together are four times the angle ADB, and the three angles DAB, DBA, ADB are five times the angle ADB. Therefore the angle ADB is the fifth part of two right angles; that is, the tenth part of four right angles. Therefore, ten such angles as ADB can be exactly placed about the point D (I. 17, Cor. 2). But equal angles stand on equal arcs, and equal arcs are subtended by equal straight lines; therefore AB is one side of an equilateral decagon inscribed in the circle. And because all triangles ADB, BDE are isosceles and have equal angles at D, their other angles are equal. But these angles make up the angles of the decagon; therefore the decagon is also equiangular; therefore it is regular.

Corollary. To inscribe a regular pentagon in a given circle.

Join the alternate angles of the decagon, and we have a regular pentagon. For all such triangles as ABE, EFC have their two sides and included angle at B and F equal in each; hence their third sides are equal, and their angles are equal. But the two angles AEB, CEF, taken from the angle of the decagon, will leave the angles of the pentagon. Therefore the pentagon is equilateral and equiangular.

Proposition 7.

Problem.—To inscribe a regular hexagon in a given circle.

Let ABCD be a circle; it is required to describe a regular hexagon in it.

Draw any radius ED. From the centre D and radius DE describe a circle cutting the given circle in A and C.

Join AE, CE, and produce them, and also DE, to meet the circumference in G, F and B; join the

B

extremities of these lines; the figure formed is a regular

=

D

B.

hexagon. Because DA DE AE, the triangle AED is equiangular (I. 23, Cor.). Therefore (I. 30), the angle AED is one third of two right angles. For the same reason DEC is one third of two right angles. Therefore the whole angle AEC is two thirds of two right angles, and the angle CEG is (I. 15) one third of two right angles. Hence the vertical angles AEF, FEB, BEG are also each equal (I. 17) to one third of two right angles. But equal angles stand on equal arcs, and equal arcs are subtended by equal straight lines; hence, DA, AF, FB, BG, GC, CD are all equal, and the hexagon is equilateral. It is also equiangular, for the arc DA is equal to the arc FB; add to both the semicircle BCD, and the whole arc BCA is equal to the whole arc FGD. But equal angles stand on equal arcs; hence the angle AFB is equal to the angle DAF. In the same manner all the other angles may be proved equal, and the hexagon is equiangular. It is therefore regular.

Corollary. A side of a regular hexagon is equal to the radius of the circumscribed circle.

Hence, to construct a hexagon, describe a circle, and lay down the radius six times around it.

Proposition 8.

Problem. To inscribe a regular pentedecagon in a given

circle.

Let AB be one side of an equilateral triangle inscribed in a circle (III. 26), and AC one side of a regular penta

A

gon (VI. 6, Cor.). If the whole circumference were to be divided into 15 equal parts, AB would subtend 5, and AC, 3. Therefore BC must contain 2; bisect (III. 23) BC in D, and lay off BD around the circle; we would have a regular pentedecagon.

Scholium.-If we bisect (III. 23) the arcs of a circle which are subtended by the sides of a square, we may form a regular polygon of 8 sides; from this, one of 16, and so on. Similarly from a decagon, we can construct polygons of 20, 40, etc., sides; from a hexagon, polygons of 12, 24, etc.; from a pentedecagon, figures of 30, 60, etc.

Proposition 9.

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

Let ABC be a polygon and DE its apothem; then the area of ABC is equal to

(AB+BF+FC+ CG+ GH
+HA)ED.

Join DA, DB, etc.; the area of
the triangle DAB is equal to (V.
16, Cor. 2) AB× ED. But the
apothem is the same for all sides;
hence the area of the whole polygon

or

Не

=AB× ED+BF× ED+etc.

A E B

G

(AB+BF+FC+ CG+ GH+HA)ED.

Proposition 10.

F

Theorem.—About a given circle a polygon may be described, which shall differ from a similar inscribed polygon, by a space less than any given space.

Let ABC be a circle; similar polygons may be described

about ABC, and inscribed in it, whose difference shall be less

than any given space.

Let AB be a side of a regular inscribed and EF of a similar circumscribed polygon (VI. 4, Cor. 2) having any number of sides, F

L

PM

E

H

K

are similar, and (III. 15, Sch.) OP-PH,.. ALPH=AMPO. Similarly AKAR = ^RMO; add to both ARPH; then KAHL= MAH.

Now the difference between the triangles NAB, NEF is the trapezoid ABFE. Hence the difference between the inscribed and circumscribed polygons of n sides is

n× ABFE, or 2n × A GHE.

Also the difference between the inscribed and circumscribed polygons of 2n sides is

2nx KAHL, or 2n x MAH.

Because EAM is a right angle, EM>AM. Hence (III. 15, Sch.) EM>HM. Hence (V. 1, Cor. 2) AEMA>^MAH. Also, since (I. 32) ▲MAH = ^SAH, ^AHG>^MAH. Therefore AGHE is divided into three triangles, EMA, AMH, AHG, of which MAH is the least; therefore MAH is less than one-third AGHE.

Hence

2nx MAH< 2n × AGHE.

Therefore the difference between the polygons of 2n sides is less than one-third the difference between the polygons of n

sides. For 4n sides the difference would be less than one-ninth the original difference, and so on. By increasing the number of sides sufficiently the difference may be made less than any given space.

Corollary.—A circle may be regarded as a regular polygon of an infinite number of sides, and whatever is true of one is true of the other.

For the circumference of the circle lies between the perimeters of the polygons. Since each time the number of sides is doubled, the difference between the polygons becomes less than one-third its former amount; when the number of sides becomes infinite the difference becomes infinitely small, and the three perimeters may be regarded as coinciding. From these considerations we will hereafter speak of the circumference of a circle being the perimeter of a regular polygon of an infinite number of sides; the area of a circle being the area of such an inscribed polygon, and the radius of a circle being its apothegm.

Proposition 11.

Theorem.—The circumferences of circles are proportional to their radii.

Let ABC, DEF be two circles, and AG, DH their radii; their circumferences are proportional to AG, DH.

In ABC, DEF describe the polygons ABC, DEF of the same number of sides,

same number of sides, the angles of one are equal to the angles