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But the sides about the equal angles of equal parallelograms Book VI. are reciprocally proportional : Wherefore, as AB to CD, so is CH to AG; but CH is equal to E, and AG to F: As c 14. 6. therefore AB is to CD, so is E to F. Wherefore, if four," &c. Q. E. D.

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If three straight lines be proportionals, the rectangle contained ly 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.

b 16 6.

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

Take D equal to B; and because as A to B, so is B to C, and that B is equal to D: A is a to B, as D to C: But if four a 7. 5. straight lines be proportionals, the rectan- A gle contained by the

B extremes is equal to that which is contained

D by the means b: There-C fore, the rectangle contained by A, C is equal

D

С to that contained by B, D: But the rect- A

B angle contained by B, D is the square of B; because B is equal to D: Therefore, the rectangle 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 6: Therefore, A is to B,

Book VI. as D to C; but B is equal to D; wherefore as A to B, so is

B to C. Therefore, " if three straight lines,&c. Q. E. D.

PROP. XVIII. PROB.

See N.

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

a 23. 1.

b 32. ).

Let AB be the given straight line, and CDEF the given rectilineal figure of four sides; it is required upon the given straight line AB 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 a 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": Wherefore the
triangle FCD is e-
quiangular to the
triangle GAB: A-
gain, at the points G

H
G, B in the straight

F к

E line GB make a the

L angle BGH equal

K to the angle DFE, and the angle GBH

A B с D

FDE ; equal to therefore the maining angle FED is equal to the remaining angle GHB, and the triangle FDE equiangular to the triangle GBH: Then, because the angle AGB is equal to the angle CFD, and BGH to DFE, the whole angle AGH is equal to the whole CFE: For the 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 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 æquali 4, 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 c. Wherefore, because the rectilineal

re

c 4. 6.

d 22. 5.

e l. def. 6.

figures ABHG, CDEF are equiangular, and have their sides Book VI. about the equal angles proportionals, they are similar to one another e.

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 B, H, 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 ; therefore, ex equali", GH is to HL, as FE to EK: c 4. 6.

d 22. 5. For the same reason, AB is to BL, as CD to DK: And BL is to LH, as 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 on. Which was to be done.

PROP. XIX. THEOR.

Similar triangles are to one another in the dupli. cate 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 EFa: the triangle a 12.cef.5. ABC has to the triangle DEF, the duplicate ratio of that which BC has to EF.

Take BG a third proportional to BC and EFb; so that BC b 11. 6. is to EF, as EF to BG, and join GA: Then, because as AB to BC, so is DE to EF; alternately, AB is to DE, as BC to FE: c 16. 5.

Book VI. But as BC to EF, so is EF to BG; therefored as AB to

DE, so is EF to BG: Wherefore, the sides of the triangles d 11. 5. ABG, DEF, which are about the equal angles, are reciprocally

proportional. But triangles, which have the sides about two
equal angles recipro-
cally proportional,

A
are equal to one an-
e 15. 6. other 2: Therefore,

D the triangle ABG is equal to the triangle DEF: And because as BC is to EF, so is EF to BG; and that

B G CE F if three straight lines

be proportionals, the f 10.def.5. first is said to have to the third the duplicate ratio of that

which it has to the second ; BC therefore has to BG the dupli

cate ratio of that which BC has to EF: But as BC to BG, g 1. 6.

so is 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 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 has to each 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 Book VI. to the angle GFLa, and BA is to AE, as GF to FLa : Wherefore, because the triangles ABE, FGL, have an angle a 1 Def. 6. in the one equal to an angle in the other, and their sides about these equal angles proportionals, the triangle ABE is equiangular , and therefore similar, to the triangle FGL“; where- b 6. 6. fore, the angle ABE is equal to the angle FGL: And, be. c 4. 6. cause the polygons are similar, the whole angle ABC is equal a to the whole angle FGH; therefore, the remaining angle EBC is equal to the remaining angle LGH: Now, because the triangles ABE, FGL are similar, EB is to BA, as LG to GF a; and also, because the polygons are similar, AB is to BC, as FG to GH“, therefore, ex æqualid, EB is d 22. 5. to BC as LG to GH; that is, the sides about the equal angles EBC, LGH are proportionals; therefore, the triangle EBC is equiangular

A to the triangle

М.

F LGH,

and similar to it . E

B. For the same

L

G reason, the triangle ECD likewise is si

D C K H milar 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 e of that which the side e 19. 6. 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 f is the triangle BEC to the triangle GLH. Again, be- f 11. 5. cause 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

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