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If Ihree-straight lines be proportionals, the rectangle contained by the extremes is equal to the square of the mean: And if the rectangle con
tained 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 contained by A, C is equal to the be proportionals, viz. as A to B, so B square of B. to C; the rectangle contained by A, And if the rectangle contained by C is equal to the square of B. A, C be equal to the square of B; À
Take D equal to B; and because is to B, as B to C. as A to B, so B to C, and that B is The same construction being made, equal to D: A is (7. 5.) to B, as D because the rectangle contained by to C: But if four straight lines be A, C is equal to the square of B, and proportionals,
the square of B is equal to the recthe rectangle
tangle contained by B, D, because contained by
B is equal to D; therefore the recthe extremes
tangle contained by A, C is equal to is equal to that
that contained by B, D: But if the which is con
rectangle contained by the extremes tained by the
be equal to that contained by the means:(16.6.)
means, the four straight lines are proTherefore the rectangle contained by portionals : (16. 6.) Therefore A is A, C is equal to that contained by B, to B, as D to C; but B is equal to D: But the rectangle contained by D; wherefore as A to B, so B to C: B, D is the square of B; because B Therefore, if three straight lines, &c. is equal to D: Therefore the rectangle Q. E. D.
PROP. XVIII. PROB.
Upon a given straight line to describe a rectilineal figure similar, and
similarly siluated to a given rectilineal figure.
Join DF, and at the points A, B in (22. 1.) the angle BGH equal to the the straight line AB, make (23. 1.) angle DFE, and the angle GBH equal the angle BAG equal to the angle at to FDE ; therefore the remaining C, and the angle ABG equal to the angle FED is equal to the remaining argle CDF; therefore the remaining angle GHB, and the triangle FDE angle CFD is equal to the remaining equiangular to the triangle GBH: angle AGB: (32. 1.) Wherefore the Then, because the angle AGB is triangle FCD is equiangular to the equal to the angle CFD, and BGH triangle GAB: Again, at the points to DFE, the whole angle AGH is G, B' in the straight line GB 'make equal to the whole CFE: For the
same reason, the angle ABH is equal und at the points 2, 8, in the straight to the angle CDE; also the angle at line BH, make the angle HBL equal A is equal to the angle at C, and the to the angle EDK, and the angle angle GHB to FED: Therefore the BHL equal to the angle DEK; thererectilineal figure ABHG is equiangular fore the remaining angle at K is equal to CDEF: But likewise these tigures to the remaining angle at L: and behave their sides about the equal an- cause the tigures ABHG, CDEF are gles proportionals : Because the tri- similar, the angle GHB is equal to angles GAB, FCD being equiangular, the angle FED, and BHL is equal to BA is (4. 6.) to AG, as DC to CF; DEK; wherefore the whole angle and because AG is to GB, as CF to GHL is equal to the whole angle FD; and as GB to GH, so, by reason FEK: For the same reason the angle of the equiangular triangles BGH, ABL is equal to the angle CDK: DFE, is FD to FE: therefore, ex Therefore the five sided figures æquali, (22. 5.) AG is to GH, as AGHLB, CFEKD are equiangular; CF to FE: In the same manner it and because the figures AGHB, may be proved that AB is to BH, as CFED are similar, GH is to HB, as CD to 'DE: And GH is to HB, as FE to ED; and as HB to HL, so is FE to ED. (4. 6.) Wherefore, be- ED to EK; (4. 6.) therefore, ex cause the rectilineal figures ABHG, æquali, (22. 5.) GH is to HL, as FE CDEF are equiangular, and have to EK: For the same reason, AB is their sides about the equal angles to BL as CD to DK: And BL is to proportionals, they are sitnilar to one LH, as (4. 6.) DK to KE, because another. (1. def. 6.)
the triangles BLH, DKE are equiNext, let it be required to describe angular: Therefore, because the five upon a given straight line AB, a rec- sided figures AGHLB, CFEKD are tilineal figure similar, and similarly equiangular, and have their sides situated to the rectilineal figure about the equal angles proportionals, CDKEF.
they are similar to one another : And Join DE, and upon the given in the same
a rectilineal straight line AB describe the recti- figure of six or more sides may be lineal figure ABHG similar, and described upon a given straight line similarly situated to the quadrilateral similar to one given, and so on. figure CDEF, by the former case; Which was to be done.
PROP. XIX. THEOR.
Similar triangles are to one another in the duplicate ratio of their homo
Let ABC, DEF be similar triangles, Then, because as AB to BC, so DE having the angle B equal to the angle to EF; alternately, (16. 5.) AB is to E, and let AB be to BC, as DE to DE, as BC to EF : But as BC to EF, so that the side BC is homologous EF, so is EF to BG; therefore (11. to ÉF : (12. def. 5.) the triangle ABC 5.) as AB to DE, so is is EF to BG: has to the triangle DEF, the dupli- Wherefore the sides of the triangles cate ratio of that which BC has to ABG, DEF, which are about the EF.
equal angles, are reciprocally proTake BG a third proportional to portional. But triangles which have BC, EF; (11. 6.) so that BC is to the sides about two equal angles reEF, as EF to BG. and join GA: ciprocally proportional, are equal to
one another : (16. 6.). Therefore the triangle ABC to the triangle ABG. triangle ABG is equal to the triangle Therefore the triangle ABC has to DEF : And because as BC is to EF, the triangle ABG the duplicate ratio 80 EF to BG; and that if three of that which BC har to EF: But straight lines be proportionals, the the triangle ABG is equal to the tri
angle DEF; wherefore also the triangle ABC has to the triangle DEF the duplicate ratio of that which BC has to EF. Therefore similar tri.
angles, &c. Q. E. D. B G
Cor. From this it is manifest, that first is said (10 def. 5.). to have to if three straight lines be proportionals, the third the duplicate ratio of that as the first is to the third, so is any which it has to the second ; BC triangle upon the first to a similar, therefore has to BG the duplicate and similarly described triangle upon ratio of that which BC has to EF: the cecond. But as BC to BG, so is (1. 6.) the
PROP, XX. THEOR.
Similar polygons may be divided into the same number of similar tri
angles, having the same ratio to one another that the polygons huve; and the polygons have to one another the duplicate ratio of that which their homologous sides have.
Let ABCDE, FGHKL be similar 6.) to the whole angle FGH; therepolygons, and let AB be the homo- fore the remaining angle ERC is logous side to FG: The polygons equal to the remaining angle LGH: ABCDE, FGHKL may be divided And because the triangles ABE, into the same number of similar tri- FGL are similar, EB is to BA, as angles, whereof each to each has the LG to GF; (1. def. 6.) and also, besame ratio which the polygons have; cause the polygons are similar, AB is and the polygon ABCDE has to the to BC, as FG to GH, (1. def. 6.) polygon FGHKL the duplicate ratio therefore, ex æquali, (22. 5.) EB is to of that which the side AB has to the BC, as LG to GH; that is, the sides side FG.
about the equal angles EBC, LGH Join BE, EC, GL, LH: And be- are proportionals; therefore, (22. 5.) cause the polygon ABCDE is similar the triangle EBC is equiangular to to the polygon FGHKL, the angle the triangle LGH, and similar to it. BAE is equal to the angle GFL, (1, (4. 6.) For the same reason, the tridef. 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: angle ECD likewise is similar to the (6. 6.), Wherefore the angle ABE triangle LHK: therefore, the similar is equal to the angle FGL: And polygons ABCDE, FGHKL are di
ecause the polygons are similar, vided into the sarne number of similar the whole angle ABC is equal (1. def. triangles.
Also, these triangles have, each to 5.), Wherefore, as the triangle ABE each, the same ratio which the poly- to the triangle FGL, so is the polygon gons have to one another, 'the ante: ABCDE to the polygon FGHKL: cedents being ABE, EBC, ECD, and But the triangle ABE has to the trithe consequents FGL, LGH, LHK: angle FGL, the duplicate ratio of that And the polygon ABCDE has to the which the side AB has to the homoly polygon ÉGHKL the duplicate ratio gous side FG. Therefore, also the of that which the side AB has to the polygon ABCDE has to the polygon homologous side FG.
FGHKL the duplicate ratio of that Because the triangle ABE is similar which AB has to ihe bomologous side to the triangle FGL, ABE has lo FGL FG. Wherefore sirailar polygons, the duplicate ratio (19. 6.) of that &c. Q. E. D. which the side BE has to the side GL: Cos. 1. In like manner, it may le For the same reason, the triangle BEC proved, that similar four sided figures, has to GLH the duplicate ratio of that or of any number of sides, are one to which BE has to GL: Therefore, as another in the duplicate ratio of their the triangle ABE to the triangle FGL, homologous sides, and it has already so (11. 5.) is the triangle BÈC to the been proved in triangles Therefore, triangle GLH. Again, because the universally, similar rectilineal figures triangle EBC is similar to the triangle are to one another in the Juplicate raLGH, EBC has to LGH the duplicate tio of their hornologous sides. ratio of that which the side EC has to Cor. 2. And if to AB, FG, two of the side LH: For the sanie reason, the homologous sides, a third proporthe triangle ECD has to the triangle tional M be taken, AB has (10. def. 5.) LHK, the duplicate ratio of that to M the duplicate ratio of that which which EC has to LH: As therefore AB has to FG: But the four sided the triangle EBC to the triangle LGH, figure or polygon upon AB, has to the so is (11. 5.) the triangle ECD to the four sided ligure, or polygo: upon triangle LHK: But it has been FG, likewise the duplicate ratio of proved, that the triangle EBC is like that which AB has to FG: Therefore, wise to the triangle LGH, as the tri- as AB is to M, so is the figure upon angle ABE to the triangle FGL. AB to the figure upon FG, which was Therefore, as the triangle ABE is to also proved in triangles. (Cor. 19. 6.) the triangle FGL, so is the triangle Therefore, universally, it is manifest, EBC to triangle LGH, and triangle that if three straight lines be proporECD to triangle LHK: And there- tionals, as the first is to the third, so fore, as one of the antecedents to one is any rectilineal figure upon the first, of the consequents, so are all the an- to a similar and similarly described tecedents to all the consequents. (12. rectilineal figure upon the second.
PROP. XXI. THEOR,
Rectilineal figures, which are similar to the same rectilineal figure, are
also similar to one another.
Let each of the rectilineal figures about the equal A,B be similar to the rectilineal figure angles propor. C: The figure A is similar to the tionals. (1. Def. tigure B.
6.) Again, beBecause A is similar to C, they are cause Big siequiangular, and also have their sides milar to C, they
are equiangular, and have their sides C proportionals. Wherefore the recabout the equal angles proportion- tilineal figures A and B are equials (1. Def. 6.): Therefore the figures angular, (1. Ax. 1.) and have their A, B, are each of them equiangular sides about the equal angles proporto C, and have the sides about the tionals. (11. 5.) Therefore A is simi. equal angles of each of them and of lar (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, situated to either of the figures MF, EF, GH be proportionals, viz. AB to NH: Then, because as AB to CD, so 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
R third proportional 0: And because is EF to PR, and that upon AB, CD AB is to CD, as EF to GH, and that are described the similar and similarly CD is (11. 5.) to X, as GH to 0; situated rectilineals KAB, LCD, and wherefore, ex æquali, (22. 5.) as AB upon EF, PR, in like manner, the sito X, so EF to 0 : But as AB to X, milar rectilineals MF, SR; KAB is to so is (2 Cor. 20. 6.) the rectilineal KAB LCD, as MF to SR; but, by the hyto the rectilineal LCD; and as EF to pothesis, KAB is to LCD, as MF to 0, so is (2 Cor. 2. 6.) the rectilineal NH; and therefore the rectilineal MF MF to the rectilineal NH: There- having the same ratio to each of the fore, as KAB to LCD so (11. 5.) is two ÑH, SR, these are equal (9.5.) MF to NH.
to one another: They are also similar, And if the rectilineal KAB be to and similarly situated; therefore GII LCD, as MF to NH; the straight is equal to PR: And because as AB line AB is to CD, as EF to GH. to CD, so is EF to PR, and that PR
Make (12. 6.) as AB to CD, so EF is equal to GH; AB is to CD, as EF to PR, and upon PR describe the rec- to GH. If therefore four straight tilineal figure SR similar and similarly lines, &c. Q. E. D.
Equiangular parallelograms have to one another the ratio which is comú
pounded of the ratios of their sides.
Let AC, CF be equiangular paral- equal to the angle ECG: the ratio of lelograms, having the angle BCD the parallelogram AC to the paralles