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aquares GB, HC; and through A draw AL parallel to BD or Book I. CE, and join AD, FC ; then, because each of the angles BAC,

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equal to the whole FBC; and

e 2. Ax.

because the 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 equal to the base FC, and the tri- f 4. 1. angle ABD to the triangle FBC i now the parallelogram BL is double g of the triangle ABD, because they are upon the same g 41. 1. base BD, and between the same parallels BD, 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 one another: h 6. AF. 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. E. D.

PROP. XLVIII. THEOR.

2

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.

G

Book I.

11. 1.

b 47. 1.

€ 8.1.

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.

b.

D

From the point A draw 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 DC is equal b 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 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 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.

B

C

THE

ELEMENTS OF EUCLID.

BOOK II.

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DEFINITIONS

I.

EVERY right angled parallelogram is said to be contained Book II. 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 the

H

E

D

F

K

two complements, is called A
a gnomon.
Thus the pa-
rallelogram HG, together
'with the complements AF,
'FC, is the gnomon, which
'is more briefly expressed
'by the letters AGK, or
EHC, which are at the
opposite angles of the pa-
'rallelograms which make the gnomon.”

B G

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.

a 11. 1.

b.3. 1. c. 31. 1.

4 34, 1,

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

the straight lines A, BC is equal B
to the rectangle contained by A,

BD, together with that contain-
ed by A, DE, and that contain-
ed by A, EC.

From the point B draw a BF
at right angles to BC, and make G
BG equal to A; and through
G draw GH parallel to BC;
and through D, E, C drawe DK,FI
EL, CH parallel to BG; then

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D E c

K

L

H

A

the rectangle BH is equal to the rectangles BK, DL, EH ; and BHis 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, because DK, that is 4, 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.

46. 1.

b 31. 1.

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 into A any two parts in the point C; the rectangle contained by AB, BC, together with the rectangle AB, AC, shall be equal to the square of AB.

*

Upon AB describe a the square 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, D

C B

FE

*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.

AC; for it is contained by DA, AC, of which AD is equal to AB; and CE is contained by AB, BC, for BE is equal to AB; therefore the rectangle contained by AB, AC, together with the rectangle AB, BC, is equal to the square of AB. If therefore a straight line, &c. Q. E. D.

PROP. III. THEOR.

IF a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the foresaid part.

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Let the straight line AB be divided into two parts in the point C; the rectangle AB, BC is equal to the rectangle AC, CB, together with the square of BC. Upon BC describe a the square A CDEB, and produce ED to F, and through A drawb AF parallel to CD or BE; then the rectangle AE is equal to the rectangles AD, CE; and AE is the rectangle contained by AB, BC, for it is contained by AB, BE, of which BE is equal to BC; and AD is contained by AC,

CB, for CD is equal to BC; and F

D

E

DB is the square of BC; therefore the rectangle AB, BC is equal to the rectangle AC, CB, together with the square of BC. If therefore a straight line, &c. Q. E. D.

B

a 46. 1.

b 31: 1.

PROP. IV. THEOR.

IF a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts.

Let the straight line AB be divided into any two parts in Cr the square of AB is equal to the squares of AC, CB, and to twice the rectangle contained by AC, CB.

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