to each of these equals ; therefore the angles cbe, eb d are equal (2 ax.) to the three angles cba, a be, ebd. Again, because the angle d ba is equal to the two angles dbe, eba, add to these equals the angle a b c, therefore the angles d ba, a b c are equal to the three angles d be, e ba, a b c : but the angles cbe, e bd 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 d ba, a b c are together equal to two right angles. Wherefore, the angles which one straight line, &c. Q. E. D. PROPOSITION XIV.—THEOREM. 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 upon the opposite sides of a b, make the a adjacent angles a b c, a b d equal together For, if b d be not in the same straight line with it ; therefore, because the straight a C b d cbe, upon one side of it, the angles a b c, a be are together equal (i. 13) to two right angles; but the angles a b c, abd are likewise together equal to two right angles; therefore the angles c ba, a be are equal to the angles cba, a bd: 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 b c. 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. shall be equal. с deb, these also are together equal (i. 13) a b to two right angles ; and cea, a ed, have been demonstrated to be equal to two right d angles; wherefore the angles cea, a ed, are equal to the angles a ed, de b. Take away the common angle a ed, and the remaining angle cea is equal (3 ax.) to the remaining angle de b. In the same manner it can be demonstrated, that the angles ceb, a ed 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. a 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 c ba, bac. Bisect (i. 10) acin e, join be and produce it to f, and make e f equal to e be; join also fc, and produce a c to g. a Because a e is equal to ec, and be to ef; a e, e b are equal to ce, e f, each to each ; and the angle a e bis equal (i. 15) to the angle cef, because they are opposite vertical angles; therefore the base a b is equal (i. 4) to the base cf, and the triangle a eb to the triangle cef, and the remaining angles to the remaining angles, each to each, to which the equal sides b d opposite ; wherefore the angle ba e is equal to the angle ecf; but the angle e cd is greater than the angle e cf; g therefore the angle a cd is greater than bae. In the same manner, if the side bc 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. are PROPOSITION XVII.-THEOREM. a angles together are less than two right angles. Produce bc to d; and because a cd is the exterior angle of the triangle a b c, acd is greater (i.16) than the interior and opposite angle a bc; to each of these add the angle acb; therefore the angles a cd, a cb are greater than the angles a b c, acb; but a cd, acb are together equal (i. 13) to two right angles; therefore the angles a b c, b ca are d less than two right angles. In like manner it may be demonstrated, that bac, a cb, as also cab, a b c, are less than two right angles. Therefore any two angles, &c. Q. E. D. PROPOSITION XVIII.-THEOREM. The greater side of every triangle is opposite to the greater angle. LET a b c be a triangle, of which the side ac is greater than the side a b; the angle a b c is also greater than the angle b ca. Because a c is greater than ab, make (i. 3) . d ad equal to a b, and join bd; and because adb is the exterior angle of the triangle b bd c, it is greater (i. 16) than the interior and opposite angle dcb; but a db is equal (i. 5) a to a b d, because the side a b is equal to the side a d; therefore the angle abd is likewise greater than the angle a cb. Much greater then is the angle a b c than acb. Therefore the greater side, &c. Q. E. D. PROPOSITION XIX. - THEOREM. The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it. LET a b c be a triangle, of which the angle a bc is greater than the angle bca; the side a c is likewise greater than the side a b. For, if it be not greater, ac must either be equal to a b, or less than it; it is not equal, because then the angle a b c would be equal (i. 5) to the angle acb; but it is not ; therefore a c is not equal to a b; neither is it less; because then the angle a b c would be less than the angle a c b (i. 18); but it is not; b therefore the side a c is not less than a b; and it has been shewn that it is not equal to a b; therefore a c is greater than a b. Wherefore the greater angle, &c. Q. E. D. PROPOSITION XX.—THEOREM. a Any two sides of a triangle are together greater than the third side. LET a b c 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, b c greater than ac; and b c, ca greater than a b. Produce ba to the point d, and make (i. 3) a d equal to ac; and join dc. Because d a is equal to a c, the angle a d cis likewise equal to a cd (i. 5); but the angle bed is greater than the angle acd; therefore the angle bod 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 op d 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 b c. In the same manner it may be demonstrated, that the sides a b, bc, are greater than ca, and b с b c, ca greater than a b. Therefore any two sides, &c. Q. E. D. PROPOSITION XXI.—THEOREM. a be. e 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 b c of the triangle a bc, to the point d within it; bd and dc are less than the other two sides ba, a c of the triangle, but contain an angle bdc greater than the angle bac. 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 a be are greater than To each of these add ec; therefore the sides ba, a c are greater than be, d 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 b ba, a c than bd, dc. Again, because the exterior angle of a triangle is greater than the interior and opposite angle, the exterior angle bd c of the triangle cde is greater than ced; for the same reason, the exterior angle ceb of the triangle a be is greater than bac; and it has been demonstrated that the angle b dc 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 с are greater than the third, viz. a and b greater than c; a and c greater than b; and b and с 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 g h equal to c; and from the centre f, at the distance fd, describe (3 post.) the circle d kl; and from the centre g, at the distance g h, 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 1k h, gh is equal (15 def.) to g gk; but g h is equal to c; therefore also gk is equal to c; and fg is equal to b; therefore the three straight lines kf, fg, g k, 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 a b be the given straight line, and a the given point in it, and dce the given rectilineal a 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 trif angle afg, the sides of which shall be equal to the three straight d lines cd, de, ec, so that cd be equal to a f, ce to ag, and de to fg; and because de ce AA |