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Let ab be the given straight line, and ABCDE the given

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Problem. To construct a triangle equal to a given polygon. Let ABCDEF be the given polygon.

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SOLUTION. Take three consecutive vertices, as B, C, D. Draw BD and CG || BD, meeting ED produced in G. C Join BG.

Then ▲ BGD=BCD, since they have the same base, BD, and are between the

same parallels, BD and CG (2, Cor. 3). Hence the polygon ABGEF

less side.

= ABCDEF, and it has one

By taking the vertices B, G, E, a polygon may be constructed equal to ABGEF, and having one side less.

By

repeating this process, we shall finally obtain a triangle equal to the given polygon.

Complete the construction.

XLIX.

Problem. To construct a square equal to a given triangle. Let ABC be a given triangle, BC its base, and AD its altitude. It is required to construct a square equal to ABC.

A

B

D

E

SOLUTION. Find a mean proportional, E, between AD and BC (43). The square described upon E will be equal to ABC; for

E2=AD X BC= ABC (7, Cor.).

SCHOLIUM. By this and the preceding problem a square may be constructed equal to any given polygon.

L.

Problem. To construct a square equal to the sum of two or more given squares, or to the difference of two given squares. 1. Let P, Q, R, S, &c., be the sides of given squares. It is required to find a line, EC, such that ECP+Q2+ R2 + S2.

SOLUTION. Take ABP, and from one extremity, B, draw BC AB, and =Q. Join AC; then

Draw AD

AC P2+ Q2 (12).

=

=

AC, and: R. Join DC;

then DC2 AC2 + R2= P2+ Q2 + R2. By repeating this process we may find a square equal to the sum of any number of given squares.

A

B

2. It is required to construct a line, BC, such that

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B

P

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-P

Q

R

-S

AR

D

S

P. With A as

SOLUTION. Construct a right angle, ABC, and lay off AB a centre, and a radius an arc cutting BC in C.

BC AC - AB2:

=

equal to Q, draw

Join AC; then

Q2 — P2 (12, Cor. 1).

SCHOLIUM 1. By this and the preceding problems, a square may be constructed equal to the sum of any number of polygons, or to the difference of any two polygons.

SCHOLIUM 2. If P, Q, R, and S are homologous sides of similar polygons, the line EC (Fig. 1) is the homologous side of a similar polygon equal to the sum of those polygons (32). The line BC (Fig. 2) is the homologous side of a similar polygon equal to the difference of the polygons whose homologous sides are P and Q.

LI.

Problem. Upon a given straight line to construct a rectangle equal to a given rectangle.

Let b' be the given line, a the altitude, and b the base, of

the given rectangle.

SOLUTION. Find a', the fourth proportional to b', a, and b (42). The rectangle constructed on b' with the altitude a' will be

the rectangle required.

For, since : b=a: a', we have,

ax b' = a x b (7, Cor.).

LII.

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Problem. To construct a rectangle when its area and the

sum of two adjacent sides are given.

Let AB be the side of the square equal to the given area, and let CD be equal to the sum of the two sides of the required rectangle. To find a point, G, in CD, such that CG X GD : AB2. SOLUTION. Upon CD as a diameter E, describe a semicircle. From C draw CE: AB, and CD. From E draw

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EF | CD, cutting the circumference
From F draw FG LCD. Then

in F.
G is the point required.

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A

B

F

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For CG: FG FG : GD (43): hence FG2 = CG × GD ;

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Problem. To construct a rectangle when its area and the difference of two adjacent sides are given.

A

Let AB be the side of the square equal to the given

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B

Ө

CE

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area, and let CD be equal to the difference of the two sides of the required rectangle. To find two lines, EF and EG, such that EFX EG = AB2, and EF EGCD.

SOLUTION. Upon CD as a diameter describe a circle. At C draw the tangent

AB, and from E the secant EF through the centre of the circle. Then EF and EG are the base and altitude of the required rectangle; for

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EXERCISE 1. If, from a point within a parallelogram, lines be drawn to the four vertices, two opposite triangles thus formed are together equal to one-half the parallelogram.

EXERCISE 2. The triangle formed by joining the middle point of one of the non-parallel sides of a trapezoid to the extremities of the opposite side is equal to one-half the trapezoid.

EXERCISE 3. Given any triangle, to construct an isosceles triangle of equal area, whose vertical angle is an angle of the given triangle (6).

EXERCISE 4. Bisect a given triangle by a line parallel to one of its sides.

BOOK IV.

REGULAR POLYGONS AND THE CIRCLE.

I.

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

II.

Theorem. A circle may be circumscribed about, or inscribed within, any regular polygon.

HYPOTII. ABCDEF is a regular polygon.

TO BE PROVED. 1. A circumference may be made to pass through the vertices A, B, C, D, E, and F ; that is, circumscribed about the polygon.

PROOF. From G and H, the middle F points of AB and BC, erect perpendiculars they will meet in a point M (I., 13), which is equally distant from A, B, and C (I., 31). From M as a centre describe the arc ABC; draw the lines MA

E

B

D

K

C

and MD; about the line MH revolve the quadrilateral MABH until BH falls upon its equal line, HC. The line BA will take the direction CD, for the angle B⇒C; the point A will fall on D, since BA = CD; and the line MA will coincide

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