PROPOSITION XXXI.—PROBLEM. To draw a straight line through a given point parallel to a given straight line. LET a be the given point, and bc the given straight line; it is required to draw a straight line through the point a, parallel to the straight line b c. h d f In be take any point d, and join a d; and at the point a, in the straight line ad, make (i. 23) the angle da e equal to the angle a dc; and produce the straight line ea to f Because the straight line ad, which meets the two straight lines bc, ef, makes the alternate angles e a d, adc equal to one another, ef is parallel (i. 27) to bc. Therefore the straight line e af is drawn through the given point a parallel to the given straight line bc. Which was to be done. PROPOSITION XXXII.-THEOREM. If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are equal to two right angles. LET abc be a triangle, and let one of its sides bc be produced to d; the exterior angle a cd is equal to the two interior and opposite angles cab, abc, and the three interior angles of the triangle, viz. abc, bca, cab, are together equal to two right angles. Through the point c draw ce parallel (i. 31) to the straight line a b; and b a e because ab is parallel to ce, and ac meets them, the alternate angles bac, ace, are equal (i. 29). Again, because a b is parallel to ce, and bd falls upon them, the exterior angle e cd is equal to the interior and opposite angle dabc; but the angle ace was shewn to be equal to the angle bac; therefore the whole exterior angle acd is equal to the two interior and opposite angles ca b, abc; to these equals add the angle a cb, and the angles a cd, a cb are equal to the three angles cba, bac, acb; but the angles acd, a cb are equal (i. 13) to two right angles: therefore also the angles cba, bac, a cb, are equal to two right angles. Wherefore, if a side of a triangle, &c. Q. E. D. COR. 1. All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides. e d For any rectilineal figure abcde, can be divided into as many triangles as the figure has sides, by drawing straight lines from a point f within the figure to each of its angles. And, by the preceding proposition, all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure and the same angles are equal to the angles of the figure, together with the angles at the point f, which is the common vertex of the triangles: that is (i. 15, cor. 2), together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides. a h COR. 2. All the exterior angles of any rectilineal figure are together equal to four right angles. Because every interior angle a bc, with its adjacent exterior a bd, is equal (i. 13) to two right angles; therefore all the interior, together with all the exterior angles of the figure, are equal to twice as many right angles as there are sides of the figure; that is, by the foregoing corollary, they are equal to all the interior angles of the figure, together with four right angles; therefore all the exterior angles are equal to four right angles. d b PROPOSITION XXXIII.—THEOREM. The straight lines which join the extremities of two equal and parallel straight lines towards the same parts are also themselves equal and parallel. LET ab, cd be equal and parallel straight lines, and joined towards the same parts by the straight lines ac, bd; ac, bd are also equal and parallel. Join be; and because a bis parallel to cd, and bc meets them, the alternate angles abc, bcd are equal (i. 29); and because ab is equal to cd, and bc common to the two triangles a bc, dcb, the two sides a b, bc, are equal to the two dc, cb: and the angle abc is equal to the angle bcd; therefore a b d the base ac is equal (i. 4) to the base bd, and the triangle abc to the triangle bcd, and the other angles to the other angles, (i. 4) each to each, to which the equal sides are opposite; therefore the angle a cb is equal to the angle cbd; and because the straight line bc meets the two straight lines a c, bd, and makes the alternate angles a cb, cbd equal to one another, a c is parallel (i. 27) to bd; and it was shewn to be equal to it. Therefore straight lines, &c. Q. E. D. PROPOSITION XXXIV. THEOREM. The opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them, that is, divides them into two equal parts. A parallelogram is a four-sided figure, of which the opposite sides are parallel ; and the diameter is the straight line joining two of its opposite angles. Let a cd b be a parallelogram, of which bc is a diameter; the opposite sides and angles of the figure are equal to one another; and the diameter b c bisects it. a Because a b is parallel to cd, and bc meets them, the alternate angles c b abc, bcd are equal (i. 29) to one another; and because a c is parallel to bd, and bc meets them, the alternate angles a cb, cbd, are equal (i. 29) to one another; wherefore the two triangles abc, cbd d have two angles a bc, bca in one, equal to two angles bcd, cbdin the other, each to each, and one side bc common to the two triangles, which is adjacent to their equal angles; therefore their other sides shall be equal, each to each, and the third angle of the one to the third angle of the other (i. 26), viz. the side ab to the side cd, and ac to bd, and the angle bac equal to the angle bdc and because the angle abc is equal to the angle bcd, and the angle cbd to the angle a cb, the whole angle abd is equal to the whole angle acd: and the angle bac has been shewn to be equal to the angle bdc; therefore the opposite sides and angles of parallelograms are equal to one another; also, their diameter bisects them; for a b being equal to cd, and bc common, the two ab, bc are equal to the two dc, cb, each to each; and the angle abc is equal to the angle bcd; therefore the triangle abc is equal (i. 4) to the triangle b cd, and the diameter bc divides the parallelogram a cd b into two equal parts. Q. E. D. PROPOSITION XXXV.-THEOREM. Parallelograms upon the same base, and between the same parallels, are equal to one another. LET the parallelograms abcd, ebcf (see the second figure) be upon the same base bc, and between the same parallels af, bc; the parallelogram abcd shall be equal to the parallelogram ebc f If the sides a d, df of the parallelograms abcd, dbcf, opposite to the base bc, be terminated in the same point d; it is plain that each of the parallelograms is double (i. 34) of f the triangle bdc; and they are therefore equal to one another. a b But, if the sides ad, ef (see second and third figures), opposite to the base bc of the parallelograms a b c d, ebcf, be not terminated in the same point d; then, because abcd is a parallelogram, ad is equal (i. 34) to bc; for the same reason ef is equal to bc; wherefore ad is equal (1 ax.) to ef; and de is common; therefore the whole, or the remainder, ae is equal (2 or 3 ax.) to the whole, or the remainder df; ab also is equal to dc; and the two ea, ab are therefore equal to the two fd, d c, each to each; and the exterior angle fdc is equal (i. 29) to the interior e a b, therefore the base e b is equal to the base fc, and the triangle e ab equal (i. 4) to the triangle fd c. Take the triangle fdc from the trapezium a bc f, and from the same trapezium take the triangle e ab: the remainders therefore are equal (3 ax.), that is, the parallelogram a bcd is equal to the parallelogram e bcf. fore parallelograms upon the same base, &c. Q. E. D. There Parallelograms upon equal bases, and between the same parallels, are equal to one another. LET abcd, efgh, be parallelograms upon equal bases bc, fg, and between the same parallels ah, bg; the parallelogram abcd is equal to efgh. b f g Join be, ch; and because bc is equal to fg, and fg to eh (i. 34), bc is equal to eh; and they are parallels, and joined towards the same parts by the straight lines be, ch. But straight lines which join equal and parallel straight lines towards the same parts, are themselves equal and parallel (i. 33); therefore e b, c h, are both equal and parallel, and e bch is a parallelogram; and it is equal (i. 35) to a bcd, because it is upon the same base bc, and between the same parallels bc, ad. For the like reason, the parallelogram efgh is equal to the same ebch. Therefore also the parallelogram abcd is equal to efgh Wherefore parallelograms, &c. Q. E. D. PROPOSITION XXXVII.-THEOREM. Triangles upon the same base, and between the same parallels, are equal to one another. LET the triangles abc, dbc, be upon the same base bc and between e b the same parallels ad, bc. The triangle abc is equal to the triangle dbc. C Produce ad both ways to the points e, f, and through b draw (i. 31) be parallel to ca; and through c draw cf parallel to bd; therefore each of the figures ebca, dbcf is a parallelogram; and ebca is equal (i. 35) to dbcf, because they are upon the same base bc, and between the same parallels bc, ef; and the triangle abc is the half of the parallelogram ebca, because the diameter a b bisects it (i. 34); and the triangle dbc is the half of the parallelogram dbcf, because the diameter dc bisects it: but the halves of equal things are equal (7 ax.); therefore the triangle abc is equal to the triangle dbc. Wherefore triangles, &c. Q. E. D. PROPOSITION XXXVIII.-THEOREM. Triangles upon equal bases, and between the same parallels, are equal to one another. LET the triangles abc, def be upon equal bases bc, ef, and between the same parallels bf, a d. The triangle abc is equal to the triangle def g a d h Produce ad both ways to the points g, h, and through b draw bg parallel (i. 31) to ca, and through f draw fh parallel to ed. Then each of the figures gbca, defh, is a parallelogram; and they are equal (i. 36) to one another, because they are upon equal bases bc, ef, and between the same parallels bf, gh; and the triangle abc is the half (i. 34) of the parallelogram gba, because the diameter ab bisects it; and the triangle def is the half (i. 34) of the parallelogram defh, because the diameter df bisects it: but the halves of equal things are equal (7 ax.); therefore the triangle a b c is equal to the triangle def. Wherefore triangles, &c. Q. E. D. b с е f PROPOSITION XXXIX.—THEOREM. Equal triangles upon the same base, and upon the same side of it, are between the same parallels. LET the equal triangles abc, dbc be upon the same base bc, and upon the same side of it; they are between the same parallels. Join ad; ad is parallel to be; for, if it is not, through the point a draw |