Sidebilder
PDF
ePub

remaining leg of the other. For, as was seen in the preceding corollary, the squares of BC, CA are equal to the squares of EF, FD; from which, if the unequal squares of BC, EF be taken, the remaining square of AC is less than the remaining square of DF, and therefore AC is less than DF.

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, the angle contained by those two sides is a right angle.

A

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

From the point A draw (I. 11.) AD perpendicular to AC, and equal to BA ; and join CD. Then, because DA is equal to AB, the squares of DA, AB are (I. 46. cor. 2.) equal. To each of these add the square

of AC;

therefore the squares of DA, AC are equal to the squares of BA, AC. But (I. 47.) the square of DC is equal to the squares of DA, AC, because DAC is a right angle; and the square of BC is equal (hyp.) to the squares BA, AC; therefore the square of DC is equal (I. ax. 1.) to the square of BC; and therefore also the side DC is equal" (I. 46.) cor. 3.) to the side BC. And because the side DA is equal to AB, and AC common to the two triangles DAC, BAC, and the base DC equal to the base BC; therefore (I. 8.) the angles DAC, BAC are equal : but (const.) DAC is a right angle; wherefore also BAC is a right angle. Therefore, if the square, &c.

B

of

[ocr errors]

BOOK II.*

A

[ocr errors]

D

E

F

K

G

C

DEFINITIONS. 1. A RECTANGLE is said to be contained by the two straight lines which are about any of the right angles. +

2. In every parallelogram, the figure which is composed of either of the parallelograms about a diagonal, and of the two complements, is called a Gnomon. I “ Thus the « parallelogram HE together with the comple“ ments BK, KD, is a gnomon, which is more

briefly expressed by the letters BHF, or “ GED, which are at the opposite angles of “ the parallelograms that make up the gnomon.

For the sake of brevity, the rectangle contained by AB and “ CD is often expressed simply by AB.CD, a point being placed “ between the letters denoting the sides of the rectangle : and the square of a line AB is often written simply AB2.

« In Algebra, the sign t, called Plus, (more by) placed between “ the names of two magnitudes, is used to denote that these “ magnitudes are added together; and the sign — "called minus,

(less by) placed between them, to signify that the latter is taken “ from the former. The sign =, which is read equal to, signifies “ that the quantities between which it stands, are equal to one “ another. These signs will be occasionally used for brevity, " in what follows.”

66

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 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

• This book treats of right-angled parallelograms, and in particular of squares.

+ That is, by any two adjacent sides, or by its length and its breadth. The rectangle contained by two straight lines is sometimes called simply the rectangle under those lines, or the rectangle of them.

A gnomon might be more simply defined to be the part of a parallelogram which remains, when either of the parallelograms about one of the diagonals is taken away.

The term, however, might be totally disused, as it is seldom employed either by Euclid or any other writer, and scarcely any circumlocution would arise from laying it aside.

B

D

Е с

G

K

L H

[ocr errors]

F

A

into any parts in the points D, E; the rectangle contained by A and BC is equal to the rectangles contained by A and BD, A and DE, and A and EC.

From B draw (I. 11.) BF at right angles to BC, and (I. 3.) make BG equal to A : through G draw (1. 31.) GH parallel to BC; and through D, E, C, draw DK, EL, CH parallel to BG. Then the rectangle BH is equal (I. ax. 10.) to the rectangles BK, DL, EH. But BH is contained by A and BC, for (IÍ. def. 1.) it is contained by GB, BC, and GB is equal (const.) to A ; and BK is contained by A and BD, for it is contained by GB, BD, of which GB is equal to A. Also DL is contained by A and DE, because DK, that is (I. 34.) BG, is equal to A : and in like manner it is shown that EH is contained by A and EC. Therefore the rectangle contained by A and BC is equal to the several rectangles contained by A and BD, by A and DE, and by A and EC. * Wherefore, if there be two straight lines, &c.

PROP. II. THEOR.

If a straight line be divided into any two parts, the rectangles contained by the whole and each of the

arts, are together equal to the square of the whole line.

Let the straight line AB be divided into any two parts in the point C; the rectangles AB.AC, and AB.BC are equal to the

square of AB.

А

CB

FE

Upon A B describe (I. 46.) the square A E, and through C draw (I. 31.) CF parallel to AD or BE. Then AE is equal (I. ax. 10.) to the rectangles AF and CE. But AE is the square of AB, and AF is the rectangle contained by BA, AC: for (II. def. 1.) it is contained by DA, AC, of which DA is equal to AB: and CE is contained by AB, BC, for BE is equal to AB; therefore the rectangles under AB, AC, and AB, BC are equal to the square of AB.t If, therefore, &c.

* This demonstration depends on the principle that the whole is equal to all its parts, or rather that it is identical with them : for BH is equal to BK, DL, and EH; but (I. def. I.) BH is equal to the rectangle under A and BC, and BK, DL, and EH are equal to those contained by BD and A, DE and A, and EC and A.

+ This proposition may also be demonstrated in the following manner :

Take a straight line D equal to AB. Then (II. 1.) the rectangles AC. D and BC.D are together equal to AB.D. But since D is equal to AB, the rectangle A B. D is equal (1. def. 27.) to the square of AB, and the rectangles AC.D and BC. D are respectively equal to AC. A B and BC. AB; wherefore the rectangles AC. A B and BC. A B are together equal to the square of AB.

In a manner similar to this, several of the following propositions may be

A

с в

[ocr errors]

D

PROP. III. THEOR.

B

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 square of that part, together with the rectangle contained by the two parts.

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

Upon BC describe (I. 46.) the square CE; prodace ED to F; and through A draw (I. 31.) AF parallel to CD or BE. Then the rectangle AE is equal (I. ax. 10.) to the rectangles CE, AD. But 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 CB; also DB is the square of BC. Therefore the rectangle AB.BC is equal to the square of BC, together with the rectangle AC.CB. * If, therefore, &c.

[ocr errors]

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 their rectangle. +

Let the straight line AB be divided into any two parts in C; the square of AB is equal to the squares of AC and CB, together with twice the rectangle under AC and CB.

Upon AB describe (I. 46.) the square AE, and join BD; through C draw (I. 31.) 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, the exterior angle BGC is equal (I. 29.) to the interior and remote angle ADB; but ADB is equal (1.5.) to ABD, because BA and AD are equal, being sides of a square: wherefore (I. ax. 1.) the angle CGB is

A

H

к

D

[ocr errors]

demonstrated. Such proofs, though perhaps not so easily understood at first by the learner, are shorter than those given by Euclid ; and they have the advau. tage of forming a mutually dependent chain of consecutive truths, instead of being established by continual appeals to original principles.

* Otherwise. Take a line D equal to CB. Then (II. 1.) the rectangle AB.D is equal to the rectangles BC.D and AC.D; that is, (const. and I. def. 27.) the rectangle A 6.BC is equal to the square of BC together with the rectangle AC.CB.

+ The enunciation may also be thus expressed :-- The square of the sum of two straight lines is equal to the sum of their squares, together with twice their rectungle.

А с

B

D

E

с

B

H Н

K

D

[ocr errors]

equal to GBC; and therefore the side BC is equal (I. 6.) to the side CG. But (const.) the figure CK is a parallelogram; and since CBK is a right angle, CK (I. def. 26. and 27.) is a square, and it is upon the side CB. For the same reason HF also is a square, and it is upon

the side HG, which is equal (1. 34.) to AC; therefore HF, CK are the squares of AC, CB. And because (1. 43.) the complements AG, GE are equal, and that AG is the rectangle contained by AC, CB, for GC is equal to CB; therefore GE is also equal to the rectangle AC.CB; wherefore AG, GE are equal to twice the rectangle AC.CB. The four figures, therefore, HF, CK, AG, GE are equal to the squares of AC, CB, and twice the rectangle AC.CB. But HF, CK, AG, GE make up the whole figure ADEB, which is the square of AB: therefore the square of AB is equal to the squares of AC, CB, and twice the rectangle AC.CB. * Wherefore, &c.

Cor. 1. From the demonstration, it is manifest that the parallelograms about the diagonal of a square are likewise squares.

Cor. 2. Hence the square of a straight line is equal to four times the square of its half: for the straight line being bisected, the rectangle of the parts is equal to the square of one of them.

PROP. V. THEOR.

If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.

Let the straight line AB be bisected in C, and divided unequally in D: the rectangle AD.DB, together with the square of CD, is equal to the square of CB.

Upon CB describe (I. 46.) the square CF, join BE, and through D draw (1. 31.) DHG parallel to CE or BF; also through H draw KLM parallel to CB or EF; and through A draw AK parallel to CL or BM. Then, because (I. 43.) the complements CH, HF are equal, to each add DM; therefore the whole CM is equal to the whole DF. But CM is equal (1.36.) to AL, because AC is equal to CB; therefore also AL is equal to DF. To each of these add CH, and the whole AH is equal to DF and CH, that is to the gnomon CMG. To each of these add LG, and (I. ax. 2.) the gnomon CMG, together with LG, is equal to AH together

A

с

DB

к

L

H

E

G F

Otherwise.- AB’ = AB.AC + AB.CB (II. 2.). But (II. 3.) AB. AC = AC? + AC.CB, and AB.CB = CB? + AC.BC. Hence (I. ax. 2.) AB.AC + AB.BC, or ABP = AC? + CB: + 2AC.CB.

« ForrigeFortsett »