85. Alternate proportion is, when antecedent is compared with antecedent, and consequent with consequent.-As, if 1: 2 :: 3:6; then, by alternation, oi permutation, it will be 1: 3 :: 2 : 6.

86. Compound ratio is, when the sum of the antecedent and consequent is compared, either with the consequent, or with the antecedent. — Thus, if 1 : 2 :: 3 : 6, then, by composition, 1+2 : 1 :: 3+ 6 : 3, and 1 + 2 : 2 :: 3 + 6:6.

87. Divided ratio is, when the difference of the antecedent and consequent is compared, either with the antecedent or with the consequent. Thus, if 1:2 :: 3:6, then, by division, 2-1:1::6 — 3:3, and 2—1:2 :: 6 — 3 : 6. Note. The term Divided, or Division, here means subtracting, or parting; being used in the sense opposed to compounding, or adding, in def. 86.

THEOREM LXVI.

Equimultiples of any two quantities have the same ratio as the quantity them

selves.

Let A and B be any two quantities, and mA, mB, any equimultiples of them, m being any number whatever: then will mA and mB have the same ratio as A and B, or A: B:: mA: mB.

Corol. Hence, like parts of quantities have the same ratio as the wholes; because the wholes are equimultiples of the like parts, or A and B are like parts of mA and mB.

THEOREM LXVII.

If four quantities, of the same kind, be proportionals; they will be in propor- . tion by alternation or permutation, or the antecedents will have the same ratio as the consequents.

Let A: B:: mA: mB; then will A: mA :: B: mB.

Otherwise. Let A: B:: C: D; then shall B: A:: C: D.

A C

For, let = =r; then A = Br, and C = Dr: therefore B =

B D

C (ax. 1), or B : A :: D: C.

1

Whence it is evident that

In a similar manner may most of the other theorems be demonstrated.

THEOREM LXVIII.

If four quantities be proportional; they will be in proportion by inversion, or inversely.

Let A: B:: mA: mB; then will B: A :: mB : mA.

THEOREM LXIX.

If four quantities be proportional; they will be in proportion by composition

; and
A mB+mA B÷A

Corol. It appears from hence, that the sum of the greatest and least of four proportional quantities, of the same kind, exceeds the sum of the other two. For since A: A+B::mA::mAemB --- A: A+B::mA+mB, where A is the least, and mA+mB the greatest; then m+1.A+mB, the sum of the greatest and least, exceeds m+1.A+B, the sum of the two other quantities.

THEOREM LXX.

If, of four proportional quantities, there be taken any equimultiples whatever of the two antecedents, and any equimultiples whatever of the two consequents ; the quantities resulting will still be proportional.

Let A: B:: mA: mB; also, let pA and pmA be any equimultiples of the two antecedents, and qB and qmB any equimultiples of the two consequents; then will pA : qB :: pmA : qmB.

For

qmB

pmA

If there be four proportional quantities, and the two consequents be either augmented or diminished by quantities that have the same ratio as the respective antecedents; the results and the antecedents will still be proportionals.

Let A: B:: mA : mB, and nA and nmA any two quantities having the same ratio as the two antecedents; then will A: B+nA :: mA : mB+ nmA. mB+nmA B+nA

For

mA

=

both the same ratio.

THEOREM LXXII.

If any number of quantities be proportional, then any one of the antecedents will be to its consequent, as the sum of all the antecedents, is to the sum of all the consequents.

Let A : B :: mA : mB:: nA : nB, &c.; then will A : B :: A+ mA + nA : B +mBnB, &c.

If a whole magnitude be to a whole, as a part taken from the first, is to a part taken from the other; then the remainder will be to the remainder, as the whole to the whole.

If any quantities be proportional; their squares, or cubes, or any or roots, of them, will also be proportional.

Let A: B::mA; mB; then will A": B" :: m"A" : m”B”.

THEOREM LXXV.

If there be two sets of proportionals; then the products or rectangles of the corresponding terms will also be proportional.

If four quantities be proportional; the rectangle or product of the two extremes, will be equal to the rectangle or product of the two means. And the

converse.

Let A: B:: mA: mB;

then is AmB = B × mA mAB, as is evident.

THEOREM LXXVII.

If three quantities be continued proportionals; the rectangle or product of the two extremes, will be equal to the square of the mean. And the converse.

If any number of quantities be continued proportionals; the ratio of the first to the third, will be duplicate or the square of the ratio of the first and second; and the ratio of the first and fourth will be triplicate or the cube of that of the first and second; and so on.

Triangles, and also parallelograms, having equal altitudes, are to each other as their bases.

Let the two triangles ADC, DEF, have the same altitude, or be between the same parallels AE, IF; then is the surface of the triangle ADC, to the surface of the triangle DEF, as the base AD is to the base DE. Or AD: DE: : the triangle ADC: the triangle DEF.

CK

ABDGHE

For, let the base AD be to the base DE, as any one number m (2), to any other number n (3); and divide the respective bases into those parts, AB, BD, DG, GH, HE, all equal to one another; and from the points of division draw the lines BC, GF, HF, to the vertices C and F. Then will these lines divide the triangles ADC, DEF, into the same number of parts as their bases, each equal to the triangle ABC, because those triangular parts have equal bases and altitude (cor. 2, th. 25); namely, the triangle ABC equal to each of the triangles BDC, DFG, GFH, HFE. So that the triangle ADC, is to the triangle DFE, as the number of parts m (2) of the former, to the numbern (3) of the latter, that is, as the base AD to the base DE (def. 79),

In like manner, the parallelogram ADKI is to the parallelogram DEFK, as the base AD is to the base DE; each of these having the same ratio as the number of their parts, m to n. Q. E. D.

THEOREM LXXX. N

Triangles, and also parallelograms, having equal bases, are to each other as their altitudes.

Let ABC, BEF, be two triangles having the equal bases AB, BE, and whose altitudes are the perpendiculars CG, FH; then will the triangle ABC: the triangle BEF:: CG: FH.

For, let BK be perpendicular to AB, and equal to CG; in which let there be taken BL FH; drawing AK and AL.

C K

H E

Then triangles of equal bases and heights being equal (cor. 2, th. 25), the triangle ABK is = ABC, and the triangle ABL = BEF. But, considering now ABK, ABL, as two triangles on the bases BK, BL, and having the same altitude AB, these will be as their bases (th. 79), namely, the triangle ABK : the triangle ABL :: BK : BL.

=

But the triangle ABK ABC, and the triangle ABL= BEF, also BK == CG and BL FH.

Therefore, the triangle ABC: triangle BEF :: CG : FH.

And since parallelograms are the doubles of these triangles, having the same bases and altitudes, they will likewise have to each other the same ratio as their altitudes.

Q. E. D.

Corol. Since, by this theorem, triangles and parallelograms, when their bases are equal, are to each other as their altitudes; and by the foregoing one, when their altitudes are equal, they are to each other as their bases; therefore, universally, when neither are equal, they are to each other in the compound ratio, or as the rectangle or product of their bases and altitudes.

THEOREM LXXXI.

If four lines be proportional; the rectangle of the extremes will be equal to the rectangle of the means. And, conversely, if the rectangle of the extremes,

of four lines, be equal to the rectangle of the means, the four lines, taken alternately, will be proportional.

Let the four lines A, B, C, D, be proportionals, or A: B :: C: D; then will the rectangle of A and D be equal to the rectangle of B and C; or the rectangle A. D= B. C.

QC

B

A

R

D P

A

B

For, let the four lines be placed with their four extremities meeting in a common point, forming at that point four right angles; and draw lines parallel to them to complete the rectangles P, Q, R, where P is the rectangle of A and D, Q the rectangle of B aud C, and R the rectangle of B and D.

Then the rectangles P and R, being between the same parallels, are to each other as their bases A and B (th. 79); and the rectangles Q and R, being between the same parallels, are to each other as their bases C and D. But the ratio of A to B, is the same as the ratio of C to D, by hypothesis; therefore the ratio of P to R, is the same as the ratio of Q to R; and consequently the rectangles P and Q are equal. Q. E. D.

Again, if the rectangle of A and D, be equal to the rectangle of B and C ; these lines will be proportional, or A: B :: C: D.

For, the rectangles being placed the same as before: then, because parallelograms between the same parallels, are to one another as their bases, the rectangle PR:: A: B, and Q: R::C: D. But as P and Q are equal, by supposition, they have the same ratio to R, that is, the ratio of A to B is equal to the ratio of C to D, or A: B:: C: D. Q. E. D.

Corol. 1. When the two means, namely, the second and third terms, are equal, their rectangle becomes a square of the second term, which supplies the place of both the second and third. And hence it follows, that when three lines are proportionals, the rectangle of the two extremes is equal to the square of the mean; and, conversely, if the rectangle of the extremes be equal to the square of the mean, the three lines are proportionals.

Corol. 2. 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, 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