For, because AE is parallel to BC we have (Prop. XVI. B. IV.),

DE: EC:: DA : AB.

But DE is equal to EC; therefore DA is equal to AB.

PROBLEM XXII.

To describe a square that shall be equivalent to a guer parallelogram, or to a given triangle.

C

First. Let ABDC be the given parallelogram, AB its base, and CE its altitude. Find a mean proportional between AB and CE (Prob. XIX.), and represent it by X; the square described on X will be equiva- A E lent to the given parallelogram ABDC.

B

D

For, by construction, AB: X::X: CE; hence X' is equal to ABX CE (Prop. I., Cor., B. II.). But AB X CE is the measure of the parallelogram; and X' is the measure of the square. Therefore the square described on X is equivalent to the given parallelogram ABDC.

Secondly. Let ABC be the given triangle, BC its base, and AD its altitude. Find a mean proportional between BC and the half of AD, and represent it by Y. Then will the square described on Y be equivalent to the_triangle ABC.

2

B

A

D C

For, by construction, BC: Y :: Y:¦ AD; hence Y2 is equivalent to BC × AD. But BCX AD is the measure of the triangle ABC; therefore the square described on Y is equivalent to the triangle ABC.

PROBLEM XXIII.

Upon a given line, to construct a rectangle equivalent to a given rectangle.

Let AB be the given straight G line, and CDFE the given rectangle. It is required to con

H

E

F

struct on the line AB a rectan

gle equivalent to CDFE.

Find a fourth proportional A

Prob. XVIII.) to the three lines AB, CD, CE, and let AG be that fourth proportional. The rectangle corstricted or the lines AB. AG will be equivalent to CDFE

For, because AB: CD::CE: AG, by Prop. I., B. II.. ABXAG=CDXCE. Therefore the rectangle ABHG is equivalent to the rectangle CDFE; and it is constructed upon the given line AB.

PROBLEM XXIV.

To construct a triangle which shall be equivalent to a given polygon.

Let ABCDE be the given polygon; it s required to construct a triangle equivaent to it.

Draw the diagonal BD cutting off the E triangle BCD. Through the point C, diaw CF parallel to DB, meeting AB produced in F. Join DF; and the polygon AFDE will be equivalent to the polygon ABCDE.

A

B

For the triangles BFD, BCD, being upon the same base BD, and between the same parallels BD, FC, are equivalent. To each of these equals, add the polygon ABDE; then will the polygon AFDE be equivalent to the polygon ABCDE; that is, we have found a polygon equivalent to the given polygon, and having the number of its sides diminished by

one.

In the same manner, a polygon may be found equivalent to AFDE, and having the number of its sides diminished by one; and, by continuing the process, the number of sides may be at last reduced to three, and a triangle be thus obtain ed equivalent to the given polygon.

PROBLEM XXV.

To make a square equivalent to the sum or difference of twi given squares.

C

First. To make a square equivalent to the surr. of twe given squares. Draw two indefinite lines AB, BC at right angles to each other. Take

AB equal to the side of one of the given squares, and BC equal to the side of the other. Join AC; it will be the side of the A required square.

B

For the triangle ABC, being right-angle: at. B. the squar

on AC will be equivalent to the sun of the squares upon Af and BC (Prop. XI., B. IV.).

square.

Secondly. To make a square equivalent to the difference of two given squares. Draw the lines AB, BC at right an gles to each other; and take AB equal to the side of the less Then from A as a center, with a radius equal to the side of the other square, describe an arc intersecting BC in C; BC will be the side of the square required; because the square of BC is equivalent to the difference of the squares of AC and AB (Prop. XI., Cor. 1, B. IV.).

Scholium. In the same manner, a square may be made equivalent to the sum of three or more given squares; for the same construction which reduces two of them to one will reduce three of them to two, and these two to one.

PROBLEM XXVI.

Upon a given straight line, to construct a polygon simila to a given polygon.

Lo ABCDE be the given polygon, and FG be the given straight line; it E s required upon the line. FG to construct a polygon similar to ABCDE.

A

D

K

F

Ꮐ

Draw the diagonals BD, BE. At the point F, in the straight line FG, make the angle GFK equal to the angle BAE; and at the point G make the angle FGK equal to the angle ABE. The lines FK, GK will intersect in K, and FGK will be a triangle similar to ABE. In the same manner, on GK construct the triangle GKI similar to BED, and on GI construct the triangle GIH similar to BDC. The polygon FGHIK will be the polygon required. For these two polygons are composed of the same number of triangles, which are similar to each other, and similarly situated; therefore the polygons are similar (Prop. XXV., Ccr., B. IV.)

PROBLEM XXVII.

Given the area of a rectangle, and the sum of two adjacent sides, to construct the rectangle.

Let AB be a straight line equal to the sum of the sides of he required rectangle.

C

Upen AB as a diameter, describe a semicircle. At the point A erect the perpendicular AC, and make it equal to the side of a square having the given area. Through C draw the line CD par- A

E B

allel to AB, and let it meet the circumference in D; and from D draw DE perpendicular to AB. Then will AE and EB be the sides of the rectangle required.

For, by Prop. XXII., Cor., B. IV., the rectangle AE×EB is equivalent to the square of DE or CA, which is, by construction, equivalent to the given area. Also, the sum of the sides AE and EB is equal to the given line AB.

Scholium. The side of the square having the given area, must not be greater than the half of AB; for in that case the line CD would not meet the circumference ADB.

PROBLEM XXVIII.

Given the area of a rectangle, and the difference of two adjacent sides, to construct the rectangle.

Let AB be a straight line equal to the difference of the sides of the required rectangle.

F

Upon AB as a diameter, describe a circle; and at the extremity of the diameter, draw the tangent AC equal to the side of a square having the given area. Through the point C and the center F draw the secant CE; then will CD, CE be the adjacent sides of the rectangle required.

For, by Prop. XXVIII., B. IV., the rectangle CD×CE is equivalent to the 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.

E

BOOK VI.

REGULAR POLYGONS, AND THE AREA OF THE CIRCLE.

Definition.

A regular polygon is one 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.

Let ABCDEF, abcdef be two regular polygons of the same 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 (Prop. XXVIII., B. I.); 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, &c.

Moreover, since the polygons are regular, the sides AB, BC, CD, &c., are equal to each other (Def.); so, also, are the sides ab, bc, cd, &c. Therefore AB: ab :: BC: bc:: CD: cd, &c. Hence the two polygons have their angles equal, and their homologous sides proportional; they are consequently similar (Def. 3, B. IV.). Therefore, regular polygons, &c.

Cor. The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides