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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 angle B equal to the angle E, and let AB be to BD, as DE to EF, so that the side BC is homologous to EF (12. des. 5.); the triangle 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 GA; then, because as AB to"BC, so DE to EF, alternately (16. 5), AB is to DE, as BC to EF: but as BC to " EF; so is EF to BG; therefore (11. 5.), 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, are equal to one another (15. 6): therefore the triangle ABG is equal to the triangle

A e // ID DEF: and because as BC ...”/ is to EF, so EF to BG ; and that if three straight o be proportionals, the A rst is said (10. def. 5.) to have to the third the # B G C E F cate 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 triangles 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: 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, whereof each to each has the same ratio which the polygons have; and the polygon ABCDP

has to the polygon FGHKL the duplicate ratio of that which the side AB has to the side FG. Join 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 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; 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, AB is to BC, as FG to GH (1. des. 6.); therefore ea aequali (22.5.), EB is to BC, as LG to GH; that is, the sides about the equal angles EBC, LGH are proportionals; therefore (22. 5.) the triangle EBC is equiangular to the tri- A M angle LGH, and si- F milar to it (4.6.). For the same rea-E son, the triangle G ECD likewise is si- L milar to the triangle LKH; therefore the similar po- D C K H lygons 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, so (11. 5.) 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 to LH ; as therefore the triangle EBC to the triangle LGH, so is (11. 5.) 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 is 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 (12.5.). 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 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. 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 AB has (10, def 5,) 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 which are similar to the same rectilineal figures, are also similar to one another.

Let each of the rectilineal figures, A, B 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 A 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 (11. 5.). Therefore A is similar (1. def. 6) to B. Q. E. D.

PROP. XXII. THEOR.

It 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 KAB is to LCD, as MF to NH. To AB, CD take a third proportional (11.6.) X; and to EF, GH a third proportional O: and because AB is to CD, as EF to GH, and that CD is (11.5) to X, as GH to O; wherefore, ex æquali (22.5), as AB to X, so is EF to O: but as AB to X, so is (2 Cor. 20. 6.) the rectilineal KAB to the rectllineal 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 (11. 5.) 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 describe (18, 6.) the rectilineal figure SR similar and similarly situated to

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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 (9.5) 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.

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 angle

* See Note.

BCD equal to the angle ECG : the ratio of the parallelogram AC to the parallelogram CF, is the same with the ratio which is compounded of the ratios of their sides.

Let BG, 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 (12.6.) L to M: therefore the ratios of K to L, and L to M, are the same with the ratios of the sides, viz. of BC to CG, and DC to CE. But the ratio of K to M is that which is said to be compounded (A. def. 5.) of the ratios of K to L, and L to M : wherefore also K has to M the ratio compounded of the ratios of the sides; and because as BC to CG, so A D H is the parallelogram AC to the parallelogram CH (1.6.); but as BC to CG, so is K to L; therefore K is (11. 5.) to L, as the parallelogram AC to the parallelogram GH: 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 Lis (11. 5.) to M, as the parallelogram CH to the parallelogram CF: therefore since it has been proved, that as K to L, so is the parallelogram AC to the parallelogram CH; K. L. M. E F and as L to M, so the parallelogram CH to the parallelogram CF; ea aequali (22.5.), K is to M, as the parallelogram AC to the parallelogram CF: but K has to M the ratio which is compounded of the ratios of the sides; therefore also the parallelogram AC has to the parallelogram CF the ratio which is compounded of the ratios of the sides. Wherefore, equiangular parallelograms, &c. Q. E. D.

PROP. XXIV. THEOR.

THE parallelograms about the diameter of any parallelogram are similar to the whole and to one another.”

Let ABCD be a parallelogram, of which the diameter is AC; and EG, HK the parallelograms about the diameter: the parallelograms EG, HK are similar both to the whole parallelogram ABCD, and to one another.

Because DC, GF are parallels, the angle ADC is equal (29. 1.) to the angle AGF: for the same reason, because BC, EF are parallels, the angle ABC is equal to the angle AEF; and each of the angles BCD, EFG is equal to the opposite angle DAB (34. 1.), and therefore are equal to one another; wherefore the parallelograms ABCD, AEFG are equiangular; and because the angle ABC is equal to the angle AEF; and the angle BAC common to the two triangles BAC, EAF, they are equiangular to one another; therefore (4.6.) as AB to BC, so is AE to EF: and because the

* See Note.

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