Book I.

Therefore (by def. 30.) it is a fquare; and it hath been defcribed upon the ftraight line AB. Which was to be done.

PRO P. XLVII.

In right angled triangles, the fquare of the fide fubtending the right angle, is equal to the fquares of the fides containing the right angle.

Let ABC be a right angled triangle, having the angle BAC a right angle: I fay that the fquare of BC is equal to the fquares of BA, AC.

For (by prop. 46.) let the fquare BDEC be described upon BC; and the fquares GB, HC upon BA, AC, and through the point A let AL be drawn parallel either to BD or CE; and let AD, FC be joined.

F

G

H

A

K

B

C

Ꭰ

LE

And because each of the angles BAC, BAG is (by conft. and fup.) a. right angle; to a certain line BA, and to the point A in it, two ftraight lines. AC, AG, not lying towards the fame parts, make the adjacent angles equal to two right angles; therefore (by prop.14.) AC is in a ftraight line with AG: certainly for the fame reafon also AB is in a straight line with AH: and fince the angle DBC is equal to the angle FBA (by com. not.10.), for eacli of them is a right angle, let the common angle ABC be added; therefore (by com. not. 2.) the whole angle DBA is equal to the whole angle FBC; and because the two ftraight lines DB, BA are equal to the two CB, BF, each to each (being fides of the fame Squ re) and the angle DBA is equal to the angle FBC; therefore (by prop. 4.) the bafe AD is equal to the bafe FC, and the triangle ABD is equal to the triangle FBC; and the parallelogram BL is double of the triangle ABD (by prop. 41.); for they have the fame base BD and are between the fame parallels BD, AL: but the square GB is double of the triangle FBC; for again, they have

the

the fame base FB, and are between the fame parallels FB, GC; Book I. but the doubles of equal magnitudes are equal to one another (by com. not 6.) wherefore alfo the parallelogram BL is equal to the fquare GB: Certainly in the fame manner it will be demonftrated, having joined AE, BK, that the parallelogram CL is equal to the fquare HC; therefore the whole square DBCE is equal to the two fquares GB, HC; and BDEC is the fquare described upon BC; and GB, HC the Squares defcribed upon BA, AC: wherefore the fquare BE described upon the fide BC is equal to the fquares upon the fides BA, AC.

Wherefore in right angled triangles, the fquare of the fide fubtending the right angle is equal to the fquares of the fides, containing the right angle. Which was to be demonftrated.

If the fquare of one of the fides of a triangle be equal to the fquares of the two remaining fides of the triangle; the angle contained by the two remaining fides of the triangle is a right angle.

For let the fquare of BC one fide of the triangle ABC be equal to the fquares of the fides BA, AC: I fay that the angle BAC is a right angle.

For from the point A let the straight line AD be drawn at right angles to AC; and make AD (by prop. 3.) equal to BA; and join DC.

And because DA is equal to AB, the B fquare of AD is equal to the fquare of AB; let the common fquare of AC be added; therefore (by com. not. 2.) the fquares of

A

D

C

DA and AC are equal to the fquares of BA and AC: but the squares of DA and AC are (by prop. 47.) equal to the square of DC; for DAC is (by conft.) a right angle; and the square of BC is equal to the fquares of BA and AC, for this is fuppofed: wherefore (by com. not. 1.) the fquare of DC is equal to the fquare of BC; so that also the fide DC is equal to BC; and fince AD is equal to AB and AC common; certainly the two AD, AC are equal

to

Book I, to the two AB, AC, and the bafe DC is equal to the bafe BC; therefore the angle DAC is equal to the angle BAC; but DAC is a right angle; therefore the angle BAC is a right angle.

Wherefore if the fquare of one of the fides of a triangle be equal to the fquares of the two remaining fides of the triangle; the angle contained by the two remaining fides of the triangle is a right angle. Which was to be demonstrated.

THE

ELEMENTS

O F

EUCLID.

BOOK II.

1.

DEFINITION S.

EVERY right angled parallelogram is faid to be contained

by the two straight lines containing the right angle.

2. Of every parallelogram fpace, let any one of the parallelograms about the diameter of it, together with the two complements, be called a GNOMON,

PROP. I.

If there be two ftraight lines, and one of them be cut into any number of fegments; the rectangle contained by the two ftraight lines is equal to the rectangles contained by the undivided line and each of the fegments of the divided line.

Let there be two ftraight lines A, BC, and let BC be cut as it may happen in the points D and E: I fay that the rectangle con

Book II.

Book II. tained by A and BC is equal to the rectangle contained by A and BD, and to that contained by A and DE, and befides to that contained by A and EC.

For let BF be drawn from the point B at right angles to BC; and let BG be made (by prop. 3. B. 1.) equal to A; and let GH be drawn, through the point G parallel to BC (by prop. 31. B. 1.); and let DK, EL, CH be drawn through the points D, E, C parallel to BG.

B

DEC

G

H

K

L

A

Certainly the rectangle BH is equal to the rectangles BK, DL, EH; and BH is the rectangle contained by A and BC; for (by def. 1. B. 2.) it is contained by GB, BC; but BG is (by conft.) equal to A : and BK is a rectangle contained by A and BD; for (by def. 1.) it is contained by F GB, BD; and GB is equal to A: but the rectangle DL is contained by A and DE; for DK, that is BG (by prop. 34. B. 1.) is equal to A: and alfo in like manner EH is a rectangle contained by A and EC, therefore the rectangle contained by A and BC is equal to the rectangles contained by A and BD; and A and DE and alfo A and EC.

Wherefore if there be two straight lines, and one of them be cut into any number of fegments; the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line and each of the segments of the divided line. Which was to be demonstrated.

PROP. II.

If a straight line be cut as it may happen, the rectangles contained by the whole and each of the fegments are equal to the fquare of the whole line.

For let the ftraight line AB be cut as it may happen in the point C I say that the rectangle contained by AB and BC; together with the rectangle contained by BA and AC is equal to the fquare of AB.

For let the fquare ADEB be defcribed upon AB (by 46. 1.); and let CF be drawn (by 31. 1.) parallel either to AD or BE.

Certainly