PROPOSITION 14. PROBLEM.

To circumscribe a circle about a given regular pentagon.

E

Let ABCDE be the given regular pentagon :
it is required to circumscribe a circle about it.
BCD, CDE by CF, DF intersecting at F.
Join FB, FA, FE.

Bisect the

1. 9.

Hyp.

But the

.. also the

CBF the CDF.

CDF is half an angle of the regular pentagon : CBF is half an angle of the regular pentagon : that is, FB bisects the ABC.

So it may be shewn that FA, FE bisect the $ at A and E. Now the FCD, FDC are each half an angle of the given regular pentagon;

.. the FCD the FDC,

.. FC FD.

IV. Def.
I. 6.

Similarly it may be shewn that FA, FB, FC, FD, FE are all equal.

From centre F, with radius FA describe a circle: this circle must pass through the points A, B, C, D, E, and therefore is circumscribed about the given pentagon.

Q. E. F.

In the same way a circle may be circumscribed about any regular polygon.

PROPOSITION 15. PROBLEM.

To inscribe a regular hexagon in a given circle.

B

H

Let ABDF be the given circle:

it is required to inscribe a regular hexagon in it.

Find G the centre of the
and draw a diameter AGD.

ABDF;

III. 1.

From centre D, with radius DG, describe the C EGCH. Join CG, EG, and produce them to cut the

given circle at F and B.

Join AB, BC, CD, DE, EF, FA.

ce of the

Then ABCDEF shall be the required regular hexagon.

Now GE=GD, being radii of the

and DG = DE, being radii of the

ACE;

EHC:

.. GE, ED, DG are all equal, and the AEGD is equilateral. Hence the EGD = one-third of two rt. angles.

I. 32.

Similarly the DGC = one-third of two rt. angles. But the EGD, DGC, CGB together two rt. angles; 1. 13. .. the remaining CGB = one-third of two rt. angles. .. the three ▲ EGD, DGC, CGB are equal to one another. And to these angles the vert. opp.

are respectively equal:

S

S

BGA, AGF, FGE

.. the EGD, DGC, CGB, BGA, AGF, FGE are all equal; .. the arcs ED, DC, CB, BA, AF, FE are all equal;

III. 26. .. the chords ED, DC, CB, BA, AF, FE are all equal: III. 29. .. the hexagon is equilateral.

Again the arc FA= the arc DE:

to each of these equals add the arc ABCD;

Proved.

then the whole arc FABCD = the whole arc ABCDE:

hence the angles at the Oce which stand on these equal arcs are equal,

that is, the FED = the AFE.

III. 27.

In like manner the remaining angles of the hexagon may be shewn to be equal.

.. the hexagon is equiangular:

..the hexagon is regular, and it is inscribed in the

ABDF.

Q. E. F.

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

PROPOSITION 16. PROBLEM.

To inscribe a regular quindecagon in a given circle.

B

E

Let ABCD be the given circle:

it is required to inscribe a regular quindecagon in it.

In the

ABCD inscribe an equilateral triangle,

and let AC be one of its sides.

In the same circle inscribe a regular pentagon, and let AB be one of its sides.

Then of such equal parts as the whole the arc AC, which is one-third of the and the arc AB, which is one-fifth of the

Iv. 2.

IV. 11.

ce contains fifteen,

ce, contains five; e, contains three; .. their difference, the arc BC, contains two.

Bisect the arc BC at E:

III. 30.

then each of the arcs BE, EC is one-fifteenth of the ce.

.. if BE, EC be joined, and st. lines equal to them be placed successively round the circle, a regular quindecagon will be inscribed in it.

H. E.

Q. E. F.

NOTE ON REGULAR POLYGONS.

The following propositions, proved by Euclid for a regular pentagon, hold good for all regular polygons.

1. The bisectors of the angles of any regular polygon are con

current.

Let D, E, A, B, C be consecutive angular points of a regular polygon of any number of

sides.

8

Bisect the EAB, ABC by AO, BO, which intersect at O.

Join EO.

D

It is required to prove that EO bisects the DEA.

Because

For in the A" EAO, BAO,

A

EA BA, being sides of a regular polygon;
and AO is common;

and the

EAO the

BAO;

Constr.

Similarly if O be joined to the remaining angular points of the polygon, it may be proved that each joining line bisects the angle to whose vertex it is drawn.

That is to say, the bisectors of the angles of the polygon meet at the point O.

.

COROLLARIES.

Since the EAB= the ABC;

8

Q. E. D.

Hyp.

and since the OAB, OBA are respectively half of the EAB, ABC; ..the OAB the OBA.

Similarly

.. OA=OB.
OE=OA.

I. 6.

Hence The bisectors of the angles of a regular polygon are all equal: and a circle described from the centre O, with radius OA, will be circumscribed about the polygon.

Also it may be shewn, as in Proposition 13, that perpendicular drawn from O to the sides of the polygon are all equal; therefore a circle described from centre O with any one of these perpendiculars as radius will be inscribed in the polygon.

2. If a polygon inscribed in a circle is equilateral, it is also equiangular.

Let AB, BC, CD be consecutive sides of an equilateral polygon inscribed in the ADK;

then shall this polygon be equiangular. Because the chord AB= the chord DC, Hyp. ..the minor arc AB = the minor arc DC. III. 28. To each of these equals add the arc AKD:

then the arc BAKD- the arc AKDC;

.. the angles at the Oce, which stand on these equal arcs, are equal;

K

that is, the BCD the ABC. III. 27. Similarly the remaining angles of the polygon may be shewn to be equal:

.. the polygon is equiangular.

Q. E.D.

3. If a polygon inscribed in a circle is equiangular, it is also equilateral, provided that the number of its sides is odd.

[Observe that Theorems 2 and 3 are only true of polygons inscribed in a circle.

The accompanying figures are sufficient to shew that otherwise a polygon may be equilateral without being equiangular, Fig. 1; or equiangular without being equilateral, Fig. 2.]

NOTE. The following extensions of Euclid's constructions for Regular Polygons should be noticed.

By continual bisection of arcs, we are enabled to divide the circumference of a circle,

by means of Proposition 6, into 4, 8, 16,..., by means of Proposition 15, into 3, 6, 12,..., by means of Proposition 11, into 5, 10, 20,..., by means of Proposition 16, into 15, 30, 60,...,

2.2",

equal parts; 3.2", equal parts; 5.2",... equal parts; 15. 2",... equal parts.

Hence we can inscribe in a circle a regular polygon the number of whose sides is included in any one of the formulæ 2. 2", 3 . 2′′, 5. 2′′, 15. 2", n being any positive integer. In addition to these, it has been shewn that a regular polygon of 2+1 sides, provided 2" +1 is a prime number, may be inscribed in a circle.