doubles of equals, are equal. In the same manner, it is proved that all the angles of the polygon FDE, etc., are equal, and that all its sides are equal. Therefore, it is a regular polygon.

REGULAR POLYGONS SIMILAR.

449. Theorem.—Regular polygons of the same number of sides are similar.

Since the polygons have the same number of sides, the sum of all the angles of the one is equal to the sum of all the angles of the other (423). But all the angles of a regular polygon are equal (439). Dividing the equal sums by the number of angles (7), it follows that an angle of the one polygon is equal to an angle of the other.

Again all the sides of a regular polygon are equal. Hence, there is the same ratio between a side of the first and a side of the second, as between any other side of the first and a corresponding side of the second. Therefore, the polygons are similar (433).

450. Corollary. The areas of two regular polygons of the same number of sides are to each other as the squares of their homologous lines (436).

451. Corollary.-The ratio of the radius to the side of a regular polygon of a given number of sides, is a constant quantity. For a radius of one is to a radius of any other, as a side of the one is to a side of the other (434). Then, by alternation (19), the radius is to the side of one regular polygon, as the radius is to the side of any other regular polygon of the same number of sides.

452. Corollary.-The same is true of the apothem and side, or of the apothem and radius.

PROBLEMS IN DRAWING.

453. Problem.-To inscribe a square in a given circle. Draw two diameters perpendicular to each other. Join their extremities by chords. These chords form an inscribed square.

For the angles at the center are equal by construction (90). Therefore, their intercepted arcs are equal (197), and the chords of those arcs are the sides of a regular polygon (446).

454. Problem.-To inscribe a regular hexagon in a circle.

Suppose the problem solved and the figure completed. Join two adjacent angles with the center, making the triangle ABC.

A

B

Now, the angle C, being measured by one-sixth of the circumference, is equal to one-sixth of four right angles, or one-third of two right angles. Hence, the sum of the two angles, CAB and CBA, is two-thirds of two right angles (256). But CA and CB are equal, being radii; therefore, the angles CAB and CBA are equal (268), and each of them must be one-third of two right angles. being equiangular, is equilateral (276). inscribed regular hexagon is equal to the radius of the circle. The solution of the problem is now evident-apply the radius to the circumference six times as a chord.

Then, the triangle ABC,
Therefore, the side of an

455. Corollary-Joining the alternate vertices makes an inscribed equilateral triangle.

456. Problem.-To inscribe a regular decagon in a given circle.

Divide the radius CA in extreme and mean ratio, at the point B. Then BC is equal to the side of a regular inscribed decagon. That is, if we apply BC as a chord, its arc will be one-tenth of the whole circumference.

Take AD, making the chord AD equal to BC. Then join DC and DB.

Then, by construction, CA: CB :: CB: BA.

Substituting for CB its equal DA,

CA DA DA: BA.

Then the triangles CDA and BDA are similar, for they have those sides proportional which include

the common angle A (317). But the triangle CDA being isosceles, the triangle BDA is the same. Hence, DB is equal to DA, and also to BC. Therefore, the angle C is equal to the angle BDC (268). But it is also equal to BDA. It follows that the angle

B

JA

CDA is twice the angle C. The angle at A being equal to CDA, the angle C must be one-fifth of the sum of these three angles; that is, one-fifth of two right angles (255), or one-tenth of four right angles. Therefore, the arc AD is one-tenth of the circumference (207); and the chord AD is equal to the side of an inscribed regular decagon.

45%.—Corollary.-By joining the alternate vertices of a decagon, we may inscribe a regular pentagon.

458. Corollary.-A regular pentedecagon, or polygon of fifteen sides, may be inscribed, by subtracting the arc subtended by the side of a regular decagon from the arc subtended by the side of a regular hexagon. The remainder is one-fifteenth of the circumference, for-16=18.

459. Problem.-Given a regular polygon inscribed in a circle, to inscribe a regular polygon of double the number of sides.

Divide each arc subtended by a given side into two equal parts (194). Join the successive points into which the circumference is divided. The figure thus formed is the required polygon.

460. We have now learned how to inscribe regular polygons of 3, 4, 5, and 15 sides, and of any number that may arise from doubling either of these four.

The problem, to inscribe a regular polygon in a circle by means of straight lines and arcs of circles, can be solved in only a limited number of cases. It is evident that the solution depends upon the division of the circumference into any number of equal parts; and this depends upon the division of the sum of four right angles into aliquot parts.

461. Notice that the regular decagon was drawn by the aid

of two isosceles triangles composing

a third, one of the two being similar to the whole. Now, if we could combine three isosceles triangles in this manner, we could draw a regular polygon of fourteen, and then one of seven sides.

However, this can not be done by means only of straight lines and arcs of circles.

The regular polygon of seventeen sides has been drawn in more than one way, using only straight lines and arcs of circles. It has also been shown, that by the same means a regular polygon of two hundred and fifty-seven sides may be drawn. No others are known where the number of the sides is a prime number.

462. Problem.-Given a regular polygon inscribed in a circle, to circumscribe a similar polygon.

The vertices of the given polygon divide the circumference into equal parts. Through these points draw tangents. These tangents produced till they meet, form the required polygon (448).

EXERCISES.

463.-1. First in right angles, and then in degrees, express the value of an angle of each regular polygon, from three sides up to twenty.

2. First in right angles, and then in degrees, express the value of an angle at the center, subtended by one side of each of the same polygons.

3. To construct a regular octagon of a given side.

4. To circumscribe a circle about a regular polygon.

5. To inscribe a circle in a regular polygon.

6. Given a regular inscribed polygon, to circumscribe a similar polygon whose sides are parallel to the former.

7. The diagonal of a square is to its side as the square root of 2 is to 1.

A PLANE OF REGULAR POLYGONS.

464. In order that any plane surface may be entirely covered by equal polygons, it is necessary that the figures be such, and such only, that the sum of three or more of their angles is equal to four right angles (92).

Hence, to find what regular polygons will fit together so as to cover any plane surface, take them in order according to the number of their sides.

Each angle of an equilateral triangle

is equal to one-third of two right angles. Therefore, six such angles exactly make up four right angles; and the equilateral triangle is such a figure as is required.

465. Each angle of the square is a right angle, four of which make four right angles. So

that a plane can be covered by equal squares.

One angle of a regular pentagon is the fifth part of six right angles. Three of these are less than, and four exceed

four right angles; so that the regular pentagon is not such a figure as is required.

466. Each angle of a regular hexagon is one-sixth of eight right angles. Three such make up four right angles. Hence, a plane may be covered with equal regular hexagons. This combination is remarkable as being the one adopted by bees in forming the honeycomb.

467. Since each angle of a regular polygon evidently increases when the number of sides increases, and since three angles of a regular hexagon are equal