and all the three inward angles of every triangle are equal to two right angles. Which was to be demonítrated. * Eudemus fays, the Pythagoreans first of all demonstrated, that the three angles of a triangle are equal to two right angles, after the following manner: A D E B Let there be a triangle A B C, and through the point a [by 31. 1.] draw the right line D E parallel to в с. Then because [by 29. 1.] the alternate angles D AB, ABC are equal to one another, if the equal angles EAC, ACв be added (for these are alternate an C gles); the two angles DAB, EAC [by axiom 2.] will be equal to the two angles ABC, ACB. And fo, adding the common angle BAC, the three angles DAB, BAC, CA will be equal to the three angles ABC, BAC, ACB. But the angles DAB, BAC, CAE are equal to two right angles; as is evident (by prop. 13. 1.]: Therefore the three angles A BC, BAC, ACB are equal to two right angles. Which was to be demonftrated. PROP. XXXIII. THEOR. Two right lines, which join two equal and parallel right lines, towards the same parts, are also themSelves equal and parallel1. Let the right lines A C, B D join towards the fame parts the equal and parallel right lines A B, CD: I say, AC, в р are equal and parallel. A For join в с. B Then because A B is parallel to CD, and the right line B c falls upon them, the alternate angles A BC, BCD [by prop. 29.] are equal. Again, because AB is equal to c D, and Bc is common, the two fides A B, B C are equal to the two fides CD, Bc, and the angle ABC equal to the angle BCD: Therefore [by prop. 4.] the base A c is equal to the base B D, and the triangle A B C equal to the triangle в CD. And the remaining angles of the one triangle will be equal to the remaining angles of the other, each to each, which C D are are opposite to the equal fides. Therefore the angle ACB is equal to the angle CBD. And because the right line в с, falling upon the two right lines AC, BD, makes the alternate angles ACB, CBD equal to one another; the right line A c [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 demonftrated. If two right lines do not the fame way join the equal parallel right lines, that is, do not join the answerable extremes A, C, and B, D, of the equal and parallel right lines AB, 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 seldom; 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 extensive 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 demonstration does. I also think, the proposition, as generally taken, might have been more clearly expressed; thus: If the extremes of two equal and parallel right lines be joined by the extremes of two other right lines, not so as that these lines cut one another, these right lines will also be equal and parallel. PROP. XXXIV. THEOR. The opposite fides and angles of any parallelogram or figure bounded by parallel lines, are equal to one another, and a diameter bisects it ". Let there be a parallelogram ACDB, and its diameter BC: I say, the opposite sides 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 в с falls A upon them, the alternate angles ABC, BCD [by prop. 29.] are equal to one another. Again, because a c is parallel to C B D BD, and the right line BC falls upon them, the alternate angles ACB, CBD are equal to one another: Therefore there are two triangles ABC, CBD, which have two angles ABC, BCA equal to two angles BCD, CBD of the other, each to each; and one fide of the one equal to one fide of the other, viz. the side BC between the equal angles, which is common to both. Therefore [by prop. 26.] the remaining fides shall be equal to the remaining fides, each to each; 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 BAC to the angle B D C. And because the angle ABC is equal to the angle BCD, and the an le CBD equal to the angle ACB; the whole angle ABD will be equal to the whole angle ACD. has been also demonftrated, that the angle BAC is equal to the angle BDC. It Therefore the oppofite fides and angles of any parallelogram [or four-fided figure bounded by parallel lines] are equal. I say also, the diameter bisects it. For because A B is equal to c D, and в с is common; the two fides AB, 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 so the triangle ABC will be equal to the triangle B CD. Therefore the diameter B C cuts the parallelogram ACDB into two equal parts. Which was to be demonftrated. m Some have thought, that Euclid should 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 unnecessarily increase the number of his definitions. In effect, he himself has told us in the proposition what it is, viz. a parallel-lined space, i. e. a space contained under parallel lines, paralleloi signifying, in Greek, parallels, and gramma a line: so that a Grecian, at least, could scarcely be at a loss to know the fignification of the word parallelogram, after he knew what were lines, and what were parallels. But whether it be so 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 spaces only parallelograms. PROP THEOR. PROP. XXXV. Parallelograms constituted upon the same base, and between the same parallels, are the one equal to the other. Let the parallelograms ABCD, EBCF be constituted upon the fame base Bc, and between the fame parallels AF, BC: Ifay, the parallelogram A B C D is equal to the parallelogram E BCF. For because ABCD is a parallelogram [by prop. 34.] AD is equal to BC; and for the fame reason EF also is equal to BC: wherefore AD is also equal to E F, and D E is common to both. Wherefore the whole A E A D EF is [by ax. 2.] equal to the whole DF. But AB [by prop. 34.] is equal to DC; therefore the two fides EA, A B of the triangle AEB G C are equal to the two fides FD, DC B of the triangle CDF, 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 EB is equal to the base Fc, 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 BEC, the whole parallelogram ABCD will be equal to the whole parallelogram E B CF. Therefore parallelograms constituted upon the same base, and between the same parallels, are the one equal to the other. Which was to be demonftrated. PROP. XXXVI. THEOR. Parallelograms constituted upon equal bases, and between the same parallels, are equal the one to the other. Let the parallelograms ABCD, EFGH be constituted upon the equal bases BC, FG, and between the fame parallels rallels AH, BG: I say, the parallelograms A B CD, EFGH are equal to one another. A D EH G For join BE, Cн. Then because BC is equal to FG, and FG equal to EH; therefore will в с be equal to E H. They are parallel too, and BE, CH joins them. But right lines which join equal and parallel right lines towards the fame parts, are equal and parallel too [by prop. 33.]: B C F Therefore EB, сн 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 same base в с, and is conftituted between the fame parallels, BC, AH. By the like way of reasoning, the parallelogram E FGHIS equal to this parallelogram EBCH: therefore the parallelogram ABCD is equal to the parallelogram E F G H too. Wherefore parallelograms conftituted upon equal bases, and between the fame parallels, are equal the one to the other. Which was to be demonstrated. PROP. XXXVII. THEOR. Triangles constituted upon the same base, and between the same parallels, are equal to one another. Let the triangles ABC, DBC be constituted upon the fame base в с, and between the fame parallels AD, BC: I fay, the triangle A B C is equal to the triangle D B C. For continue out A D both ways to the points E and F; and through B [by prop. 31.] draw the right line в E parallel to the right line CA; and through c draw cr 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 stand both upon the fame base в с, and are between the same pa rallels BC, EF. But the triangle |