to each of these equals ; therefore the angles cbe, eb d are equal (2 ax.) to the three angles cba, abe, ebd. Again, because the angle dba is equal to the two angles dbe, eba, add to these equals the angle a bc, therefore the angles dba, abc are equal to the three angles d be, eba, abc but the angles cbe, ebd have been demonstrated to be equal to the same three angles; and things that are equal to the same are equal (1 ax.) to one another; therefore the angles cbe, ebd are equal to the angles dba, abc; but cbe, ebd are two right angles; therefore dba, a b c are together equal to two right angles. Wherefore, the angles which one straight line, &c. Q. E. D. PROPOSITION XIV.-THEOREM. If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line. At the point b in the straight line a b, let the two straight lines bc, bd b a e d upon the opposite sides of a b, make the adjacent angles abc, abd equal together to two right angles, b d is in the same straight line with c b. For, if b d be not in the same straight line with cb, let be be in the same straight line with it; therefore, because the straight line ab makes angles with the straight line cbe, upon one side of it, the angles a b c, abe are together equal (i. 13) to two right angles; but the angles a bc, abd are likewise together equal to two right angles; therefore the angles cba, a be are equal to the angles c ba, abd: take away the common angle cba, the remaining angle a be is equal (3 ax.) to the remaining angle a b d, the less to the greater, which is impossible; therefore be is not in the same straight line with bc. And in like manner it may be demonstrated, that no other can be in the same straight line with it but b d, which therefore is in the same straight line with cb. Wherefore, if at a point, &c. Q. E. D. PROPOSITION XV.—THEOREM. If two straight lines cut one another, the vertical, or opposite, angles shall be equal. LET the two straight lines a b, cd cut one another in the point e; the angle a ec shall be equal to the angle de b, and ceb to aed. Because the straight line a e makes with c d the angles cea, aed, these angles are together equal (i. 13) to two right angles. Again, because the straight line de makes with ab the angles aed, deb, these also are together equal (i. 13) to two right angles; and cea, a ed, have been demonstrated to be equal to two right angles; wherefore the angles cea, aed, are equal to the angles aed, deb. Take away the common angle a e d, and the remaining angle ce a is equal (3 ax.) to the remaining angle de b. In the same manner it can be demonstrated, that the angles ceb, aed are equal. Therefore, if two straight lines, &c. Q. E. D. COR. 1. From this it is manifest, that, if two straight lines cut one another, the angles they make at the point where they cut, are together equal to four right angles. COR. 2. And consequently that all the angles made by any number of lines meeting in one point, are together equal to four right angles. PROPOSITION XVI.-THEOREM. If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles. LET a b c be a triangle, and let its side b c be produced to d, the exterior angle a cd is greater than either of the interior opposite angles cba bac. Bisect (i. 10) a c in e, join be and produce it to f, and make e f equal to be; join also fc, and produce a c to g. Because ae is equal to ec, and be to ef; a e, eb are equal to ce, ef, each to each; and the angle a e b is equal (i. 15) to the angle cef, because they are opposite vertical angles; therefore the base ab is equal (i. 4) to the base cf, and the triangle aeb to the triangle cef, and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite; wherefore the angle bae is equal to the angle ecf; but the angle ecd is greater than the angle ecf; therefore the angle a cd is greater than b a c d g bae. In the same manner, if the side b c be bisected, it may be demonstrated that the angle bcg, that is, (i. 15) the angle a cd, is greater than the angle a bc. Therefore, if one side, &c. Q. E. D. PROPOSITION XVII.—THEOREM. Any two angles of a triangle are together less than two right angles. LET abc be any triangle; any two of its angles together are less than two right angles. Produce bc to d; and because acd is the exterior angle of the triangle abc, acd is greater (i. 16) than the interior and opposite angle abc; to each of these add the angle acb; therefore the angles a cd, acb are greater than the angles a b c, a cb; but a cd, acb are together equal (i. 13) to two right angles; therefore the angles abc, bca are b a с d 14 less than two right angles. In like manner it may be demonstrated, that bac, acb, as also cab, abc, are less than two right angles. Therefore any two angles, &c. Q. E. D. b PROPOSITION XVIII.-THEOREM. The greater side of every triangle is opposite to the greater angle. d с LET abc be a triangle, of which the side ac is greater than the side ab; the angle abc is also greater than the angle bca. Because a c is greater than ab, make (i. 3) cis a d equal to a b, and join bd; and because adb is the exterior angle of the triangle bd c, it is greater (i. 16) than the interior and opposite angle dcb; but a db is equal (i. 5) to a bd, because the side a b is equal to the side ad; therefore the angle abd is likewise greater than the angle acb. angle a bc than a cb. Therefore the greater side, &c. Q. E. D. Much greater then is the PROPOSITION XIX.-THEOREM. The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it. LET abc be a triangle, of which the angle abc is greater than the angle bca; the side a c is likewise greater than the side a b. a For, if it be not greater, ac must either be equal to ab, or less than it; it is not equal, because then the angle abc would be equal (i. 5) to the angle a cb; but it is not; therefore a c is not equal to ab; neither is it less; because then the angle a b c would be less than the angle a cb (i. 18); but it is not; therefore the side a c is not less than ab; and it has been shewn that it is not equal to ab; therefore ac is greater than a b. Wherefore the greater angle, &c. b Q. E. D. PROPOSITION XX. THEOREM. Any two sides of a triangle are together greater than the third side. LET abc be a triangle; any two sides of it together are greater than the third side, viz. the sides ba, ac greater than the side bc; and a b, bc greater than ac; and bc, ca greater than a b. Produce ba to the point d, and make (i. 3) ad equal to ac; and join d c. Because da is equal to a c, the angle a dc is likewise equal to a c d (i. 5); but the angle bcd is greater than the angle acd; therefore the angle op d a bcd is greater than the angle adc; and because the angle bed of the triangle dcb is greater than its angle bdc, and that the greater (i. 19) side is posite to the greater angle, therefore the side db is greater than the side bc; but db is equal to ba and a c; therefore the sides ba, a c are greater than bc. In the same manner it may be demonstrated, that the sides a b, bc, are greater than ca, and bc, ca greater than a b. Therefore any two sides, &c. Q. E. D. b PROPOSITION XXI.-THEOREM. If from the ends of the side of a triangle there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle. LET the two straight lines bd, cd be drawn from b, c, the ends of the side bc of the triangle a bc, to the point d within it; bd and dc are less than the other two sides ba, ac of the triangle, but contain an angle bdc greater than the angle bac a e a Produce bd to e; and because two sides of a triangle are greater than the third side, the two sides ba, a e, of the triangle abe are greater than be. To each of these add ec; therefore the sides ba, ac are greater than be, ec. Again, because the two sides ce, ed of the triangle ced are greater than cd, add db to each of these; therefore the sides ce, eb are greater than cd, db; but it has been shewn that ba, a c, are greater than be, ec, much greater then are ba, ac than bd, dc. b Again, because the exterior angle of a triangle is greater than the interior and opposite angle, the exterior angle bdc of the triangle cde is greater than ced; for the same reason, the exterior angle ceb of the triangle abe is greater than bac; and it has been demonstrated that the angle bdc is greater than the angle ceb; much greater then is the angle bdc than the angle bac. Therefore, if from the ends of, &c. Q. E. D. PROPOSITION XXII.-PROBLEM. To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third (i. 20.). LET a, b, c be the three given straight lines, of which any two whatever are greater than the third, viz. a and b greater than c; a and c greater than b; and b and c than a It is required to make a triangle of which the sides shall be equal to a, b, c, each to each. Take a straight line de terminated at the point d, but unlimited towards e, and make (i. 3) df equal to a, fg to b, and gh equal to c; and from the centre f, at the distance fd, describe (3 post.) the circle dkl; and from the centre g, at the distance gh, describe (3 post.) another circle hlk; and join kf, kg; the triangle kfg has its sides equal to the three straight lines a, b, c. Because the point fis the centre of the circle d kl, fd is equal (15 def.) to fk; but fd is equal to the straight line a; therefore fk is equal to a: again, because g is the centre of the circle 1kh, gh is equal (15 def.) to gk; but gh is equal to c; therefore also gk is equal to c; and fg is equal to b; therefore the three straight lines kf, fg, gk, are equal to the three a, b, c and therefore the triangle kfg has its three sides k f, fg, gk, equal to the three given straight lines a, b, c. Which was to be done. PROPOSITION XXIII.-PROBLEM. At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. LET ab be the given straight line, and a the given point in it, and dce the given rectilineal angle; it is required to make an angle at the given point a in the given straight line a b, that shall be equal to the given rectilineal angle dce. Take in cd, ce any g points d, e, and join de; and make (i. 22) the triangle afg, the sides of which shall be equal to the three straight lines cd, de, ec, so that cd be equal to af, ce to ag, and de to fg; and because dc, ce |