Page images
PDF
EPUB

Additions to the First Book.

PROP. I. THEOR. if a right line, and two points on the same side 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 least of all, when they make equal angles with the given right line.

Let the given right line be A B, and the given points ont the same lide of it c, d, and let the point e be so taken in the given right line

D AB, that drawing two right lines CE, DE to it from the given points C C, D, the angles A E C, BED which they make with A B, be equal to one

H I another: I say, the right

E

B A lines CE, DE taken together, will be less than any other two right lines cH, DH taken together, drawn froin the given points C, D to any other point H taken in the right line A B

For from one of the given points, as D, [by 12. 1.] draw the right line di perpendicular to AB, which continue out to meet ce continued at G. Join CH, G H. Theni 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 EC; therefore the angles D E B, GEB are equal to one another. And because [by constr.) DI is perpendicular to A B, the angle A ID is a right angle. And so will the angle A I G be a right angle. Wherefore the two triangles E DI, GE I have two angles D EI, DIE equal to two angles GEI, GIE, each to cach, and the

lide

E 3

fide Er common. Therefore [by 26. 1.] the rest of their fides will be equal, and the remaining angles. Wherefore the side e D will be equal to the side e G, and the side ID to the side I G;' and adding ce to both, CE, E D 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 E IG, the side D I equal to the side i G, and the fide H I common; the triangles DHI, GH I [by 4. 1.] will have their sides D H, H G equal: and adding C H to each of them, DH, C H together, will be equal to CH, H G together. But [by 20. 1.] C G is less than CH, HG together; that is, fince çg has been proved to be equal to cE, ED, and CH, H G equal to CH, HD; CE, ED together will be less than c H, HD together,

Therefore if a right line and two points on the fame side 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 least of all, when they make equal angles with the given right line. Which was to be demonstrated.

in F.

PRO P. II. THEOR. The two diagonals of any right-lined quadrilateral

figure, are together less 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].

Let A B C D be a right-lined quadrilateral figure, and the right lines A C, B D its diagonals, intersecting one another Let E, be any assumed point [different from F] and let

the four right lines A E, BE, с

CE, De be drawn from a to B

the angles A, B, C, D, of the quadrilateral figure A B CD: I say, the diagonals A C, B D taken together, are less than

the four right lines A E, BE, E

CE, D E taken together.

For fince [by 20. 1.] the А A

I two sides of any triangle,

howfo

howsoever taken, are greater than the third fide, the fide AC of the triangle ACE is less than the sides A E, E C. So likewise the lide B D of the triangle B E D is less than the sides B E, ED; therefore the sides A C, B D put toge. ther, are less than the sides A E, E C, B E, E D put together; that is, the two diagonals AC, BD together, are less than the four right lines A E, E C, B E, E D to ether.

Therefore the two diagonals of any right-lined quadrilateral figure, are together less than the tour right lines that are drawn from any point whatsoever to the iour angles of that figure [except it be the intersection of the diagonals]. Which was to be demonstrated.

SCHOLIU M. It may be demonstrated after the same manner, that any thret sides of any quadrilateral figure, are greater than the third fade ; and that all the four sides 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 baš fides.

For any right-lined figure, by drawing diagonals, may be resolved or divided into as many 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 ma, be divided into three triangles; if it has six sides, it may be divided into four triangles, and so on. But since [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 ail 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 sides.

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 sides. Which was to be demonstrated.

Otherwise thus : From any point within the figure draw right lines to all the angles of the figure, which will resolve or divide the figure into as many triangles as the figure has sides. Then since 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 assumed point within the figure, make twice as many right angles as the figure has fides. But because [by 13. 1.) two right lines mutually cutting one another, make four right angles at the point of section, 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 sides. 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 sides, 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 demonstrated.

Corol. 1. Hence, if the sides of any right-lined figure be

continued out, all the angles without the figure made by each side, with the continuation of that next to it, will be equal to four right angles. For fince each inward angle, and its correspondent outward one here spoken of, is equal to two right angles, all the angles of the figure, together with these outward angles, are

equal

H

equal to twice as many right angles as the figure has angles, that is, has fides. But since 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 sides; therefore all the inward angles, together with four right angles, is equal to all the inward and outward angles here spoken of; and taking away from both all the inward angles, there will remain all the outward angles here fpoken of equal to four right

angles. Corol. 2. Hence all right-lined figures of the fame species

have the sums of all their inward angles equal; and so are the sums of all their angles fpoken of in the first corollary. Moreover, if the number of the sides of the several species 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 sums of all the inward angles of each species of figures, will be the even numbers, 2, 4, 6, 8, 10, &c.

SCHOLIUM. If a right-lined figure has outward angles as well as in. ward ones, the proposition does

B not then hold true. As in the quadrilateral figure A B C D, which has one ontward angle D, as well as three inward

D ones A, B, C; or in the irregular pentagon ABCDE, hav

А ing two outward angles B, E, and three inward angles

H Η. A, C, D; or in the irregular heptagon ABCDEFG, having three outward angles G, E, D, and four inward ones A, B,

D C, F, &c. unless any one should B objed, that this distinction I

E have made between an inward and outward angle is unnecef A

fary

[ocr errors]
« PreviousContinue »