16. 6.

23. 1.

+ 32. 1.

I 23. 1.

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; 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.

PROPOSITION 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 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* the angle BAG equal to the angle at C, and the angle ABG equal to the

H

angle GAB again, at the points G, B, in the straight line GB, make‡ the angle BGH equal to the angle DFE, and the angle GBH equal to FDE; therefore the remaining 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 || 4. 6. 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§, AG is to GH, as CF § 22. 5. 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¶.

Wherefore, because the rectilineal figures ABHG, CDEF are equiangular, and have their sides about the equal angles proportionals, they are similar to one another*.

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 Lt:

T 4.6.

* 1 Def. 6.

+ 32. 1.

And because the figures ABHG, CDEF are similar‡, the ‡ Constr. 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:

§ 22. 5.

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 æquali§, GH is to HL, as FE to EK: 4.6. For the same reason, AB is to BL as CD to DK: And BL is to LH, as DK to KE, because the triangles 4. 6. 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.

P

12 Def. 5.

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PROPOSITION 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 BC, as DE to EF, so that the side BC is homologous to EF*: 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†, so that BC is to EF, as EF to BG, and join GA:

Then, because, as AB to BC, so DE to EF; alternately‡, AB is to DE, as BC to EF: but as BC to EF, so is EF to BG; therefore||, 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:

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But triangles, which have the sides about two equal angles reciprocally proportional, are equal to one another§; therefore the triangle ABG is equal to the triangle DEF:

And because as BC is to EF, so EF to BG; and that if 10 Def. 5. 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; BC therefore has to BG the duplicate ratio of that which BC has to EF:

1.6.

But as BC to BG, 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.

PROPOSITION 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*, and BA is to AE, as GF to FL†:

* 1 Def. 6. † 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, and therefore similar, to the triangle FGL|| ; † 4. 6. wherefore the angle ABE is equal to the angle FGL:

6.

And because the polygons are similar, the whole angle ABC is equal to the whole angle FGH; therefore the § 1 Def. 6. 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 GFT; and also, because the polygons ¶ 1 Def. 6. are similar, AB is to BC, as FG to GH*; therefore, ex 1 Def. 6. æqualif, EB is to BC, as LG to GH; that is, the sides + 22. 5.

about the equal angles EBC, LGH are proportionals; therefore the triangle EBC is equiangular to the triangle 1 6. 6. LGH, and similar to it; for the same reason, the tri- || 4.6. angle ECD likewise is similar to the triangle LHK: therefore the similar polygons ABCDE, FGHKL are divided into the same number of similar triangles.

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T 11. 5.

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§ 10 Def. 5.

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 FGHIKL, 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§ 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 is to the triangle FGL, so¶ 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 LHt:

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 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; 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§ to M the dupli