ABC has to the triangle DEF the duplicate ratio of that which BC has to EF.

Take BG a third proportional to BC, EF (11. 6.): so that BC is to EF, as EF to BG, and join G, A: then, because, as AB to BC, so DE to EF; alternately (4. Pr.), AB is to DE, as BC to EF: but as BC to EF, so is EF to BG; therefore (5. Cor. Def. Pr.) as AB to DE, so is EF to BG; wherefore the sides of the triangles ABG, DEF

which are about the equal angles are reciprocally proportional: but triangles which have the sides about two equal angles reciprocally proportional equivalent to one another (15.

are

6.) therefore the triangle ABG is equivalent to the triangle DEF and because as BC is to EF, so EF to BG; and that if three straight lines be proportionals, the first is said (17. Def. Pr.) 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 as BC to BG, so is (1. 6.) the triangle ABC to the triangle ABG. Therefore the triangle ABC has to the triangle ABG the duplicate ratio of that which BC has to EF: but the triangle ABG is equivalent to the triangle DEF: wherefore also the triangle ABC has to the triangle DEF the duplicate ratio of that which BC has to EF. Therefore similar triangles, &c. Q. E. D.

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, and similarly described triangle upon the second.

PROP. XX. THEOR.

SIMILAR polygons may be divided into the same number of similar triangles, having the same ratio to one another that the polygons have; and the polygons have to one another the duplicate ratio of that which their homologous sides have.

Let ABCDE, FGHKL be similar polygons, and let AB be the homologous side to FG: the polygons ABCDE, FGHKL may be divided into the same number of similar triangles, of which each to each has the same ratio which the polygons have; and the polygon ABCDE has to the polygon FGHKL the duplicate ratio of that which the side AB has to the side

FG.

M

F

Draw BE, EC, GL, LH: and because the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL (1. Def. 6.) and BA is to AE, as GF to FL (1. Def. 6.): wherefore, be

E.

B

L

G

cause the triangles A BE, FGL have an angle in one equal to an angle in the other, and their sides about these equal angles proportionals, the triangle ABE is equiangular (6. 6.), and therefore similar to the triangle FGL (4. 6.); wherefore the angle ABE is equal to the angle FGL; and, because the polygons are similar, the whole angle ABC is equal (1. Def. 6.) to the whole angle FGH; therefore the remaining angle EBC is equal to the remaining angle LGH: and because the triangles ABE, FGL are similar, EB is to BA, as LG to GF (1. Def. 6.); and also, because the polygons are similar, A B is to BC, as FG to GH (1. Def. 6.); therefore, ex æquali (10. Pr.), EB is to BC, as LG to GH; that is, the sides about the equal angles EBC, LGH are proportionals; therefore the triangle EBC is equiangular to the triangle LGH and similar to it (4. 6.). For the same reason, the triangle ECD likewise is similar to the triangle LHK: therefore the similar polygons ABCDE, FGHKL are divided into the same number of similar triangles.

Also these triangles have, each to each, the same ratio which the polygons have to one another, the antecedents being ABE, EBC, ECD, and the consequents FGL, LGH, LHK: and the polygon ABCDE has to the polygon FGHKL the duplicate ratio of that which the side AB has to the homologous side FG.

Because the triangle ABE is similar to the triangle FGL, ABE has to FGL the duplicate ratio (19. 6.) of that which the side BE has to the side GL: for the same reason, the triangle BEC has to GLH the duplicate ratio of that which BE has to GL: therefore, as the triangle ABE to the triangle FGL, 80 (5. Cor. Def. Pr.) is the triangle BEC to the triangle GLH. Again, because the triangle EBC is similar to the triangle LGH, EBC has to LGH the duplicate ratio of that which the side EC has to the side LH. For the same reason, the triangle ECD has to the triangle LHK the duplicate ratio of that which EC has LH: as therefore the triangle EBC to the triangle LGH, so is (5. Cor. Def. Pr.) the triangle ECD to the triangle LHK: but it has been proved that the triangle EBC is likewise to the triangle LGH, as the triangle ABE to

the triangle FGL: therefore, as the triangle ABE to the triangle FGL, so is triangle EBC to triangle LGH, and triangle ECD to triangle LHK: and therefore, as one of the antecedents to one of the consequents, so are all the antecedents to all the consequents (9. Pr.). Wherefore, as the triangle ABE to the triangle FGL, so is the polygon ABCDE to the polygon FGHKL: but the triangle ABE has to the triangle FGL the duplicate ratio of that which the side AB has to the homologous side FG. Therefore also the polygon ABCDE has to the polygon FGHKL the duplicate ratio of that which AB has to the homologous side FG. Wherefore

similar polygons, &c. Q. E. D.

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

N. B. Squares are similar figures, and therefore they are to one another in the duplicate ratio of their sides: hence, similar plane figures are to one another as the squares of their homologous sides. The square of a line as AB, is, for the sake of brevity, represented by AB2.

COR. 2. And if to AB, FG, two of the homologous sides, a third proportional м be taken, AB has (17. Def. Pr.) to M the duplicate ratio of that which AB has to FG; but the four-sided figure or polygon upon AB has to the four-sided figure or polygon upon FG likewise 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 in triangles (Cor. 19. 6.). Therefore, universally, it is manifest that if three straight lines be proportionals, as the first is to the third, so is any rectilineal figure upon the first, to a similar, and similarly described rectilineal figure upon the second.

PROP. XXI. THEOR.

RECTILINEAL figures of any kind, which are similar to the same rectilineal figure, are also similar to one another.

Let each of the rectilineal figures A, B, which may be of any kind, be similar to the rectilineal figure c: the figure A is similar to the figure B.

Because A is similar to C, they are equiangular, and also have their sides about the equal angles proportionals

(1. Def. 6.). Again, because B is similar to C, they are equiangular, and have their sides about the equal angles proportionals (1. Def. 6.). Therefore the figures

B

A, B are each of them equiangular to C, and have the sides about the equal angles of each of them and of c proportionals. Wherefore the rectilineal figures A and B are equiangular (1. Ax. 1.), and have their sides about the equal angles proportionals (4. 6.). Therefore A is similar to B. Q. E. D.

PROP. XXII. THEOR.

IF four straight lines be proportionals, the similar rectilineal figures similarly described upon them shall also be proportionals: and if the similar rectilineal figures similarly described upon four straight lines be proportionals, those straight lines shall be proportionals.

Let the four straight lines AB, CD, EF, GH be proportionals, viz. AB to CD, as EF to 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: the rectilineal figure K AB is to LCD, as MF to NH.

K

To AB, CD take a third proportional (11. 6.) x, and to EF, GH a third proportional o: and because AB is to

L

EF to O: but as AB to X, so is (2. Cor. 20. 6.) the rectilineal KAB to the rectilineal LCD, and as EF to O, so is (2. Cor. 20. 6.) the rectilineal MF to the rectilineal NH: therefore, as KAB to LCD, so (5. Cor. Def. Pr.) is MF to NH.

And if the rectilineal KAB be to LCD, as MF to NH; the straight line AB is to CD, as EF to GH.

Make (12. 6.) as AB to CD, so EF to PR, and upon PR de

scribe (18. 6.) the rectilineal figure SR similar, and similarly situated to either of the figures MF, NH: then, because as AB to CD, so is EF to PR, and that upon AB, CD are described the similar, and similarly situated rectilineals KAB, LCD, and upon EF, PR in like manner the similar rectilineals. MF, SR; KAB is to LCD, as MF to SR; but by the hypothesis KAB is to LCD, as MF to NH; and therefore the rectilineal MF having the same ratio to each of the two NH, SR, these are equal (4. Cor. Def. Pr.) to one another: they are also similar, and similarly situated; therefore GH is equal to PR:: and because as AB to CD, so is EF to PR, and that PR is equal to GH; AB is to CD, as EF to GH. If therefore four straight lines, &c. Q. E. D.

N. B. Squares being similar figures, if four straight lines be proportionals, the squares described upon them are proportionals: that is, if A, B, C, D be four straight lines, and if a : B :: c : D, then ao : B2 :: C2 : D2.

PROP. XXIII. THEOR.

EQUIANGULAR parallelograms have to one another the ratio which is compounded of the ratio of their sides.

Let AC, CF be equiangular parallelograms, having the angle BCD equal to the angle ECG: the ratio of the parallelogram AC to the parallelogram CF is the same as the ratio which is compounded of the ratios of their sides.

D

H

Let BC, CG be placed in a straight line; therefore DC and CE are also in a straight line (14. 1.), and complete the parallelogram DG; and taking any straight line K, make (12. 6.) as BC to CG, so K to L; and as DC to CE, so make L to м: therefore, the ratios of K to L, and L to M, are the same as the ratios of the sides, viz. of BC to CG, and DC to CE. But the ratio of K to м is that which is said to A be compounded (16. Def. Pr.) of the ratios of K to L, and L to M; wherefore also K has to м the ratio compounded of the ratios of the sides: and because as BC to CG, so is the parallelogram AC to the parallelogram CH (1. 6.); but as BC to CG, so is K to L; therefore K is (5. Cor. Def. Pr.) to L, as the parallelogram AC to the parallelogram CH. Again, because as DC to CE, so is the parallelogram CH to the parallelogram CF; but as DC to CE, so is L to M; wherefore L is (5. Cor. Def.

B

KLM