and all the three inward angles of every triangle are equal to two right angles. Which was to be demonítrated. A D E Eudemus fays, the Pythagoreans first of all demonftrated, that, the three angles of a triangle are equal to two right angles, after the following manner: Let there be a triangle ABC, and through the point A [by 31. 1.] draw the right line D E parallel to B C. Then becaufe [by 29. 1.] the alternate angles DA B, ABC are equal to one another, if the equal angles E AC, A CB be added (for these are alternate angles); the two angles DA B, EAC [by axiom 2.] will be equal to the two angles ABC, A CB. And fo, adding the common angle BAC, the three angles DA B, BAC, CAE will be equal to the three angles ABC, BAC, But the angles DA B, BA C, CA E are equal to two right angles; as is evident [by prop. 13. 1.]: Therefore the three angles A B C, BAC, A CB are equal to two right angles. Which was to be demonftrated. B A CB. C PROP. XXXIII. THEOR. Two right lines, which join two equal and parallel right lines, towards the fame parts, are also themfelves equal and paralleli. Let the right lines A C, B D join towards the fame parts the equal and parallel right lines A B, CD: I fay, A C, B D are equal and parallel. A For join B C. C B Then because A B is parallel to c D, and the right line B C falls upon them, the alternate angles A B C, B C D [by prop. 29.] are equal. Again, because AB is equal to c D, and B C is common, the two fides A B, BC are equal to the two fides CD, BC, and the angle ABC equal to the angle BC D: Therefore [by prop. 4.] the bafe A c is equal to the bafe B D, and the triangle A B C equal to the triangle B C D. And the remaining angles of the one triangle will be equal to the remaining angles of the other, each to each, which D are are oppofite to the equal fides. Therefore the angle ACB is equal to the angle C B D. And because the right line BC, falling upon the two right lines AC, BD, makes the alternate angles AC B, CBD equal to one another; the right line AC [by prop. 27.] is parallel to the right line в D. It has also been proved to be equal to it. Therefore two right lines which join two equal and parallel right lines, towards the fame parts, are themselves both equal and parallel. Which was to be demonstrated. If two right lines do not the fame way join the equal parallel right lines, that is, do not join the answerable extremes ▲, c, and B, D, of the equal and parallel right lines A B, CD; but the alternate ones A, D; B, C, croffing one another. The latter part of the propofition will never be true, and the former one but feldom; therefore Euclid was obliged to say, that the right lines muft join the equal and parallel ones the fame way. This propofition of Euclid is more extenfive than it is generally thought to be. For it neither confines the number of the right lines joining the equal and parallel right lines to two, nor the equal parallels to two, though the demonftration does. I alfo think, the propofition, as generally taken, might have been more clearly expreffed; thus: If the extremes of two equal and parallel right lines be joined by the extremes of two other right lines, not fo as that thefe lines cut one another, these right lines will alfo be equal and parallel. PROP. XXXIV. THEOR. The oppofite fides and angles of any parallelogram or figure bounded by parallel lines, are equal to one another, and a diameter bifects it ". Let there be a parallelogram ACDB, and its diameter BC: I fay, the oppofite fides and angles of the parallelogram ACDB are equal, and a diameter divides it into two equal parts. For because A B is parallel to CD, and the right line B c falls upon them, the alternate angles ABC, BCD [by prop. 29.] are equal to one another. Again, because A c is parallel to A B D D 4 BD, BD, and the right line BC falls upon them, the alternate angles ACB, CBD are equal to one another: Therefore there are two triangles A B C, C B D, which have two angles ABC, BC A equal to two angles B C D, CBD of the other, each to each; and one fide of the one equal to one fide of the other, viz. the fide BC between the equal angles, which is common to both. Therefore [by prop. 26.] the remaining fides fhall be equal to the remaining fides, each to cach; and the remaining angle to the remaining angle. Wherefore the fide A B is equal to the fide c D, and the fide AC to BD, and the angle B A C to the angle B DC. And because the angle ABC is equal to the angle BC D, and the an le CBD equal to the angle AC B; the whole angle A B D will be equal to the whole angle A CD. It has been alfo demontrated, that the angle B A C is equal to the angle B DC. Therefore the oppofite fides and angles of any parallelogram [or four-fided figure bounded by parallel lines] are equal. I fay alfo, the diameter bifects it. For becaufe A B is equal to c D, and B C is common; the two fides A B, BC are each equal to the two fides DC, CB, and the angle ABC is equal to the angle BCD: Therefore the base A C [by prop. 4.] is equal to the base B D; and fo the triangle ABC will be equal to the triangle B C D. Therefore the diameter B C cuts the parallelogram ACDB into two equal parts. Which was to be demonftrated, m Some have thought, that Euclid fhould have particularly defined a parallelogram at the beginning of his Elements; but he did not think fit to do it; because, perhaps, he had not a mind to unneceffarily increase the number of his definitions. In effect, he himself has told us in the propofition what it is, viz. a parallel-lined fpace, i. e. a fpace contained under parallel lines, paralleloi fignifying, in Greek, parallels, and gramma a line fo that a Grecian, at least, could scarcely be at a lofs to know the fignification of the word parallelogram, after he knew what were lines, and what were parallels. But whether it be fo right to leave out the word space, as it is done all along afterwards throughout the whole Elements, I cannot well tell, unless it be for the fake of brevity, to call parallelogram fpaces only parallelograms. : PROP. PROP. XXXV. THEOR. Parallelograms conftituted upon the fame bafe, and between the fame parallels, are the one equal to the other. Let the parallelograms ABCD, E B C F be conftituted upon the fame bafe BC, and between the fame parallels AF, BC: Ifay, the parallelogram A B C D is equal to the parallelogram E BCF. A DE F C For because A B C D is a parallelogram [by prop. 34.] AD is equal to BC; and for the fame reafon EF alfo is equal to BC: wherefore AD is also equal to E F, and DE is common to both. Wherefore the whole A E is [by ax. 2.] equal to the whole DF. But AB [by prop. 34.] is equal to DC; therefore the two fides E A, AB of the triangle A EB are equal to the two fides F D, DC B of the triangle C D F, each to each; and the angle FDC [by prop. 29.] equal to the angle E A B, the outward one to the inward one. Therefore [by prop. 4.] the base E B is equal to the base F C, and the triangle E A B equal to the triangle FDC. If the triangle DGE, which is common to both, be taken away, there will remain the trapezium ABGD [by ax. 3.] equal to the trapezium remaining EGCF. If to both these be added the common triangle BGC, the whole parallelogram ABCD will be equal to the whole parallelogram E B C F. Therefore parallelograms conftituted upon the same base, and between the fame parallels, are the one equal to the other. Which was to be demonstrated. PROP. XXXVI. THEOR. Parallelograms conftituted upon equal bases, and between the fame parallels, are equal the one to the other. Let the parallelograms ABC D, EFGH be conftituted upon the equal bafes BC, FG, and between the fame pa rallels rallels AH, BG: I fay, the parallelograms A B C D, E F G H are equal to one another. For join BE, CH. A E F H G But Then because B C is equal to FG, and FG equal to E H; therefore will B C be equal to E H. They are parallel too, and BE, CH joins them. right lines which join equal and parallel right lines towards the fame parts, are equal and parallel too [by prop. 33.]: Therefore EB, CH are both equal and parallel; and fo EBCH is a parallelogram, which [by prop. 35.] is equal to the parallelogram ABCD; for it has the fame base B C, and is conftituted between the fame parallels, BC, A H. By the like way of reafoning, the parallelogram EFGH is equal to this parallelogram EBCH: therefore the parallelogram ABCD is equal to the parallelogram E F G H too. B Wherefore parallelograms conftituted upon equal bafes, and between the fame parallels, are equal the one to the other. Which was to be demonstrated. PROP. XXXVII. THEOR. Triangles conftituted upon the fame bafe, and between the fame parallels, are equal to one another. Let the triangles ABC, DBC be conftituted upon the fame bafe B C, and between the fame parallels AD, BC: I fay, the triangle A B C is equal to the triangle D BC. For continue out A D both ways to the points E and F ; and through B [by prop. 31.] draw the right line B E parallel to the right line CA; and through c draw C F pa rallel to B D. Then EBCA, DBCF are F both parallelograms, and the parallelogram E BCA [by prop. 35.] is equal to the parallelogram DBCF, for they ftand both upon the fame base в C, and are between the fame parallels BC, EF. But the triangle |