Book F. the right line HF falls upon the parallels AH, EF, the angles AHF, HFE [by prop. 29.] are equal to two right angles; and fo BHG, GFE are less than two right angles. But right lines infinitely produced from angles lefs than two right angles [by ax. 11.] will meet one another. Let them be produced [by poft. 2.] meeting at K, and thro' K draw [by prop. 31.] KL, parallel to E A or F H. And continue out A H, G B to the points L, M.

Then is HLKF a parallelogram, whofe diameter is HK, and the parallelograms about H K are A G, ME; and these which are called the complements are L B, BF: therefore [by prop. 43.] the parallelogram L B is equal to B F. But BF is equal to the triangle c: wherefore L B will be equal to the triangle c too. And becaufe [by prop. 15.] the angle G B E is equal to the angle A B M, and alfo G B E is equal to the angle D; the angle A B M will be therefore equal to the angle D.

Wherefore at the given right line AB is applied the parallelogram L B equal to the triangle c, and having one angle A B M equal to the angle D. Which was to be done.

PROP. XLV. PROBL. To conftitute a parallelogram equal to a given rightlined figure, having one of its angles equal to a given angle.

Let the given right-lined figure be A B C D, and the given right-lined angle be E: It is required to constitute a parallelogram equal to the given right-lined figure AB CD, having one angle equal to the angle E.

Draw D B, and conftitute [by prop. 42.] the parallelo

D

gram FH equal to

A F

GL the triangle A DB,

having one angle

to the angle E. Then because the angle E is equal to either H KF, or GHM; the angle H K F will be equal to the angle G H M. Let K HG be added to both; then the angles FKH, KH G are equal to the angles K HG, G H M. But the angles FKH, KHG [by prop. 29.] are equal to two right angles. Therefore the angles K HG, GHM will alfo be equal to two right angles. And fo because at the given point н, in the right line G H, two right lines KH, HM not lying the fame way, make the adjacent angles equal to two right angles; therefore [by prop. 14.] K H and H M make but one right line. And because the right line GH falls upon the parallels KM, FG, the alternate angles MHG, HGF [by prop. 29.] are equal. Add HGL to both; then are the angles MHG, HGL equal to the angles HGF, HG L. But [by prop. 29.] MHG, HGL are equal to two right angles: wherefore alfo the angles HGF, HGL will be equal to two right angles. Wherefore F G and GL are both in the fame right line. And because K F is equal to H G, and parallel to it; and HG alfo to ML; KF will be [by ax. 1. and prop. 30.] equal and parallel to M L. But the right lines K м, FL join them; therefore the right lines K M, F L [by prop. 33.] are equal and parallel wherefore K F L M is a parallelogram. And because the triangle A B D is equal to the parallelogram HF, and the triangle A B C to the parallelogram G M, the whole right-line figure A B C D will be equal to the whole parallelogram K FLM.

Therefore the parallelogram K F L M is constituted equal to the given right-lined figure A B C D, having one angle FK M, which is equal to the given angle E. Which was to be done.

PROP, XLVI. PROBL.

To defcribe a fquare upon a given right line.

Let A B be the given right line upon which it is required to describe a square,

From the given point A in the right line A B [by prop. 11.] draw the right line A c at right angles to AB; and [by prop. 3.] put A D equal to AB. From the point D [by prop. 31.] draw D E parallel to A B; and from the point B draw B E parallel to A D.

E

Then

IC

E

Then is A DEB a parallelogram; and fo DE is equal to A B, and AD to BE: But A B is allo equal to A D. Therefore the four right lines BA, AD, DE, E B are equal to one another. Wherefore the parallelogram A DEB is equilateral: I fay, it is alfo right-angled. For because the right line AD falls upon the two parallels AB, DE? the angles BAD, A DE [by prop. 29.] are equal to two right angles. But B [by confir.] BAD is a right angle: Therefore will A D E be a right angle too. But [by prop 34.] the oppofite fides and angles of any parallelogram are equal wherefore each of the oppofite angles ABE, BED is a right angle; and fo A D E B is right-angled. It has also been proved to be equilateral. Therefore it is neceffarily a fquare, and is defcribed upon the right line a B. Which was to be done.

Some have thought that Euclid should have demonstrated, that the fquares defcribed upon equal lines are equal; and that if the fquares are equal, the lines upon which they are defcribed are alfo equal. But he thought there was no occafion for this, it being too evident to require a demonftration; for if the fquares are fuppofed to be laid upon one another, they will agree together, and fo will all their fides and angles; that is, they will be all equal.

Proclus demonftrates it more at large; but I think he might as well have let it alone.

PROP. XLVII.

THEOR.

In right-angled triangles, the fquare defcribed upon the fide oppofite to the right angle, is equal to both the fquares defcribed upon the fides containing the right angle P.

Let ABC be a right-angled triangle, having the right angle BA C: I fay, the fquare defcribed upon the right line B C, is equal to both the fquares defcribed upon the fides B A, AC,

For upon B C defcribe the fquare B DEC; and upon BA, A C, the fquares GB, HC; and through A draw A L parallel to B D or CE, and draw A D, F C.

Then

F

В

D

H

LE

Then because the angles BAC, BAG are each right an`gles; and two right lines AC, AG not lying the fame way, make the adjacent angles, at the point A therein, equal to two right angles; the right lines CA, AG are in the fame direction. And for the same reason AB and Aн make but one right line. And becaufe the angle DBC [by ax. 10.] is equal to the angle F B A, for they are both right angles. Add the angle ABC, which is common; then will the whole angle DBA be equal to the whole angle F BC. But because the two fides DB, BA are equal to the two fides CB, BF, each to each; and the angle DBA is equal to the angle FBC; the base A D [by prop. 4.] will be equal to the bafe FC; and the triangle ABD to the triangle F B C. But [by prop. 41.] the parallelogram BL is double to the triangle ABD; for they have the fame base B D, and are between the fame parallels B D, AL: but the fquare G B is double to the triangle FBC, fince they have the fame base FB, and are between the fame parallels FB, GC. But the doubles of equal things are themselves equal: therefore the parallelogram B L is equal to the fquare GB. After the fame manner, by drawing AE, BK, we demoftrate, that the parallelogram CL is alfo equal to the fquare H C. Therefore the whole fquare B DEC is equal to both the fquares GB, HC. And BDEC is the fquare defcribed upon the right lines B C; but the squares GB, HC are described upon the fides BA, AC. Therefore the fquare described upon the fide BC, is equal to both the fquares described upon the fides BA, AC.

Therefore in right-angled triangles, the fquare defcribed upon the fide oppofite to the right angle, is equal to both the fquares defcribed upon the fides containing the right angle. Which was to be demonftrated.

P This famous, ufeful, and elegant propofition is reckoned to have Pythagoras for its inventor. It may be demonftrated feveral other ways. See Clavius, Schouten, and others; but there is none fo fimple and elegant as that of Euclid. The propofition is only a particular cafe of prop. 31. lib. 6.

As by this propofition it is easy to find a right line whofe fquare

E 2

Book I. fquare fhall be equal to the fquares of two given right lines, fo it is as difficult to find a right line, whofe cube fhall be equal to the cubes of two given right lines. Indeed, by a right line and circle this is impoffible to do at all. But, granting the defcription of the conic fections, or the way of finding two mean proportionals between two given right lines, it may be done with tolerable ease.

PROP. XLVIII. THEOR. If the fquare defcribed upon one fide of a triangle be equal to both the fquares defcribed upon the other two fides of that triangle; the angle contained by thefe two remaining fides will be a right angle,

For let the fquare defcribed upon one fide B C of the triangle ABC be equal to both the fquares defcribed upon the remaining fides BA, AC of that triangle: I fay, the angle B A C is a right angle.

For [by prop. 11.] draw A D from the point a at right Dangles to A C. Put A D equal

to B A, and draw D C.

Then because A D is equa

B C,

to AB; the fquare defcribed upon D A will be equal to the fquare defcribed upon A B. Add the common fquare defcribed upon A C. Then the fquares described upon D A, A c are equal to the fquares described upon в A, A c. But [by prop, 47.] the fquare defcribed upon DC is equal to both the fquares described upon DA, A C, for the angle DAC is a right one: but the fquare described upon is put equal to both the fquares defcribed upon B A, a C. Therefore the fquare described upon DC is equal to the fquare defcribed upon BC; and fo the fide DC is equal to the fide B C. And because A D is equal to A B, and A C is common, the two fides A D, AC are equal to the two fides BA, A C, and the base D c is equal to the base C B : Therefore [by prop. 8.] the angle DA C is equal to the angle B A C. But the angle D A c is a right angle: Wherefore B A C will be a right angle too.

If therefore the fquare which is defcribed upon one fide of a triangle be equal to both the fquares defcribed upon the other two fides, the angle contained by these other two ides will be a right angle. Which was to be demonftrated.

Additions