA, are equal to EB, BC and the included angle EBC; therefore the angle EAB is equal (I. 4) to ECB. But EAB is one half an angle of the polygon, therefore ECB is also. In the same way a line from E to any angle of the polygon will bisect that angle. Therefore, in the same way that EF, EG were proved equal, all lines drawn from E perpendicular to the sides of the polygon are equal to one another, and a circle described with the centre E and radius equal to one of them, will pass through the extremities of the others, and the lines of the polygon will touch this circle (III. 14, Cor. 2). Hence the circle will be inscribed in the polygon (III. Def. 12).

Corollary. From this it is manifest that the apothem is the same, no matter what side it is drawn to, and bisects the angle made by lines from the centre to the angles of the polygon.

Proposition 4.

Problem.-To describe a circle about a given regular polygon.

Let ABC be a regular polygon; it is required to describe a circle about it.

F

B

Bisect two adjacent angles, A and B, by the lines EA, BE meeting in E; join ED; because the equal angles at A and B are bisected, therefore EAB is equal to EBA, and (I. 23) EA to EB. Also, because in the two triangles EBA, EBD the two sides EB, BA and included angle EBA are equal to EB, BD and included angle EBD, the side EA is equal (I. 4) to the side ED. In the same manner all lines

drawn from E to the angles of the polygon are equal, and a circle from the centre E, at the distance of any one of them, will pass through the extremities of the others, and will be described about the polygon.

Corollary 1.-From this and the last proposition it is evident that, The centres of the circles inscribed in a regular polygon and described about it, are in the same point.

Corollary 2.-Also, that if the radii EA, EB, etc. be produced, they will also bisect the angles of a circumscribed polygon of the same number of sides, which will touch the circle at F, G, etc., the extremities of radii bisecting the angles at the centre. The sides of the two polygons being parallel, the angles will be equal and the polygons similar. Hence, To draw a circumscribed polygon similar to a given inscribed polygon, bisect the angles at the centre, and from the extremities of the radii draw lines touching the circle.

Proposition 5.

Problem. To describe an isosceles triangle having each of the angles at the base double the third angle.

Draw any circle BDE, and in it any radius AB; divide (V. 23) AB in C in extreme and mean ratio; lay down BD equal to AC, and join AD and CD; ABD will be the triangle required.

Because AB is divided in C in extreme and mean ratio, and BD is equal to AC, therefore,

AB BD BD : BC;

:

hence the two triangles ABD, DBC, having a common angle B, have the sides about that angle proportional; they are therefore similar (V. 6); hence the angle DCB is equal to the

E

angle ADB, or ABD, and (I. 23) CD is equal to BD or AC. Because CD is equal to CA, the angle CDA is equal to the angle CAD; but BCD is equal (I. 30) to CAD and CDA; therefore BCD is double CAD; but BCD is equal to ABD or ADB; hence each of the angles ABD, ADB is double BAD.

Proposition 6.

C

Problem. To inscribe a regular decagon in a given circle. Let ABC be a given circle; it is required to describe a regular decagon in it.

Construct (VI. 5) the isosceles triangle DAB on the radius DA,

having each of the angles at the base double of the third angle. Because each of the angles DAB,

DBA is double ADB, the two

E

A