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PROP. XVII. THEOR.

If three straight lines be proportionals, the rectangle contained by the extremes is equal to the square of the mean: And if the rectangle contained by the extremes be equal to the square of the mean, the three straight lines are proportionals.

Let the three straight lines A, B, C be proportionals, viz. as A to B, so B to C; the rectangle contained by A, C is equal to the square of B.

C

Take D equal to B; and because as A to B, so B to C, and that B is equal to D: A is (7. 5.) to B, as D to C: But if four straight lines be proportionals, A the rectangle B. contained by D the extremes is equal to that which is contained by the means:(16.6.) Therefore the rectangle contained by A, C is equal to that contained by B, D: But the rectangle contained by B, D is the square of B; because B is equal to D: Therefore the rectangle

A

B

contained by A, C is equal to the square of B.

And if the rectangle contained by A, C be equal to the square of B; A is to B, as B to C.

The same construction being made, because the rectangle contained by A, C is equal to the square of B, and the square of B is equal to the rectangle contained by B, D, because B is equal to D; therefore the rectangle contained by A, C is equal to that contained by B, D: But if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportionals: (16. 6.) Therefore A is to B, as D to C; but B is equal to D; wherefore as A to B, so B to C: Therefore, if three straight lines, &c. Q. E. D.

PROP. XVIII. PROB.

Upon a given straight line to describe a rectilineal figure similar, and similarly situated to a given rectilineal figure.

Let AB be the given straight line, and CDEF the given rectilineal figure of four sides; it is required upon the given straight line A B to describe a rectilineal figure similar, and similarly situated to CDEF. ·

Join DF, and at the points A, B in the straight line AB, make (23. 1.) the angle BAG equal to the angle at C, and the angle ABG equal to the angle CDF; therefore the remaining angle CFD is equal to the remaining angle AGB: (32. 1.) Wherefore the triangle FCD is equiangular to the triangle GAB: Again, at the points G, B in the straight line GB make

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same reason, the angle ABH is equal to the angle CDE; also the angle at A is equal to the angle at C, and the angle GHB to FED: Therefore the rectilineal figure ABHG is equiangular to CDEF: But likewise these figures have their sides about the equal angles proportionals: Because the triangles GAB, FCD being equiangular, BA is (4. 6.) to AG, as DC to CF; and because AG is to GB, as CF to FD; and as GB to GH, so, by reason of the equiangular triangles BGH, DFE, is FD to FE: therefore, ex quali, (22. 5.) AG is to GH, as CF to FE: In the same manner it may be proved that AB is to BH, as CD to DE: And GH is to HB, as FE to ED. (4. 6.) Wherefore, because the rectilineal figures ABHG, CDEF are equiangular, and have their sides about the equal angles proportionals, they are similar to one another. (1. def. 6.)

Next, let it be required to describe upon a given straight line AB, a rectilineal figure similar, and similarly situated to the rectilineal figure CDKEF.

Join DE, and upon the given straight line AB describe the rectilineal figure ABHG similar, and similarly situated to the quadrilateral figure CDEF, by the former case;

and at the points 3, A, in the straight line BH, make the angle HBL equal to the angle EDK, and the angle BHL equal to the angle DEK; therefore the remaining angle at K is equal to the remaining angle at L: and because the figures ABHG, CDEF are similar, the angle GHB is equal to the angle FED, and BHL is equal to DEK; wherefore the whole angle GHL is equal to the whole angle FEK: For the same reason the angle ABL is equal to the angle CDK: Therefore the five sided figures AGHLB, CFEKD are equiangular; and because the figures AGHB, CFED are similar, GH is to HB, as FE to ED; and as HB to HL, so is ED to EK; (4. 6.) therefore, ex æquali, (22. 5.) GH is to HL, as FE to EK: For the same reason, AB is to BL as CD to DK: And BL is to LH, as (4. 6.) DK to KE, because the triangles BLH, DKE are equiangular: Therefore, because the five sided figures AGHLB, CFEKD are equiangular, and have their sides about the equal angles proportionals, they are similar to one another: And in the same manner a rectilineal figure of six or more sides may be described upon a given straight line similar to one given, and so Which was to be done.

PROP. XIX. THEOR.

on.

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 BC, as DE to EF, so that the side BC is homologous to EF: (12. def. 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 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

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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 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 ABCDE 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: (1. 6.) Wherefore the angle ABE is equal to the angle FGL: And ecause the polygons are similar, the whole angle ABC is equal (1. def.

6.) to the whole angle FGH; therefore the remaining angle ERC 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. def. 6.) therefore, ex æquali, (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 triangle LGH, and similar to it. (4. 6.) For the same reason, the tri

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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 like wise to the triangle LGH, as the triangle ABE to the triangle FGL. Therefore, as the triangle ABE is to the triangle FGL, so is the 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 homolo gous 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 Le 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 figure, are

also similar to one another.

angles propor

Let each of the rectilineal figures about the equal
A, B be similar to the rectilineal figure
C: The figure A is similar to the_tionals. (1. Def.
figure B.

Because A is similar to C, they are equiangular, and also have their sides

6.) Again, be

cause B is similar to C, they

A

A

are equiangular, and have their sides about the equal angles proportionals (1. Def. 6.): 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 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.

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 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 0; wherefore, ex æquali, (22. 5.) as AB to X, so EF to 0: 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. 2. 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 the rectilineal figure SR similar and similarly

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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 GI 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 com pounded of the ratios 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 paralle

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