* 11. 1.

* 47. 1

8. 1.

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.

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.

Let 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 shall be a right angle.

B

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 equals 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 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 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, each to each; and the base DC has been proved 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.

49

THE

ELEMENTS OF EUCLID.

BOOK II.

DEFINITIONS.

I.

Every right-angled parallelogram is called a rectangle, and is said to be 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

the two complements, is called a

gnomon. 'Thus the parallelogram

HG, together with the comple

ments AF, FC, is the gnomon,

'which is more briefly expressed B G

'by AGK, or EHC, the letters at the opposite angles

' of the parallelograms, which make the gnomon.'

F

11. 1.

* 3. 1.

* 31. 1. * 31. 1.

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.

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 shall be equal to the rectangle contained by A, BD, together with that contained by A, DE, and that contained by A, EC.

B

D

E C

K

A

From the point B draw* BF at right angles to BC, and make BG equal* to A; and through G draw * GH parallel 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, and EH; but BH is contained by A, B,C, 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 * 34. 1. DK, that is * 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 rectangles contained by A, BD, by A, DE, and by A, EC. Wherefore, if there be two straight lines, &c. Q. E. D.

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 any two

a

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

A

C

46. 1.

F

E

Upon AB describe the square ADEB, and through C draw* CF, parallel to AD or BE; then AE is equal to the rectangles AF, CE; but 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 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 aforesaid part.

Let the straight line AB be divided into any two parts in the point C; the rectangle AB, BC shall be equal to the rectangle AC, CB, together with the square of BC.

Upon BC describe the square A C CDEB, and produce ED to F, and through A draw* AF parallel to CD or BE; then the rectangle AE is equal to the rectangles AD, CE;

but AE is the rectangle contained F D

* 31. 1.

B

* 46. 1.

E

by AB, BC, for it is contained by AB, BE, of which BE is equal to BC; and AD is contained by AC, CB,

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.

* 31. 1.

for CD is equal to CB; and BD 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, &c. Q. E. D.

46. 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 C; the square of AB is equal to the squares of AC, CB, and to twice the rectangle contained by AC, CB.

Upon AB describe* the square ADEB, and join BD, * 31. 1. and through C draw* CGF parallel to AD or BE, and through G draw HK parallel to AB, or DE. Then, because CF is parallel to AD, and BD falls upon them, * 29. 1. the exterior angle BGC is equal to the interior and

5. 1.

*6. 1.

B

opposite angle ADB; but ADB is equal to the angle ABD, because BA is equal to AD, being sides of a square; wherefore the angle CGB is equal to the angle CBG; and therefore the side CB is equal to the side CG: but CB is * 34. 1. equal✶ also to GK, and CG to BK; H therefore the figure CGKB is equilateral; it is likewise rectangular; for since KBC is a right angle, by construction; therefore all the angles of the parallelogram CGKB are right angles;* therefore CGKB is rectangular: but it is also equilateral, as was demonstrated; therefore it is a square, and it is upon the side CB; for the same reason HF is also a square, and it is upon the 34. 1. side HG, which is equal to AC: therefore HF, CK are

* Cor. 46 1.

D