The First Six, and the Eleventh and Twelfth Books of Euclid's Elements

PROP. XIX. THEOR.- Similar triangles are to one another in the duplicate ratio of their homologous sides.

Let ABC, DEF be similar triangles, having the angles B and E equal, and AB: BC:: DE: EF, so that (V. def. 13) the side BC is homologous to EF. The triangle ABC has to the triangle DEF, the duplicate ratio of that which BC has to EF.

Take (VI. 11) BG a third proportional to BC, EF, so that BC: EF:: EF: BG, and join GA.*

Then, because

AB: BC: DE EF; alternately, AB DE BC: EF; but (const.) as BC EF:: EF: BG;

B GHC

therefore (V. 11) as AB: DE :: EF: BG. The sides, therefore, of the triangles ABG, DEF, which are about the equal angles, are reciprocally proportional; and therefore (VI. 15) the triangles ABG, DEF are equal. Again, because BC: EF:: EF: BG; and that if three straight lines be proportionals, the first is said (V. def. 11) to have to the third the duplicate ratio of that which it has to the second; BC therefore has to BG the duplicate ratio of that which BC has to EF. But (VI. 1) as BC to BG, so is the triangle ABC to ABG. Therefore (V. 11) the triangle ABC has to ABG the duplicate ratio of that which BC has to EF. But the triangle ABG is equal to DEF; wherefore also the triangle ABC has to DEF the duplicate ratio of that which BC has to EF. Therefore, &c.

Cor. From this it is manifest, that if three straight lines be proportionals, as the first is to the third, so is any triangle upon the first to a similar triangle, similarly described on the second.

PROP. XX. THEOR.-Similar polygons may be divided into the same number of similar triangles: (2) these triangles have the same ratio to one another that the polygons have: and (3) the polygons have to one another the duplicate ratio of that which their homologous sides have.

Let AD, FK be similar polygons, and let the sides AB, FG be homologous: the polygons may be divided into the same number of similar triangles: (2) these triangles have, each to each, the

must have recourse to works that treat expressly on such subjects, particularly treatises on practical geometry, surveying, and the use of mathematical in

struments.

* The third proportional might be taken to EF and BC, and placed from E along EF produced; and then a triangle equal to ABC would be formed by joining D with the extremity of the produced line.

+ This proposition might also be proved by making BH equal to EF, joining AH, and drawing through Ha parallel to AC. The triangle cut off by the parallel is equal (I. 26) to DEF. But (VI. 1) the triangle ABC is to the triangle ABH, as BC to BH; and, for the same reason, the triangle ABH is to the triangle cut off by the parallel, or to DEF, as BC to BH, or (const.) as BH to BG. Therefore, ez. quo, the triangle ABC has to the triangle DEF, the same ratio that BC has to BG, or (V. def. 11) the duplicate ratio of that which BC has to EF.

same ratio which the polygons have: and (3) the polygon AD has to the polygon FK the duplicate ratio of that which the side AB has to FG.

1. Let the angles AED, FLK be equal, and the sides about them proportional; and draw the diagonals EB, EC, LG, LH. Then (VI. def. 1) because the polygons are similar, the angles A and F are equal, and BA : AE :: GF: FL; wherefore, because the triangles ABE, FGL have an angle in one equal to an angle in the other, and the sides about those angles proportionals, the triangle ABE is (VI. 6) equiangular, and therefore (VI. 4 and def. I) similar to the triangle FGL; wherefore the angles ABE, FGL are equal. Again, because the polygons are similar, the whole angle ABC is equal (VI. def. 1) to the whole FGH; therefore (I. ax. 3) the remaining angles EBC, LGH are equal: * and (VI. 4)

because the triangles ABE, FGL are equiangular, EB: BA :: LG: GF; and, also (VI. def. 1) because the polygons are similar, AB: BC: FG: GH; therefore, ex æquo, (V. 22) EB: BC:: LG: GH; that is, the sides about the equal angles EBC, LGH are proportionals; therefore (VI. 6) the triangles EBC, LGH are equiangular, and (VI. 4, and def. 1) similar. In the same manner, the triangles ECD, LHK may be shown to be similar. Therefore the similar polygons AD, FK are divided into the same number of similar triangles.

2. Because the triangles ABE, FGL are similar, ABE has to FGL the duplicate ratio (VI. 19) of that which the side BE has to GL. For the same reason, the triangle BEC has to GLH the duplicate ratio of that which BE has to GL. Therefore (V. 11) as the triangle ABE to FGL, so is the triangle BEC to GLH. Again, because the triangles EBC, LGH are similar, EBC has to LGH the duplicate ratio of that which EC has to LH. For the same reason, the triangle ECD has to LHK the duplicate ratio of that which EC has to LH. As, therefore, (V. 11) the triangle EBC is to LGH, so is the triangle ECD to LHK. But it has been proved that the triangle EBC is to LGH, as the triangle ABE to FGL. Therefore (V. 11) as the triangle ABE is to FGL, so is the triangle EBC to LGH, and ECD to LHK: and therefore (V. 12) as one of the antecedents is to its consequent, so are all the antecedents to all the consequents. Wherefore as the triangle ABE is to FGL, so is the polygon AD to the polygon FK,

3. The triangle ABE has (VI. 19) to FGL, the duplicate ratio of that which the side AB has to the homologous side FG. Therefore, also, (V. 11) the polygon AD has to the polygon FK the duplicate ratio of that which AB has to the homologous side FG. Similar polygons, therefore, &c.

* Should the angles A and F be re-entrant, so that the points A and F were on the other sides of EB and LG, the angles EBC, LGH would not be the difference, but the sum, of ABC, ABE, and of FGH, FGL.

Cor. 1. In like manner, it may be proved, that similar figures of four sides, or of any number of sides, are one to another in the duplicate ratio of their homologous sides, and the same has already been proved (VI. 19) in respect to triangles. Therefore, universally, similar rectilineal figures are to one another in the duplicate ratio of their homologous sides.

Cor. 2. If to AB, FG, two of the homologous sides a third proportional M be taken, AB has (V. def. 11) to M the duplicate ratio of that which AB has to FG. But the four-sided figure or polygon upon AB has likewise to the four-sided figure or polygon upon FG the duplicate ratio of that which AB has to FG. Therefore as AB is to M, so is the figure upon AB to the figure upon FG; which was also proved (VI. 19, cor.) in triangles. Therefore, universally, if three straight lines be proportionals, the first is to the third, as any rectilineal figure upon the first, to a similar and similarly described figure upon the second.

Cor. 3. Because all squares are similar figures, the ratio of any two squares to one another, is the same as the duplicate ratio of their sides; and hence, also, any two similar rectilineal figures are to one another (V. 11) as the squares described on their homologous sides.*

Cor. 4. The perimeters of the figures, that is, the sums of their sides, are proportional to the homologous sides. For (hyp. and VI. def. 1) the sides are proportional, and therefore, alternately, AB: FG:: BC: GH :: CD : HK ́:: DE : KL :: EA : LF wherefore (V. 12) AB : FG :: AB+BC+CD+DE+EA : FG+ GH+HK+KL+LF.

PROP. XXI. THEOR.-Rectilineal figures which are similar to the same figure, are similar to one another.

Let each of the rectilineal figures A, B be similar to C: A is similar to B.

Because A is similar to C, they are (VI. def. 1) equiangular, and have also their sides about the equal angles proportional. For the same reason, B and C are equiangular, and have their sides about the equal angles proportional. Therefore the figures A, B are each of them equiangular to C, and have the sides about the equal angles of

each of them, and of C proportional; wherefore (I. ax. 1) A and

This corollary and the preceding, afford easy means of comparing the magnitudes of similar rectilineal figures. Thus, the second enables us to find two lines having the same ratio as the figures, and the third affords, in general, the easiest means of exhibiting their ratio in numbers, or by algebraic symbols; the area of a square (I. 46, cor. 5) being found simply by multiplying a side by itself. Thus, if AB were 3 inches, and FG 2 inches, the areas of the squares of these lines would be 9 square inches and 4 square inches; and therefore the figure AD would be to FK as 9 to 4, or would be rather more than twice as great. If, again, AB were 5 feet, and FG 3 feet, AD would be to FK as 25 to 9, or would be nearly three times as great.

B are equiangular; and (V. 11) they have their sides about the equal angles proportionals. Therefore (VI. def. 1) A is similar to B. Rectilineal figures, therefore, &c.

PROP. XXII. THEOR.-If four straight lines be proportionals, the similar rectilineal figures, similarly described upon them, are also proportionals: and (2) if the similar rectilineal figures, similarly described upon four straight lines, be proportionals, those lines are proportionals.

Let AB CD EF GH; and upon AB, CD let the similar rectilineal figures KAB, LCD be similarly described; and upon EF, GH the similar rectilineal figures MF, NH in like manner: KAB is to LCD, as MF to NH.

To AB, CD take (VI. 11) a third proportional X; and to EF, GH a third proportional O. Then, because

AB CD EF : GH, and (V. 11) CD: X :: GH: 0; therefore, ex æquo, as AB: X EF: 0. But (VI. 20, cor. 2) as AB to X, so is the figure KAB to LCD; and as EF to O, so is MF to NH: therefore (V. 11) as KAB to LCD, so is MF to NH.

Again, if the rectilineal figure KAB be to LCD, as MF to NH; AB CD EF : GH.

To AB, CD, and EF, take (VI. 12) a fourth proportional PR, and on it describe (VI. 18) the figure SR similar and similarly situated to either

upon EF, PR, in like manner, the similar rectilineal figures MF, SR; KAB LCD:: MF: SR; but (hyp.) KAB LCD:: MF: NH; and therefore MF having (V. 11) the same ratio to each of the two NH, SR, these are equal (V. 9) to one another. They are also similar and similarly situated; therefore GH is equal to PR: and because as AB: CD :: EF: PR, and that PR is equal to GH; therefore AB: CD :: EF: GH; wherefore, &c.

PROP. XXIII. THEOR.-Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.

Let AC, CF be equiangular parallelograms, having the angles

BCD, ECG equal: AC has to CF the ratio which is compounded of the ratios of their sides.

D H

Let BC, CG be placed in a straight line; therefore DC, CE are also (I. 14, cor.) in a straight line. Complete the parallelogram DG. To DC, CE, CG take (VI. 12) a fourth proportional K. Then (V. def. 10) the ratio of BC to K is said to be compounded of the ratios of BC to CG and of CG to K, or (const.) of the ratios of BC to CG and DC to CE: wherefore BC has to K the ratio compounded of the ratios of the sides. Now (VI. 1) as BC to CG, so is AC to CH; and as DC to CE, so is CH to CF, or (const.) as CG to K, so is CH to CF. Therefore, since there are two ranks of magnitudes, BC, CG, K, and AC, CH, CF,

E F

which, taken two and two in order, are proportionals, ex quo BC is to K, as AC to CF. But it has been shown that BC has to K the ratio compounded of the ratios of the sides; and therefore AC has to CF the ratio compounded of the ratios of the sides; that is, of the ratios of BC to CG, and DC to CE. Therefore, equiangular parallelograms, &c.*

Cor. Triangles which have one angle of the one equal, or supplemental, to one angle of the other, have to one another the ratio which is compounded of the ratios of the sides containing those angles. For, if BD and EG be joined, the triangles BCD, ECG being (I. 34) halves of the parallelograms AC, CF, have (V. 15) the same ratio as the parallelograms; and if BD and CF be joined, the same is true regarding the triangles BCD, CEF, which have the angles BCD, CEF supplemental.

* Otherwise :-Because there are three parallelograms AC, CH, CF, the first AC (V. def. 10) has to the third CF, the ratio which is compounded of the ratio of the first AC to the second CH, and of the ratio of CH to the third CF; but AC is to CH, as the straight line BC to CG; and CH is to CF, as the straight line CD is to CE; therefore AC has to CF the ratio which is compounded of ratios that are the same with the ratios of the sides.

The demonstration in the text proceeds on the same principle as that given by Dr. Simson from the Greek of Euclid, but it is considerably simplified by using the lines BC, CG, instead of two lines proportional to them. The foregoing proof, in this Note, is given by Simson in his Notes, from Candalla. As a proof, the latter is quite satisfactory; but the other has the advantage of exhibiting in a clear manner the ratio which is compounded of the ratios of the sides, as it is simply that of BC to K.

Dr. Simson remarks in his Note on this proposition, that "nothing is usually reckoned more difficult in the elements of geometry by learners, than the doctrine of compound ratio." This distinguished geometer, however, has both freed the text of Euclid from the errors introduced by Theon or others, and has explained the subject in such a manner as to remove the difficulties that were formerly felt. According to him, "every proposition in which compound ratio is made use of, may without it be both enunciated and demonstrated;" and "the use of compound ratio consists wholly in this, that by means of it, circumlocutions may be avoided, and thereby propositions may be more briefly either enunciated or demonstrated, or both may be done. For instance, if this 23d proposition of the sixth book were to be enunciated, without mentioning compound ratio, it might be done as follows:If two parallelograms be equiangular, and if as a side of the first to a side of the second, so any assumed straight line be made to a second straight line; and as the

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The First Six, and the Eleventh and Twelfth Books of Euclid's Elements: With ...