Additions to the First Book. PROP. I. THEOR. İf a right line and two points on the same side of it be given, and two right lines be drawn from thofe points to any point in the given right line; these two right lines taken together will be the leaft of all, when they make equal angles with the given right line. C D Let the given right line be A B, and the given points or the fame fide of it C, D, and let the point E be so taken in the given right line A B, that drawing two right lines CE, DE to it from the given points C, D, the angles A E C, BED which they make with AB, be equal to one another: I fay, the right A lines CE, DE taken together, will be less than any other two right lines CH, DH taken together, drawn from the H I E B given points C, D to any other point н taken in the right line A B. For from one of the given points, as D, [by 12. 1.] draw the right line D I perpendicular to A B, which conti nue out to meet CE continued at G. Join C H, G H. Then because the vertical angles A EC, GEB [by 15: 1.] are equal to one another, and the angle DEB [by fuppofition] is equal to the angle A E C; therefore the angles D E B, GEB are equal to one another. And becaufe [by conftr.] D I is perpendicular to A B, the angle AID is a right angle. And fo will the angle A IG be a right angle. Wherefore the two triangles EDI, GE I have two angles D & I, DIE equal to two angles GEI, GIE, each to cach, and the E 3 fide fide EI common. Therefore [by 26. 1.] the rest of their fides will be equal, and the remaining angles. Wherefore the fide E D will be equal to the fide E G, and the fide ID to the fide IG; and adding CE to both, CE, ED together will be equal to CE, EG together; that is, to C G. Again, because the right angle EID is equal to the right angle EIG, the fide D I equal to the fide IG, and the fide HI common; the triangles D H I, GHI [by 4. 1.] will have their fides D H, H G equal: and adding CH to each of them, DH, CH together, will be equal to CH, H G together. But [by 20. 1.] CG is lefs than CH, HG together; that is, fince CG has been proved to be equal to CE, ED, and CH, HG equal to CH, HD; CE, ED together will be less than CH, HD together. Therefore if a right line and two points on the fame fide of it be given, and two right lines be drawn from those points to any point in the given right line; these two right lines taken together will be the leaft of all, when they make equal angles with the given right line. Which was to be demonftrated. The two diagonals of any right-lined quadrilateral figure, are together lefs than the four right lines that are drawn from any point whatsoever to the four angles of that figure [except it be the interfection of the diagonals]. Let A B C D be a right-lined quadrilateral figure, and the right lines A C, B D its diagonals, intersecting one another in F. Let E, be any affumed point [different from F] and let the four right lines A E, B E, CE, DE be drawn from E to the angles A, B, C, D of the quadrilateral figure A B C D: I fay, the diagonals A C, B D taken together, are less than the four right lines A E, BE, CE, D E taken together. For fince [by 20. 1.] the I two fides of any triangle, howfo howfoever taken, are greater than the third fide, the fide AC of the triangle A CE is lefs than the fides AE, E C. So likewife the fide B D of the triangle B E D is less than the fides BE, ED; therefore the fides A C, B D put toge ther, are less than the fides A E, E C, BE, E D put together; that is, the two diagonals A C, B D together, are lefs than the four right lines A E, E C, B E, ED together. Therefore the two diagonals of any right-lined quadrilateral figure, are together lefs than the four right lines that are drawn from any point whatsoever to the four angles of that figure [except it be the intersection of the diagonals]. Which was to be demonftrated. SCHOLI U M. It may be demonftrated after the fame manner, that any three fides of any quadrilateral figure, are greater than the third fide; and that all the four fides are greater than twice the diagonals. PROP. III. THEOR. All the angles of any right-lined figure, when the angles are all inward, are equal to twice as many right angles, wanting four, as that figure has fides. For any right-lined figure, by drawing diagonals, may be refolved or divided into as ma ny triangles as the figure has fides, wanting two. As if a right-lined figure has four fides, it may be divided into two triangles; if a figure has five fides, it may be divided into three triangles; if it has fix fides, it may be divided into four triangles, and so on. But fince [by 32. 1.] the angles of all these triangles are equal to twice as many right angles as there are triangles, every triangle having two, and the angles of all the triangles are equal to all the inward angles of the right-lined figure; therefore all the inward angles of the figure are equal to twice E 4 as many right angles as there are triangles; that is, equal to twice as many right angles, wanting the angles of two triangles; that is, wanting four right angles, as the figure has fides. Therefore all the angles of any right-lined figure, when the angles are all inward, are equal to twice as many right angles, wanting four, as that figure has fides. Which was to be demonftrated. Otherwife thus: From any point within the figure draw right lines to all the angles of the figure, which will refolve or divide the figure into as many triangles as the figure has fides. Then fince all the angles of every triangle [by 32. 1.] are equal to two right angles, the angles of all the triangles together will make twice as many right angles as the figure has fides: therefore all the angles of the figure, together with all the angles about the affumed point within the figure, make twice as many right angles as the figure has fides. But because [by 13. I.] two right lines mutually cutting one another, make four right angles at the point of fection, and these four angles are equal to any number of angles about that point (for every whole is equal to all its parts taken together) therefore all the angles of a right-lined figure, together with four right angles, are equal to twice as many right angles as the figure has fides. And taking away four right angles from each, there will remain all the angles of the figure equal to twice as many right angles as the figure has fides, wanting four right angles. Therefore all the angles of any right-lined figure, when the angles are all inward, are equal to twice as many right angles, wanting four, as that figure has fides. Which was to be demonftrated. Corol. 1. Hence, if the fides of any right-lined figure be continued out, all the angles without the figure made by each fide, with the continuation of that next to it, will be equal to four right angles. For fince each inward angle, and its correfpondent outward one here fpoken of, is equal to two right angles, all the angles of the figure, together with these outward angles, are equal equal to twice as many right angles as the figure has angles, that is, has fides. But fince all the inward angles together, with four right angles, has been proved to be equal to twice as many right angles as the figure has fides; therefore all the inward angles, together with four right angles, is equal to all the I inward and outward angles here fpoken of; and taking away from both all the inward angles, there will remain all the outward angles here spoken of equal to four right angles. Carol. 2. Hence all right-lined figures of the fame fpecies have the fums of all their inward angles equal; and fo are the fums of all their angles fpoken of in the first corollary. Moreover, if the number of the fides of the feveral fpecies of right-lined figures beginning at the triangle runs on according to the natural numbers 3, 4, 5, 6, 7, &c. the answerable numbers of the right angles equal to the fums of all the inward angles of each fpecies of figures, will be the even numbers, 2, 4, 6, 8, 10, &c. SCHOLIUM. A D B If a right-lined figure has outward angles as well as in. ward ones, the propofition does not then hold true. As in the quadrilateral figure A B C D, which has one ontward angle D, as well as three inward ones A, B, C ; or in the irregular pentagon ABCDE, having two outward angles B, E, and three inward angles A, C, D; or in the irregular heptagon ABCDEFG, having three outward angles G, E, D, and four inward ones A, B, C, F, &c. unless any one should object, that this diftinction I have made between an inward and outward angle is unnecef- A C H B D E Jary, |