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Cor. 3. If all these ratios are equal, and the magnitudes expressed numerically,
Of four proportional magnitudes, if the first be greater than the second, the third is greater than the fourth; if equal, equal; and if less, less.
LET a b c d; then if a be greater than b, c is greater than d; if equal, equal, and if less, less.
Then if a be greater than b, the fraction is less than unity, and
hence is also less than unity, or c is greater than d. If a be equal to b, the
Cor. Since (th. 69) we have a : c:: bd, the same reasoning will lead to the conclusion, that if four magnitudes of the same kind be proportionals, then the second will be less, equal to, or greater than the fourth, according as the first is less, equal to, or greater than the third.
If any equimultiples whatever of the first and third of four magnitudes be taken, and any whatever of the second and fourth; then, according as the multiple of the first be greater than, equal to, or less than that of the second, that of the third will be greater than, equal to, or less than that of the fourth.
LET a b c d; then it is to be shown that
(1) if ma be greater than nb, mc is also greater than nd;
(2) if ma be equal to nb, mc is also equal to nd;
(3) if ma be less than nb, mc is also less than nd.
For, since ab:: cd, ma: nc :: mb: nd (th. 70); and since these last are proportionals, according as ma is greater than, equal to, or less than nc, mb is greater than, equal to, or less than nd (th. 75).
If there be four magnitudes such that when any equimultiples whatever are taken of the first and third, and any whatever of the second and fourth, and if when the multiple of the first is greater than, equal to, and less than that of the second, that of the third is greater than, equal to, and less, respectively, than that of the fourth these four magnitudes will be proportional.
LET a, b, c, d, be four magnitudes, and m, n, any numbers whatever; and when ma is greater than, equal to, and less than nb, let me be greater than, equal to, and less than nd: then we have a b:: c: d.
For, if possible, let the fourth magnitude not be a fourth proportional to a, b, c, and let the fourth proportional to them be d. Then if d, = d + d', we have
a b c d + d'. Hence,
Now in this case we have ma: nb :· mc: nd + nd'.
Now the second of both of these sets of conditions can only be fulfilled by d'= 0; and of the other two of each set, the first is not necessarily fulfilled by any other value of d'. Whence that the three conditions may be fulfilled, we must have d a fourth proportional to a, b, c, d; that is, under the given circumstances, ab::c: d.
If any number of quantities be continued proportionals, then the ratio of the first to the last is that power of the ratio which expresses the number of ratios compounded.
LET there be n equal ratios a: b, b: c, c : d,
p: q compounded: then
Cor. Hence the duplicate ratio is the square of the simple ratio, the triplicate ratio is its cube, and so on.
THEOREMS DEPENDING ON RATIOS.
Parallelograms, or triangles, having equal altitudes, are to one another as their bases; those having equal bases are to one another as their altitudes; and those having neither equal bases nor altitudes, are to one in a ratio compounded of the ratios of their bases and the ratio of their altitudes.
TM BRC NP QW GF
FIRST. (1). Let BADC, GDEF be any two parallelograms having the same altitude (and, therefore, when their bases AD, DE, are in the same straight line AE, their opposite sides BC, GF, are in the same straight line BF parallel to AE): then they are to each other as their bases AD, In AD produced, take any number of parts AL, LS, each equal to AD; and any number EH, HK, KV, each equal to DE; and draw LM, ST, EN, HP, KQ, VW, all parallel to AB or DC, and meeting BF, produced, as in the figure.
Then each of the parallelograms TL, MA, is equal to BD; and there are as
many of them as there were taken lines AL, LS, equal to AD. Hence, whatever multiple the base SD is of the base DA, the same multiple is the parallelogram SC of the parallelogram AC. In like manner, whatever multiple the base DV is of the base DE, the same multiple is the parallelogram DW of the parallelogram DN, or (th. 25) of the parallelogram DF.
Again, if SD, the multiple of AD, be greater than DV, the multiple of DE, the multiple SC of AC will be greater than the multiple DW of DN or DF; if equal, equal; if less, less. Hence, (th. 77,) parallelogram AC: parallelogram DN: base AD base DE.
(2.) Let ARD, DGE, be two triangles of equal altitudes, they will be to each other as their bases, AD, DE.
For, being of equal altitudes, they are between the same parallels; and as each of the triangles ARD, DGE, is the half of the parallelograms AC, DF, having the same base and altitude, they are to one another as those parallelograms. But the parallelograms have been proved to have the same ratio as their bases; and hence triangle ARD: triangle DGE :: base AD: base DE.
Secondly. (1). Let the parallelograms AC, DF, have equal bases AD, DG: they will be to each other as their altitudes.
For, draw AK, DL, GN, perpendicular to AG, and produce BC, EF, to meet them, as in the figure. Then the lines BC, EF, being parallel to AG, are parallel to each other (th. 15); and the parallelograms AL, DN, are rectangles (th. 22, cor.1), and equal to AC, DF, respectively (th. 25); and the rectangle AM is equal to the rectangle DN. But the rectangles AM, AL, are to one another as AH, AK, by the former part of the proposition; that is, as the altitudes of the parallelograms AC, DF. Whence also we have parallelogram AC: parallelogram DF:: altitude AH: altitude AK.
(2). The triangles ABD, DFG, being the halves of the parallelograms AC, DF, are to one another in the same ratio; that is,
L B C
triangle ABD triangle DFG :: base AD: base DG. Thirdly. (1). Let ABCD, DEFG, be any two parallelograms, having neither their bases nor altitudes equal, and make the same construction as in the last case: they will be to each other in a ratio compounded of the ratio of their bases AD, DG, and the ratio of their altitudes AK, AH. For, by the first and second cases respectively, we have parellelogram AM parallelogram DN DG and parallelogram AL
: hence also
parallelogram AM АН
AK parallelogram AM parallelogram AL parallelogram AC DG АН parallelogram DN parallelogram AM parallelogram DF’ the parallelograms AC, DF, have the ratio which is compounded of the ratio of their bases AD, DG, and the ratio of their altitudes AK, AH.
(2). Since the triangles ABD, DFG, have the same bases and altitudes as the parallelograms AC, DF, they have the same ratios as the parallelograms themselves, and hence the proposition is also true respecting triangles.
Parallelograms, or triangles, which have one angle of the one equal to one angle of the other, have to one another a ratio compounded of the ratios of the sides about the equal angles.
(1). LET ABCD, FCEG, be two parallelograms, having the angles DCB, ECF equal to one another: then they shall be to one another in a ratio compounded of the ratios of DC, CE, and BC, CF.
parallelogram CH BC ; hence parallelogram CG CF parallelogram AC DC CB parallelogram CG CE CF
(2). Let BCD, FCE, be triangles having the angles at C equal; they shall be to one another in a ratio compounded of the ratio of the sides DC, CE, and BC, CF.
For the triangles BCD, FCE, being the halves of the parallelograms AC, CG, they have the same ratio as the parallelograms: that is, by the last case, the ratio compounded of the ratios of the sides.
In parallelograms, or triangles, having one angle of the one equal to one angle of the other, if the sides about the equal angles are reciprocally proportional, the parallelograms or triangles are equal; and if they be equal, the sides about the equal angles are reciprocally proportional.
FIRST, let the parallelograms AC, CG, have their angles at C equal, and the sides about C reciprocally proportional, (that is, DC: CE:: CF: CB) then they will be equal. For by the last proposition we have parallelogram AC DC CB parallelogram CG CE CF DC CB we have CE CF parallelogram CG.
; and since also
1, and therefore parallelogram AC
In like manner, under the same circumstances, the triangles BCD, FCE, being the halves of the equal parallelograms, are also equal.
Secondly. Let the parallelograms AC, CG, be equal, and have the angles at C equal: then the sides shall be reciprocally proportional, or DC: CE :: CF : CB.
; or again, finally, DC: CE :: CF: CB.
In the same way, it may be proved for the triangles BCD, FCE.
Cor. 1. If four straight lines be proportional, the rectangle of their extremes is equal to the rectangle of the means: and if the rectangle of the extremes be equal to the rectangle of the means, the four straight lines are proportional.
section, and draw lines parallel to them to complete the figure; where P is the rectangle of A, and D, Q, that of B and C, and R that of D and B. Then the figures P, Q, R, are rectangles, and the theorem is a case of the proposition, in which the alleged properties have been generally proved.
Cor. 2. If three straight lines be proportional, the rectangle of the extremes is equal to the square of the mean; and if the rectangle of the extremes be equal to the square of the mean, the three straight lines are proportional.
For in this case B = C, and the rectangle of B and C becomes the square of B or C. Whence by the last cor. the truth follows.
Since it appears, by the rules of proportion in arithmetic and algebra, that when four quantities are proportional, the product of the extremes is equal to the product of the two means; and by this theorem, that the rectangle of the extremes is equal to the rectangle of the two means; it follows, that the area or space of a rectangle is represented or expressed by the product of its length and breadth multiplied together: and, in general, the area of a rectangle in geometry is represented by the product of the measures of its length and breadth, or base and height; and a square is similarly represented by the measure of the side multiplied by itself. Hence, what is shown of such products, is to be also understood of the squares and rectangles.
If a line be drawn in a triangle parallel to one of its sides, it will cut the other two sides proportionally.
LET DE be parallel to the side BC of the triangle ABC; then
will AD: DB:: AE: EC.
For, draw BE and CD. Then the triangles DBE, DCE, are equal to each other, because they have the same base DE, and are between the same parallels DE, BC (th. 25). But the two triangles ADE, BDE, on the bases AD, DB, have the same altitude; and the two triangles ADE, CDE, on the bases AE, EC, have also the same altitude; and because triangles of the same altitude are to each other as their bases, therefore
A ADE: BDE: AD: DB, ▲ ADE: CDE:: AE: EC.
But BDE = CDE; and equals must have to equals the same ratio; therefore AD: DB:: AE EC. In a similar manner, the theorem is proved when the sides of the triangle are cut in prolongation beyond either the vertex or the base.
Cor. Hence, also, the whole lines AB, AC, are proportional to their corresponding proportional segments (cor. th. 66),