tended by the angle FCI, at the centre, and equal to the other vertical angle HCG, and to the two vertical angles DKB, and AKE; therefore, the sum of the two arcs DB, and AFIE, is equal to the sum of the two arcs HDBG, and FI, the two first being the vertical arcs subtended by the chords intersecting one another in K, and the two latter the sum of the two vertical arcs subtended by the diameters; as was to be demonstrated.

Corol. The distance between the point of intersection of the chords and the centre of the circle, is called the eccentricity of this point, and half the sum of the vertical arcs of any such point, is equal to the arc subtended by the same angle from the centre.

Scholium. If a series of equal angles be supposed around one point, the sum of which is equal to one whole circumference of a circle, (or four right angles,) and these be placed in any point of the circle without the centre, revolving around that point as centre, the sum of all the differences of the arcs which they will in. tersect by any two positions, will always be equal to the sum of the differences of the arcs which these would have described by the same revolution made from the centre of the circle; because their sum being in both cases a whole circumference, the sum of the differences, and the angle represented by it at the centre, will be equal.

CHAPTER V.

Of the proportionality of Lines and Figures.

§ 60. PROBLEM. To divide a given terminated straight line into a given number of equal parts.

APPLIC. The determinate straight line AB, (fig. 66,) is required to be divided into a certain number of equal parts, (suppose in 5.)

Solution. From A, one of the end points of the given straight line draw an indefinite straight line AC, making any acute angle with it: upon this AC, lay off in succession lines of equal length, in number equal to the number of division required, as AD=DE=EF—FG=GH; join HB, and through every one of the points, D, E, F, G, draw lines parallel to HB, as Da, Eb, Fc, Gd, cutting the given straight line AB, in the points; a, b, c, d, and this line will be divided by them into the number of equal parts required, as Aa=ab =bc=cd=dB; which was to be done.

Proof. Through the points D, E, F, G, draw the Dn, Eo, Fp, and Gq, parallel to AB, which will intersect the dividing lines g, h, and i; the parallels, aD, bE, cF, dG, and BH, are cut by the straight line AH, they make the inner and outer angles on the same side of it equal, that is, the angles ADa, DEg, EFh, FGi, and GHq, are equal; in like manner, because the parallels AB, Dn, Eo, Fp, and Gq, are cut by the AH, the angles DAa, EDg, FEh, GFi, HGq, are all equal; the lines AD, DE, &c. being all equal by construction, and included between equal angles (as just shewn,) all the triangles AaD, DgE, EhF, and so on, are equal in all respects, therefore also the sides opposite to equal angles are equal, that is Aa=Dg =Eh=Fi=Gq. On account of the parallelism of the various lines drawn by the construction, the figures Db, Ec, Fd, and GB, are parallelograms, having the opposite sides equal, and

as all these sides are equal to Aa, there will also be Aa=ab= bc-cd-dB, that is the line AB, has been divided into the number of parts required, all equal to one another; which was to be proved.

Corol. All the individual triangles formed by the foregoing construction, by any of the parallel lines with its opposite angular point, are equiangular among themselves and with the whole triangle formed by ABH, formed by the extreme lines, (that is they are all similar triangles.)

Scholium. It might already from the result of this Problem be deduced that triangles are to one another in the ratio of the squares of their sides opposite to equal angles, or what is called homologous sides. For every parallelogram that resulted being discomponible in as many triangles equal to the AaD, there will be found in each of the triangles made by any of the parallel lines as many such equal triangles as the square of the side indicates, taking the side of the triangle AaD, as unity; as for instance AcF, having three parts Aa+ab+bc, will contain nine triangles equal to AaD; and so for any other of the triangles presented by the figure.

§ 61. PROBLEM. To find the greatest common measure between two given straight lines.

APPLIC. The two straight lines AB, and CD, being given: (fig. 67,) it is required to find their greatest common measure, if they have any.

Solution. From the greater line AB, take away the smaller CD, as often as it may be taken, in the figure: Aa=ab=bc= CD; leaving the remainder cB, smaller than CD; from the smaller line CD, take away the remainder cB, as often as it may be taken, as Cd=dg=cB, and leaving gD, as a remainder, smaller than cB; then take away from the former remainder cB, this second remainder gD, as often as it may be taken. Continue this process until the remainder last obtained measures the preceding one exactly; that is, contains it a full number of times. Suppose in this case the gD, to have been contained exactly twice in cB; that is, that it would have given

1

gD=cf=ƒB, then the line gD, will be the greatest common measure between AB, and CD; which was required.

Proof. The greatest line that can be a common measure between two given lines, can be no greater than the smaller line itself, as here CD, this being taken away from the greater line AB, as often as it is contained in it whole times, and leaving cB, as remainder, smaller than CD; any line that shall measure the two given lines without remainder must measure the cB, and the CD, which together measure the AB; between these two lines cB, and CD, there can again be no greater common measure than cB, itself; this being taken away as often as admissible, and leaving the remainder, gD, the same reasoning holds good between the gD, and cB, as between the former remainder and the smaller line CD, and before between CD, and AB; hence gD, being found to measure the cB, without remainder, this gD, will also measure the Cg, a multiple of cB; thence it will also measure the cD, therefore also the Ac, a multiple of CD, and as it measures also the remainder cB, it measures the whole AB; therefore, the gD, is a common measure between the two given lines AB, and CD; and it is the greatest possible, because of all the lines successively occuring there can never be a greater common measure than the smaller, which were always the successive remainders; if any common measure is possible it must be one of the remainders successively obtained by this process, and if none is found by this method there is no common measure between them.

62. THEOREM. Triangles that are between the same parallels, (or that are of the same altitude,) are to one another in the same ratio as their bases.

APPLIC. If the triangles ABC, and DEF, (fig. 68,) have the same altitude, that is: that they are, or can be, between the same parallels; these triangles will have to one another the same ratio as their bases, whether these bases are commensurable or not; that is, there will be:

ABC: DEF=AB : DE.

Demonstr. 1. When the two bases have a common measure. Suppose this to be the Aa, and therefore the AB, and the DE, be both divided into equal parts Aa-ab-bc=cB= Dd-de-eE; join aC, bC, cC, in the triangle ABC; and dF, eF, in the triangle DEF, the resulting triangles will all be equal to one another, as AaC=ab C=bcC=cBC=DdF =deF=eEF; and the whole triangles ABC, and DEF, will be to one another as the number of these equal triangles which they contain; the number of these triangles in each being equal to the number of these equal bases, these triangles are to one another as their bases, or the proportion results;

ABC

DEF-AB: DE;

which was to be demonstrated.

Sup

2. When the two bases have no common measure. pose AB, and DE, (in figure 686,) have no common measure, the triangles ABC, and DEF, are to one another as their bases AB, and DE; suppose they were not to one another in that ratio, but that they were to another as AB, to Df, smaller (or greater) than DE; find a subdivision of AB, so that the parts be smaller than fE, and suppose them laid off in DE, until one of the subdivisions fall between f, and E, in the point g; which must always take place, because the subdivisions have been taken smaller than fE; join gF, and fF; then by preceding part of the demonstration there will be ABC: Dg F-AB: Dg; but, by the suppoABC: DEFAB : Dƒ;

sition

the triangle Dg F, must be smaller than DEF; therefore, also, the Dg, must be smaller than the Df, which is contradictory, as the point g, must fall between ƒ, and E, and in the manner inversely, if the Df, was supposed greater than DE; the Df, would be shewn smaller than Dg, while by the supposition it would be larger; therefore, as the two triangles cannot be to one another either as bases smaller or greater than AB, to DE, they must be to one another as these bases, when their bases have no common measure.

Corol. Parallelograms upon equal bases, and of the same altitude with triangles, being the double of these triangles, the paral