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Book L.

11. 1.

PROP. XLVI. PROB.

To describe a square upon a given straight line.

Let AB be the given straight line; It is required to describe a square upon AB.

E

From the point A drawa AC at right angles to AB; and 3. 1. make AD equal to AB, and through the point D draw DE 31. 1. parallel to AB, and through B draw BE parallel to AD; 34. 1. therefore ADEB is a parallelogram: whence AB is equald to DE, and AD to BE: but BA is equal to AD; therefore the four straight lines BA, AD, DE, EB, are equal to one another, and the parallelogram ADEB is equilateral, like- D wise all its angles are right angles; because the straight line AD meeting the parallels AB, DE, the angles € 29. 1. BAD, ADE are equale to two right angles: but BAD is a right angle; therefore also ADE is a right angle ; A but the opposite angles of parallelograms are equald; therefore each of the opposite angles ABE, BED is a right angle; wherefore the figure ADEB is rectangular, and it has been demonstrated that it is equilateral; it is therefore a square, and it is described upon the given straight line AB: Which was to be done.

B

COR. Hence every parallelogram that has one right angle has all its angles right angles.

46. 1.

PROP. XLVII. THEOR.

IN any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.

Let ABC be a right-angled triangle having the right angle BAC; the square described upon the side BC is equal to the squares described upon BA, AC.

On BC describes the square BDEC, and on BA, AC the

squares GB, HC; and through A drawb AL parallel to BD, Book I. or CE, and join AD, FC. Then, because each of the angles

BAC, BAG is a rightangle, the two straight lines AC, AG, upon the opposite sides of AB, make with it at the point A the adjacent angles equal to two right angles; therefore CA is in the same straight lined with AG; for the same reason, AB and AH are in the same straight line; and because the angle DBC is equal to the angle FBA, each of them being a

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angle DBA is equale to the whole FBC; and because the * 2 Ax. two sides AB, BD, are equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBC; therefore the base AD is equalf to the base FC, and the triangle 4. 1. ABD to the triangle FBC: Now the parallelogram BL is doubles of the triangle ABD, because they are upon the 41. 1. same base BD, and between the same parallels BĎ, AL; and the square GB is double of the triangle FBC, because these also are upon the same base FB, and between the same parallels FB, GC. But the doubles of equals are equal to 6 Ax. one another: Therefore the parallelogram BL is equal to the square GB: And, in the same manner, by joining AE, BK it is demonstrated, that the parallelogram CL is equal to the square HC; Therefore the whole square BDEC is equal to the two squares GB, HC; and the square BDEC is described upon the straight line BC, and the squares GB, HC upon BA, AC: Wherefore the square upon the side BC is equal to the squares upon the sides BA, AC. Therefore, in any right-angled triangle, &c. Q. É. D.

PROP. XLVIII. THEOR.

If the square described upon one of the sides of a triangle, be equal to the squares described upon the other two sides of it; the angle contained by these two sides is a right angle.

BOOK I.

a 11. 1.

If the

square described upon BC, one of the sides of the triangle ABC, be equal to the squares upon the other sides BA, AC, the angle BAC is a right angle.

D

From the point A drawa AD at right angles to AC, and make AD equal to BA, and join DC: Then, because DA is equal to AB, the square of DA is equal to the square of AB: To each of these add the square of AC; therefore the squares of DA, AC are equal to the squares of BA, AC: But the square of 47. 1. DC is equal to the squares of DA, AC,

because DAC is a right angle; and the
square of BC, by hypothesis, is equal to
the squares of BA, AC; therefore the
square of DC is equal to the square of B

A

BC; and therefore also the side DC is equal to the side BC. And because the side DA is equal to AB, and AC common to the two triangles DAC, BAC, the two DA, AC are equal to the two BA, AC; and the base DC is equal 8. 1. to the base BC; therefore the angle DAC is equal to the angle BAC; but DAC is a right angle; therefore also BAC is a right angle. Therefore, if the square, &c. Q. E. D.

THE

ELEMENTS

OF

43

EUCLI D.

BOOK II.

DEFINITIONS.

I.

EVERY right-angled parallelogram is said to be Book II. contained by any two of the straight lines which contain one of the right angles.

II.

In every parallelogram, any of the parallelograms about

a diameter, together with A

Thus

the two complements, is
called a Gnomon.
'the parallelogram HG,
'together with the com-
'plements AF, FC, is the
gnomon, which is more
'briefly expressed by the

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letters AGK, or EHC,

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K

C

'which are at the opposite angles of the parallelograms 'which make the gnomon."

PROP. I. THEOR.

If there be two straight lines, one of which is divided into any number of parts; the rectangle contained by the two straight lines, is equal to the rectangles contained by the undivided line, and the several parts of the divided line.

BOOK II.

Let A and BC be two straight lines; and let BC be divided into any parts in the points D, E; the rectangle con

tained by the straight lines A, B

BC is equal to the rectangle

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contained by A, BD, together
with that contained by A, DE,
and that contained by A, EC.
From the point B drawa G
BF at right angles to BC,
3. 1. and make BG equal to A;
31. 1. and through Gdraw GH pa- F

a 11. 1.

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rallel to BC; and through D, E, C, draw DK, EL, CH parallel to BG; then the rectangle BH is equal to the rectangles BK, DL, EH; and BH is contained by A, BC, for it is contained by GB, BC, and GB is equal to A; and BK is contained by A, BD, for it is contained by GB, BD, of which GB is equal to A; and DL is contained by A, DE, 34. 1. because DK, that isd BG is equal to A; and in like manner the rectangle EH is contained by A, EC: Therefore the rectangle contained by A, BC, is equal to the several rectangles contained by A, BD, and by A, DE: and also by A, EC. Wherefore, if there be two straight lines, &c. Q. E. D.

a 46. 1.

b

PROP. II. THEOR.

IF a straight line be divided into any two parts, the
rectangles contained by the whole and each of the
parts, are together equal to the square of the whole
line.

Let the straight line AB be divided
A
into any two parts in the point C; the
rectangle contained by AB, BC, toge-
ther with the rectangle* AB, AC, shall
be equal to the square of AB.

B

Upon AB describe the square 31. 1. ADEB, and through C drawb CF parallel to AD or BE. Then AE is equal to the rectangles AF, CE; and AE is the square of AB; and AF is the rectangle contained by BA, AC; for it is contained by DA, AC of which AD is equal to AB; and CE is contain

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F E

*N. B. To avoid repeating the word contained too frequently, the rectangle contained by two straight lines AB, AC is sometimes simply called the rectangle AB, AC.

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