square of AC, which is, by construction, equivalent to the given area. Also, the difference of the lines CE, CD is equal to DE or AB.

PROBLEM XXIX.

To find two straight lines having the same ratio as the areas of two given polygons.

Since any two polygons can always be transformed into squares, this problem requires us to find two straight lines in the same ratio as two given squares.

Draw two lines, AC, BC, at right angles with each other, and make AC equal to a side of one of the given squares, and BC equal to a side of the other given square. Join AB, and from C draw CD perpendicular to AB. Then (B. IV., Pr. 11, Cor. 2) we have AD: DB:: AC2: CB2.

D

Therefore AD, DB are in the ratio of the areas of the given polygons.

PROBLEM XXX.

To find a square which shall be to a given square in the ratio of two given straight lines.

Upon a line of indefinite length, take AB equal to one of the given lines, and BC equal to the other line. Upon AC as a diameter describe a semicircle, and at B erect the perpen-A dicular BD, cutting the circumference in D.

E

D

B

Join DA, DC; and upon DA, or DA produced, take DE equal to a side of the given square. Through the point E draw EF parallel to AC; then DF is a side of the required square.

For, because EF is parallel to AC (B. IV., Pr. 16), we have
DE: DF:: DA: DC;

whence (B. II., Pr. 11) DE2: DF2 :: DA2: DC2.

Also, because ADC is a right-angled triangle (B. IV., Pr. 11), we have DA2: DC2::AB: BC.

Therefore the square described on DE is to the square described on DF in the ratio of the two given straight lines.

PROBLEM XXXI.

To construct a polygon similar to one given polygon, and equiv alent to another given polygon.

Let P and Q be two given polygons. It is required to construct a polygon similar to P, and equivalent to Q.

Find M, the side of a square

pro

equivalent to P (Pr. 24, Schol.), and N, the side of a square equivalent to Q. Let AB be one side of P, and let CD be a fourth portional to the three lines M, N, AB. Upon the side CD homologous to AB, construct the polygon P' similar to P (Pr. 26); it will be equivalent to the polygon Q.

Hence P'Q. Therefore the polygon P' is similar to the polygon P, and equivalent to the polygon Q.

B

C

PROBLEM XXXII.

To draw a common tangent to two given circles.

a

Let C and c be the centres of the two given circles. With C as a centre, and a radius CB equal to the difference of the two given radii CA and ca, describe a circumference, and from c draw a straight line touching the circle CB in the point B (Prob. 14). Join CB, and produce it to meet the given circumference in A. Draw ca parallel to CA, and join Aa. Then Aa is the common tangent to the two given circles.

D

For, by the construction, BC-AC-ac; and also BC-ACAB; whence ac=AB, and ABca is a parallelogram (B. I., Pr. 32). But the angle B is a right angle; therefore this parallelogram is a rectangle, and the angles at A and a are right angles. Hence Aa is a tangent to both circles.

1

Since two tangents can be drawn from c to the circle BC, there are two common tangents to the given circles, viz., Aa and Dd. Scholium. Two other tangents can be drawn to the two given circles, and their points of contact will lie upon opposite sides of the line joining the centres. For this purpose CB must be taken equal to the sum of the given radii.

BOOK VI.

REGULAR POLYGONS, AND THE AREA OF THE CIRCLE.

Definition.

A regular polygon is a polygon which is both equiangular and equilateral.

An equilateral triangle is a regular polygon of three sides; a square is one of four.

PROPOSITION I. THEOREM.

Regular polygons of the same number of sides are similar figures.

F

B

Da

Let ABCDEF, abcdef be two regular polygons of the same d number of sides; then will they be similar figures.

For, since the two polygons have the same number of sides,

they must have the same number of angles. Moreover, the sum of the angles of the one polygon is equal to the sum of the angles of the other (B. I., Pr. 28); and, since the polygons are each equiangular, it follows that the angle A is the same part of the sum of the angles A, B, C, D, E, F, that the angle a is of the sum of the angles a, b, c, d, e, f. Therefore the two angles A and a are equal to each other. The same is true of the angles B and b, C

and c, etc.

Moreover, since the polygons are regular, the sides AB, BC, CD, etc., are equal to each other (Def.); so, also, are the sides ab, bc, cd, etc. Therefore AB: ab:: BC: bc:: CD: cd, etc. Hence the two polygons have their angles equal, and their homologous sides proportional; they are consequently similar (B. IV., Def. 4). Therefore, regular polygons, etc.

Cor. The perimeters of two regular polygons of the same number of sides are to each other as their homologous sides, and their areas are as the squares of those sides (B. IV., Pr. 27).

Scholium. The magnitude of the angles of a regular polygon is .determined by the number of its sides.

PROPOSITION II. THEOREM.

A circle may be described about any regular polygon, and a circle may also be inscribed within it.

Let ABCDEF be any regular polygon; a circle may be described about it, and another may be inscribed within it.

Bisect the angles FAB, ABC by the straight A lines AO, BO, and, from the point O in which they meet, draw the lines OC,OD, OE, OF to the other angles of the polygon.

F

B

H

Then, because in the triangles OBA, OBC, AB is, by hypothesis, equal to BC, BO is common to the two triangles, and the included angles OBA, OBC are, by construction, equal to each other; therefore the angle OAB is equal to the angle OCB. But OAB is, by construction, the half of FAB, and FAB is, by hypothesis, equal to DCB; therefore OCB is the half of DCB; that is, the angle BCD is bisected by the line OC. In the same manner, it may be proved that the angles CDE, DEF, EFA are bisected by the straight lines OD, OE, OF.

Now, because the angles OAB, OBA, being halves of equal angles, are equal to each other, OA is equal to OB (B. I., Pr. 11). For the same reason, OC, OD, OE, OF are each of them equal to OA. Therefore a circumference described from the centre O, with a radius equal to OA, will pass through each of the points B, C, D, E, F, and be described about the polygon.

Secondly. A circle may be inscribed within the polygon ABC DEF.

For the sides AB, BC, CD, etc., are equal chords of the same circle; hence they are equally distant from the centre O (B. III., Pr. 8); that is, the perpendiculars OG, OH, etc., are all equal to each other. Therefore, if from O as a centre, with a radius OG, a circumference be described, it will touch the side BC (B. III., Pr. 9), and each of the other sides of the polygon; hence the circle will be inscribed within the polygon. Therefore a circle may be described, etc.

Scholium 1. In regular polygons, the centre of the inscribed and circumscribed circles is also called the centre of the polygon; and the perpendicular from the centre upon one of the sides, that is, the radius of the inscribed circle, is called the .apothegm of the polygon.