Let the three ftraight lines A, B, C be proportionals; viz. as A Book VI. is to B fo let B be to C: I fay that the rectangle contained by A and C is equal to the fquare of the mean B.

Let D be made equal to B. And because it is, as A is to B fo is B to C; and B is equal to D; therefore it is, as A is to B fo is D to C: but if four straight lines be proportionals, the rectangle contained by the extremes is equal (by 16. 6.) to the rectangle contained

by the means; therefore the rectangle contained by A and C is equal to the rectangle contained by B and D : but the rectangle contained by B and D is equal to the Square of B; for B is equal to D therefore the rectangle contained by A and C is equal to the Square of B.

But let the rectangle contained by A and C be equal to the Square of B; I say that it is, as A is to B fo is B to C.

For the fame things being conftructed; because (by fupp.) the rectangle contained by A and C is equal to the Square of B; but the Square of B is equal to the rectangle contained by B and D; for B is equal to D: therefore the rectangle contained by A and C is equal to the rectangle contained by B and D; but (by 16. 6.) if the rectangle contained by the extremes be equal to the rectangle contained by the means; the four straight lines are proportionals : therefore it is, as A is to B fo is D to C; but B is equal to D; therefore as A is to B fo is B to C.

Wherefore if three ftraight lines be proportionals; the rectangle contained by the extremes is equal to the square of the mean proportional: and if the rectangle contained by the extremes be equal to the fquare of the mean term; the three ftraight lines will be proportionals. Which was to be demonftrated.

PRO P. XVIII.

Upon a given straight line to describe a rectilineal figure, fimilar and fimilarly fituated to a given rectilineal figure.

Book VI.

Let AB be the given ftraight line; and CE the given rectilineal figure: it is required, upon the given ftraight line AB to defcribe a rectilineal figure fimilar and fimilarly fituated to the rectilineal figure CE.

Let DF be joined; and with
the ftraight line AB and at the
points A, B in it, let the angle G
GAB be made equal to the angle

E

H

F

A

BC

D

at C; and the angle ABG equal
to the angle CDF; therefore (by
the 32. 1.) the remaining angle
CFD is equal to the remaining angle AGB; therefore the triangle
FCD is equiangular to the triangle GAB: therefore there is this
proportion (by 4. 6.) as FD is to GB fo is FC to GA; and fo is
CD to AB: Again, with the straight line BG, and at the points
B, G in it, let the angle BGH be made equal to the angle DFE;
and the angle GBH equal to the angle FDE; therefore the remain-
ing angle at H is equal to the remaining angle at E: therefore the
triangle FDE is equiangular to the triangle GBH; therefore there
is this proportion (by 4. 6.) as FD is to GB fo is FE to GH; and
Jo is ED to HB: But it has been also demonstrated that as FD is
to GB fo is FC to GA, and fo is CD to AB: therefore (by 11. 5.)
as FC is to GA fo is CD to AB and fo is FE to GH, and befides fo
is ED to HB And because the angle CFD is equal to the angle
AGB and DFE is equal to BGH; therefore the whole angle CFE
is equal to the whole AGH. Certainly for the fame reason also
the angle CDE is equal to ABH but also the angle at C is equal
to the angle at A; and the angle at E to the angle at H: therefore
the figure AH is equiangular to the figure CE; and has the fides.
about the equal angles in it proportionals; therefore (by 1. def. 6.)
the rectilineal figure AH is fimilar to the rectilineal figure CE.

Wherefore upon a given ftraight line AB a rectilineal figure AH hath been defcribed, fimilar and fimilarly fituated to the given rectilineal figure CE. Which was to be done.

PROP. XIX.

Similar triangles are to one another in the duplicate ratio of the fides of like ratio.

Let

Let ABC, DEF be fimilar triangles, having the angle at B equal Book VI. to the angle at E: and as AB is to BC fo let DE be to EF; fo that (by 12. def. 5.) BC may be the fide of like ratio to EF: I fay that the triangle ABC has to the triangle DEF the duplicate ratio of that which BC has to EF.

For (by 11.6.) let BG be taken a third proportional to BC, EF; fo that it may be, as BC is to EF fo is EF to BG: and let GA be joined,

A

D

B G

CE

F

Wherefore because it is, as AB is to BC fo is DE to EF; therefore alternately (by 16. 5.) it is, as AB is to DE fo is BC to EF: but (by const.) as BC is to EF fo is EF to BG: therefore alfo (by 11. 5.) as AB is to DE fo is EF BG: therefore the fides about the equal angles of the triangles ABG, DEF are reciprocally proportional: but those triangles having one angle equal to one angle, and of which the fides about the equal angles are reciprocally proportional, are equal. Therefore the triangle ABG is equal to the triangle DEF: and because it is, as BC is to EF fo is EF to BG; and if three straight lines be proportionals (by 10. def. 5.) the firft is faid to have to the third a duplicate ratio of that which it has to the fecond: therefore BC has to BG a duplicate ratio of that which BC has to EF; but as BC is to BG so (by 1. 6.) is the triangle ABC to the triangle ABG: therefore also 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; therefore (by 7. 5.) the triangle ABC has to the triangle DEF the duplicate ratio of that which BC bas to EF.

Wherefore fimilar triangles are to one another in the duplicate: ratio of the fides of like ratio. Which was to be demonftrated.

Cor. Certainly from this it is manifeft, that if three straight. lines be proportionals, it is, as the first is to the third fo is a triangle defcribed upon the first, to a fimilar and fimilarly fituated triangle defcribed upon the fecond; fince it has been demonstrated, as CB is to BG fo is the triangle ABC to the triangle ABG, that is DEF.: PROP

Book VI.

PROP.

XX.

Similar polygons are divided into fimilar triangles, and into the fame number, and have the fame ratio to one another which the whole polygons have and the one polygon has to the other polygon the duplicate ratio of that which one fide of the one has to the fide of like ratio of the other.

Let ABCDE, FGHKL be fimilar polygons, and let AB be the fide of like ratio FG: I say that the polygons ABCDE, FGHKL are divided into fimilar triangles; and into the fame number; and have the fame ratio to one another which the whole polygons have ; and that the polygon ABCDE has to the polygon FGHKL the duplicate ratio of that which AB has to FG.

Let BE, EC; GL, LH be joined.

And because the polygon ABCDE is fimilar to the polygon FGHKL, the angle BAE is equal to the angle GFL: and it is, as BA is to AE fo is GF to FL: wherefore because there are two triangles ABE, FGL having one angle equal to one angle, and the fides about the equal angles proportionals; therefore (by 6. 6.) the triangle ABE is equiangular to the triangle FGL; so that (by 4. 6.) it is also fimilar: therefore the angle ABE is equal to the angle FGL and alfo on account of the fimilarity of the polygons, the whole angle ABC is equal to the whole angle FGH; therefore the remaining angle EBC is equal to the remaining angle LGH; and because, on account of the fimilarity of the triangles ABE, FGL, it is, as EB is to BA fo is LG to GF; but also, on account of the fimilarity of the polygons, it is, as AB is to BC fo is FG to GH: therefore by equality it is, (by 22, 5.) as EB is to BC fo is LG to GH; and the fides therefore about the equal angles EBC, LGH are proportionals; therefore the triangle EBC is equiangular to the triangle LGH (by 6. 6.); fo that it is also fimilar (by 4. 6.): Certainly for the fame reafon the triangle ECD is fimilar to the triangle LHK therefore the fimilar polygons ABCDE, FGHKL are divided into fimilar triangles, and into an equal number.

I fay also that they have the fame ratio to one another which the whole polygons have: that is, fo that the triangles are proportionals; and that the antecedents are the triangles ABE, EBC,

ECD

ECD and the confequents of them, the triangles FGL, LGH, Book VI. LHK: also that the polygon ABCDE has to the polygon FGHKL the duplicate ratio of that which the fide of like ratio of the one has to the fide of like ratio of the other, that is, which AB has to FG.

A

X

E

F

M B

L

G

D

K

H

For let AC, FH be joined. And because, on account of the fimilarity of the polygons, the angle ABC is equal to the angle FGH, and it is as AB is to BC fo is FG to GH; the triangle ABC is equiangular (by 6. 6.) to the triangle FGH therefore the angle BAC is equal to the angle GFH; and BCA to GHF: and because the angle BAM is equal to GFN; and the angle ABM has been demonstrated to be equal to FGN; therefore the remaining angle AMB is equal (by 32. 1.) to the remaining angle FNG; therefore the triangle ABM is equiangular to the triangle FGN: Certainly in the fame manner we fhall demonftrate that the triangle BMC is equiangular to the triangle GNH; therefore there is this proportion, (by 4. 6.) as AM is to MB fo is FN to NG; and as BM is to MC fo is GN to NH: so that also by equality (by 22. 5.) as AM is to MC fo is FN to NH: but as AM is to MC fo (by 1. 6.) is the triangle ABM to the triangle BMC; and fo is the triangle AEM to the triangle MEC; for they are to one another as their bafes: And (by 12. 5.) as one of the antecedents is to one of the confequents fo are all the antecedents to all the confequents; therefore as the triangle AMB is to the triangle BMC fo is the triangle ABE to the triangle CBE : but as the triangle AMB is to the triangle BMC fo is AM to MC: therefore alfo (by 11. 5.) as AM is to MC fo is the triangle ABE to the triangle CBE. Certainly for the fame reafon alfo as FN is to NH fo is the triangle FGL to the triangle GLH: and it is, as AM is to MC fo is FN to NH; therefore (by 11. 5.) as the triangle ABE is to the triangle BEC fo is the triangle FGL to the triangle GHL; and alternately (by 16. 5.) as the triangle ABE is to the triangle FGL fo is the triangle BEC to the triangle GLH. Cer