For let C, D, E, &c. be the successive remainders in the process for finding the common measure (last prop.). Then C cannot measure B, otherwise B would measure A (Prop. X.), and there could be no remainder; but C is contained as often in B as B is contained in A (Def. 7.). Let then P be the greatest multiple of B which is contained in A, and let Q be an equimultiple of C, which must be the greatest contained in B; then (Prop. VI. Cor.) A: P :: B: Q, and (Prop. X.) A: B: A-P: B-Q; but A-P=C, and B-Q=D, by hypothesis; therefore A B :: C: D, or B: C::C: D; hence D cannot measure C, for if it could, C would measure B. Let now P' be the greatest multiple of C in B, and Q' the like multiple of D; then, pursuing the same course as before, there results the proportion C: D:: D: E, so that E cannot measure D; and so on for each succeeding remainder. It appears therefore that no remainder can ever measure the preceding one, consequently the process for finding the common measure of A and B will be interminable, and therefore (Prop. XVIII. Cor.) these two magnitudes are incommensurable.

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BOOK VI.

DEFINITIONS.

1. Similar figures are such as have the angles of the one respectively equal to those of the other, and the sides containing the equal angles proportional.

2. The homologous sides of two similar figures are those, which are each interjacent to two angles respectively equal. In different circles similar arcs, sectors, and segments, are those of which the arcs subtend equal angles at the centre.

3. If two sides of another figure form the extremes, and two sides of another figure form the means, of a proportion, these sides are said to be reciprocally proportional.

Note. In addition to the symbols employed in the preceding book, we shall, for the sake of brevity, occasionally express the square of a line, AB, by AB'; and for the rectangle of AB, BC, we shall often write simply AB BC.

PROPOSITION I. THEOREM.

Triangles, whose altitudes are equal, are to each other as their bases.

Let the triangles ABC, abc, have equal altitudes, then ABC: abc:: BČ: bc.

For upon the prolongation of the base BC, take any number of distances CD, DE, EF, &c., each equal to BC, and draw AD, AE, AF, &c. In like

manner

B take any

number of distances,

ed, de, ef, &c. upon the prolongation of bc, each equal to bc, and draw ad, ae, af, &c.

Then the triangles ABC, ACD, ADE, AEF are equivalent,

for they have the same altitude and equal bases (Prop. III. Cor. 2. B. II.); therefore whatever multiple BF is of BC, the same multiple the triangle ABF is of the triangle ABC. For similar reasons whatever multiple bf is of bc, the same multiple the triangle abf is of the triangle abc; and if the base BF be greater than the base bf, the triangle ABF must be greater than the triangle abf, or if the base bf be greater than the base BF, the triangle abf must be greater than the triangle ABF, for by hypothesis these triangles have equal altitudes. Now the base BF and the triangle ABF are any equimultiples of the base BC and the triangle ABC; also the base bf and the triangle abf are any equimultiples of the base bc and the triangle abc; therefore the four magnitudes BC, bc, ABC, abc, are proportional, for it has been shown that it is impossible to find any equimultiples of the antecedents, and any equimultiples of the consequents, such, that the multiple of one antecedent may be greater than that of its consequent, while the multiple of the other antecedent is not greater than its consequent (Prop. IV. B. V.).

Cor. Hence, rhomboids whose altitudes are equal, are to each other as their bases, for rhomboids are the doubles of triangles of the same base and altitude.

Scholium.

The converse of this proposition is obviously true, that is, triangles which are to each other as their bases have equal altitudes, for the base of one triangle is to the base of the other, as the former triangle to one of equal altitude upon the latter base, so that if the altitudes were unequal, the triangles could not be to each other as their bases.

Triangles, whose bases are equal, are to each other as their altitudes.

For every triangle is equivalent to a right angled triangle of equal base and altitude, and in right angled triangles, either of the perpendicular sides being considered as base, the other will be the altitude; therefore, if in two such triangles either the bases or the altitudes be equal, they will, by last proposition, be to each other as the remaining sides; therefore triangles, whose bases are equal, are to each other as their altitudes.

Cor. Therefore, rhomboids, whose bases are equal, are to each other as their altitudes. :

Scholium.

The converse of this proposition, viz. triangles which are to each other as their altitudes have equal bases, is evidently true (see preceding scholium).

PROPOSITION III. THEOREM.

If four straight lines are proportional, the rectangle contained by the extremes is equivalent to the rectangle contained by the means, and conversely, if two rectangles are equivalent, their containing sides are proportional.

Let the four lines AB, CD, DE, BF be proportional, then AF, the rectangle of the extremes, is equivalent to CE, the rectangle of the means.

Make DG equal to BF, and draw GH parallel to DC.

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Then (Prop. I. Cor.) AF CG:: AB: CD; but by hypothesis, AB: CD:: DE: BF: therefore (Prop.

II. B. V.) AF: CG:: DE: BF, or DG, but DE: DG :: CE CG; therefore (Prop. II. B. V.) AF: CG :: CE: CG, and the consequents being equal the antecedents are equal (Prop. IX. Cor. 2. B. V.), that is, AF=CG.

Conversely. Let the rectangle AF be equivalent to the rectangle CE.

Then (Prop. I. Cor.) AB: CD:: AF or CE: CG, but CE: CG:: DE: DG (Prop. I. Cor.); therefore (Prop. II. B. V.) AB: CD :: DE : DG or BF.

Cor. It follows that if three lines are in continued proportion, the rectangle of the extremes is equivalent to the square of the mean, and conversely, if a square be equivalent to a rectangle, the side of the square is a mean between the sides of the rectangle.

The rectangles contained by the corresponding lines which form two proportions are themselves proportional.

Let there be the proportions AB: BC:: CD: DE, and BF: CG: DH: EI, then also AF: BG :: CH: DI.

In CG or its production take Cf=BF, and in EI or its production take Eh DH, and through fand h draw parallels to BC and DE.

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Then (Prop. I. Cor. B. VI.) AF: Bf:: AB: BC:: CD: DE: CH: Dh, and alternately, AF: CH:: Bf: Dh. Now Bf BG:: Cf: CG :: Eh: EI :: Dh : DI; hence, alternately Bf: Dh:: BG: DI; but it has just been shown that Rf: Dh: AF: CH; therefore AF: CH :: BG: DI, or AF: BG:: CH: DI.

Cor. 1. Hence the squares of four proportional lines are proportional.

Cor. 2. If three lines are in continued proportion, the square on the first is to the square on the second, as the first line is to the third. Thus if A, B, C are three lines in continued proportion, then A: B:: B: C, and since A: B :: A: B, we have by the proposition, the square on A to the square on B, as the rectangle of A, B to the rectangle of B, C, and these rectangles are as A to C (Prop. I. Cor.).

Scholium.

The converse of this proposition is not true, for it cannot be inferred that proportional rectangles have proportional sides, since a rectangle may be transformed into an equivalent one, having a side of any given length (Prop. XVII. B. IV.).

The converse of the corollaries are true, that is, first, if four squares be proportional, their sides will be proportional; for let A, B, C, D represent the sides of four proportional squares, then if these sides are not proportional let there be the proportion A; B:: C: Q, then, by the corollary, A2 : Bo :: CQ; but by hypothesis, A2: B2: C: D2; consequently (Prop. IX. Cor. 3. B. V.), Q=D2, and therefore Q = D.

Again, if the squares on two lines are to each other as the first line is to a third, the three lines are in continued proportion, for let A2: B2 :: A : C, and let there be the continued proportion A: B :: B: Q, then, by the corollary, A2 : B2 :: A:Q; but by hypothesis, A: B:: A: C, hence (Prop. IX. Cor. 3. B. V.), Q=C.