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2. The sides of the triangles ABG, DEF, which are about the

equal angles, are reciprocally proportional :

AA

B

but triangles, which have the sides about two equal angles reciprocally proportional, are equal to one another (VI. 15); therefore

3. The triangle ABG is equal 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 to have to the third, the duplicate ratio of that which it has to the second (V. Def. 10); therefore

4. BC has to BG the duplicate ratio of that which BC has to

EF; but as BC is to BG, so is the triangle ABC to the triangle ABG (VI. 1); therefore

5. The triangle ABC has to the triangle ABG, the duplicate

ratio of that which BC has to EF; but the triangle ABG is equal to the triangle DEF; therefore also

6. 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 proporonals, as the first is to the third, so is any triangle upon the first, to a similar and similarly described triangle upon the second.

PROPOSITION 20.-Theorem. 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 side homologous to FG;

Then the polygons ABCDE, FGHKL may be divided into the same number of similar triangles, whereof each shall have to each the same ratio which the polygons have ;

and the polygon ABCDE shall have to the polygon FGHKL the duplicate ratio of that which the side AB has to the side FG.

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Construetion. Join BE, EC, GL, LH.

Demonstration. Because the polygon ABCDE is similar to the polygon FGHKL, therefore

1. The angle BAE is equal to the angle GFL (VI. Def. 1),

and BA is to AE, as GF to FL (VI. Def. 1); therefore, because the triangles ABE, FGL have an angle in one, equal to an angle in the other, and their sides about these equal angles proportionals,

2. The triangle ABE is equiangular to the triangle FGL

(VI. 6); and therefore similar to it (VI. 4); wherefore

3. The angle ABE is equal to the angle FGL; and, because the polygons are similar, the whole angle ABC is equal to the whole angle FGH (VI. Def. 1); therefore

4. The remaining angle EBC is equal to the remaining angle

LGH (I. 32 and Ax. 3) ; and because the triangles ABE, FG L are similar,

5. EB is to BA, as LG to GF (VI. 4) ; and also, because the polygons are similar, AB is to BC, as FG to GH (VI. Def. 1); therefore, ex æquali,

6. EB is to BC, as LG to GH (V. 22); that is, the sides about the equal angles EBC, LGH are proportionals; therefore

7. The triangle EBC is equiangular to the triangle LGH,

(VI. 6), and similar to it (VI. 4); for the same reason, the triangle ECD likewise is similar to the triangle LHK; therefore

8. The similar polygons ABCDE, FGHKL are divided into

the same number of similar triangles. Also, these triangles shall 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 shall have to the polygon FGHKL, the duplicate ratio of that which the side A B has to the homologous side FG. Because the triangle ABE is similar to the triangle FGL,

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9. ABE has to FGL, the duplicate ratio of that which the

side BE has to the side GL (VI. 19) ; for the same reason, the triangle BEC has to GLH the duplicate ratio of that which BE has to GL; therefore

10. As the triangle ABE is to the triangle FGL, so is the tri

angle BEC to the triangle GLH (V. 11). Again, because the triangle EBC is similar to the triangle LGII,

11. EBC has to LGH, the duplicate ratio of that which tho

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 to LH; therefore

12. As the triangle EBC is to the triangle LGH, so is the tri

angle ECD to the triangle LHK (V. 11); but it has been proved, that the triangle EBC is likewise to the triangle LGH, as the triangle ABE to the triangle FGL; therefore

13. As the triangle ABE to the triangle FGL, so is the triangle

EBC to the triangle LGH, and the triangle ECD to the

triangle LHK; and therefore, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents (V. 12); that is,

14. As the triangle ABE to the triangle FGL, so is the polygon

ABCDE to the polygon FGHKL; but the triangle A BE has to the triangle FGL, the duplicate ratio of that which the side AB has to the homologous side FG (VI. 19); therefore also

15. 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 four-sided figures, 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 (VI. 19); therefore, universally, similar rectilineal figures are to one another in the duplicate ratio of their homologous sides.

Cor. 2. And if to AB, FG, two of the homologous sides, a third proportional M be taken (VI. 11),

1. AB has to M the duplicate ratio of that which AB has to

FG (V. Def. 10); 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 (VI. 20, Cor. 1); therefore

2. As AB is to M, so is the figure upon AB to the figure upon

FG (V. 11); which was also proved in triangles (VI. 19, Cor.) ; therefore

3. 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.

PROPOSITION 21.-Theorem.

Rectilineal figures which are similar to the same rectilineal figure, are also similar to one another.

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

Then the figure A shall be similar to the figure B.

B

Demonstration. Because A is similar to C,

1. They are equiangular, and also have their sides about the

equal angles proportional (VI. Def. 1). Again, because B is similar to C,

2. They are equiangular, and have their sides about the equal

angles proportionals (VI. Def. 1). 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 proportionals. Wherefore

3. The rectilineal figures A and B are equiangular (I. Ax. 1),

and have their sides about the equal angles proportionals

(V. 11); therefore 4. A is similar to B.(VI. Def. 1).

Therefore, rectilineal figures, &c. Q.E.D.

PROPOSITION 22.-Theorem.

If four straight lines be proportionals, the similar rectilineal figures similarly described upon them shall also be proportionals; and conversely, 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.

Then the rectilineal figure KAB shall be to LCD, as MF to NH.

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Construction. To AB, CD take a third proportional X (VI. 11); and to EF, GH a third proportional 0.

Demonstration. Because AB is to CD as EF to GH, therefore CD is to X, as GH to 0 (V. 11); wherefore, ex æquali,

1. As AB to X, so is EF to 0 (V. 22); but as AB to X, so is the rectilineal figure KAB to the rectilineal figure LCD, and as EF to 0, so is the rectilineal figure MF to the rectilineal figure NH (VI. 20, Cor. 2); therefore

2. As KAB to LCD, so is MF to NH (V. 11). Next. Let the rectilineal figure KAB be to LCD, as MF to NH; Then the straight line AB shall be to CD, as EF to GH.

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