gle contained under A and C, is equal to the Rectangle under B and D, but the Rectangle under B and D is equal to the Square of D, for B is equal to D. Wherefore the Rectangle contained under A, C, is equal to the Square of B.

And if the Rectangle contained under A, C, be equal to the Square of B; I fay, as A is to B, fo is B to C.

For the fame Conftruction remaining, the Rectangle contained under A and C is equal to the Square of B; but the Square of B is the Rectangle contained under B, D, for Bis equal to D, and the Rectangle contained under A, C, fhall be equal to the Rectangle contained under B, D. But if the Rectangle contained under the Extremes, be equal to the Rectangle contained under the Means, the four Right Lines t16 of this. fhall be + Proportionals. Therefore A is to B as D is to C; but B is equal to D. Wherefore A is to B, as B is to C. Therefore if three Right Lines be proportional, the Rectangle contained under the Extremes, is equal to the Square of the Mean; and if the Rectan gle under the Extremes, be equal to the Square of the Mean, then the three Right Lines are proportional ; which was to be demonftrated.

1

23. I.

PROPOSITION XVIII.

PROBLEM.

Upon a given Right Line, to defcribe a Right-lin❜d Figure fimilar, and fimilarly fituate to a Right-lin❜d Figure given.

LE

ETAB be the Right Line given, and CE the Right-lin'd Figure. It is required to defcribe upon the Right Line AB a Figure fimilar, and fimilarly fituate to the Right Line Figure CE.

Join DF, and make* at the Points A and B, with the Line AB, the Angles GAB, ABG, each equal to the Angles C and CDE. Whence the other AnCor. 32. gle CFD is equal to the other Angle AGB; and to the Triangle FCD is equiangular to the Triangle GAB; and confequently, as FD is to GB, fo is #4 of this. FC to GA; and fo is CD to AB. Again, make

1.

the

the Angles BGH, GBH, at the Points B and G, with the Right Line BG, each equal to the Angles EFD, EDF; then the remaining Angle at E, is † † Cor.32.1. equal to the remaining Angle at H. Therefore the Triangle FDE, is equiangular to the Triangle GBH; and confequently, as FD is to GB, fo is FE to + of this GH; and fo ED to HB. But it has been proved that FD is to GB, as FC is to GA, and as CD to AB. And therefore as F C is to AG, fo is* CD to 11. 5. AB; and fo F E to GH; and fo ED to HB. And because the Angle CFD is equal to the Angle AGB; and the Angle DFE equal to the Angle BGH; the whole Angle CFE fhall be equal to the whole Angle AGH. By the fame Reafon, the An gle CDE is equal to the Angle ABH; and the Angle at C equal to the Angle A; and the Angle E equal to the Angle H. Therefore the Figure AH is equiangular to the Figure CE; and they have the Sides about the equal Angles proportional. Confequently, the Right-lin'd Figure AH will be + fimi» + Def. 1 of lar to the Right-lin'd Figure CE. Therefore there this. is defcribed upon the given Right Line A B, the Rightlin'd Figure AH fimilar, and fimilarly fituate to the given Right-lin❜d Figure CE; which was to be done.

PROPOSITION XIX.

THEOREM.

Similar Triangles are in the duplicate Proportion of their homologous Sides.

LET ABC, DEF, be fimilar Triangles, having the Angle B equal to the Angle E; and let A B be to BC as DE is to EF, fo that BC be the Side homologous to EF. I fay, the Triangle ABC, to the Triangle DE F, has a duplicate Proportion to that of the Side BC to the Side EF.

For take BG a third Proportional to BC and 11 ofthis. EF; that is, let BC be to EF, as EF is to BG, and join GA.

Then because AB is to BC, as DE is to EF; it fhall be (by Alternation) as AB is to DE, fo is BC to EF; but as BC is to EF, fo is EF to BG. There

M 4

fore

t11.5. fore as A B is to D E, fo is EF to BG; confequently, the Sides that are about the equal Angles of the Triangles ABG, DEF, are reciprocal: But those Triangles that have one Angle of the one, equal to one Angle of the other; and the Sides about the #15 of this. equal Angles reciprocal, are equal. Therefore the Triangle ABG, is equal to the Triangle DEF; and becaufe BC is to EF, as EF is to BG, and if three *Def. 10. Right Lines be proportional, the firft has a duplicate Proportion to the third, of what it has to the fecond. BC to BG fhall have a duplicate Proportion of that which BC has to EF; but the Triangle ABG is equal to the Triangle DEF. Therefore the Triangle ABC, to the Triangle DEF, fhall be in the duplicate Proportion of that which the Side BC has to the Side EF. Wherefore fimilar Triangles are in the duplicate Proportion of their homologous Sides; which was to be demonftrated.

Croll. From hence it is manifeft, if three Right Lines be proportional, then as the firft is to the third, fo is a Triangle made upon the first to a fimilar, and fimilarly defcribed Triangle upon the fecond: Becaufe it has been proved, as CB is to BG, fo is the Triangle ABC to the Triangle ABG, that is, to the Triangle DEF; which was to be demonftrated.

PROPOSITION XX.

THEOREM.

Similar Polygons are divided into fimilar Triangles, equal in Number, and homologous to the Wholes; and Polygon to Polygon, is in the duplicate Proportion of that which one homologous Side has to the other.

LFT ABCDE, FGHKL, be fimilar Polygons;

and let the Side A B be homologous to the Side FG. I fay, the Polygons ABCDE, FGHKL, are divided into equal Numbers of fimilar Triangles, and homologous to the Wholes; and the Polygon ABCDE, to the Polygon FGHKL, is in the duplicate Proportion of that which the Side AB has to the Side FG.

For let BE, EC, GL, LH, be joined.

Then

*

Then because the Polygon ABCDE is fimilar to the Polygon FGHKL, the Angle BAE is equal to the Angle GFL; and BA is to AE as GF is to FL. Now fince A BE, FGL, are two Triangles, having one Angle of the one equal to one Angle of the other, and the Sides about the equal Angles proportional; the Triangle ABE will be equiangular * 6 of this. to the Triangle F GL, and fo alfo fimilar to it. Therefore the Angle A BE, is equal to the Angle FGL; but the whole Angle ABC is equal to the whole + Def. 1 Angle F GH, because of the Similiarity of the Poly- of this. gons. Therefore the remaining Angle EBC is equal to the remaining Angle LGH: And fince (by the Similarity of the Triangles ABE, FGL) as EB is to EA, fo is LG to GF; And fince alfo (by the Similarity of the Polygons) AB is to BC, as FG is to GH; it shall be by Equality of Proportion, as ‡ 21.5. EB is to BC, fo is LG to GH, that is, the Sides about the equal Angles EBC, LGH, are proportional. Wherefore the Triangle EBC is equiangular to the Triangle LGH; and confequently alfo fimilar to it. For the fame Reafon, the Triangle ECD, is likewife fimilar to the Triangle LHK; therefore the fimilar Polygons ABCDE, FGHK L, are divided into equal Numbers of fimilar Triangles.

I fay, they are alfo homologous to the Wholes, that is, that the Triangles are proportional; and the Antecedents are ABE, EBC, ECD, and their Confequents FGL, LGH, LHK. And the Polygon ABCDE, to the Polygon FGHKL, is in the duplicate Proportion of an homologous Side of the one, to an homologous Side of the other, that is, AB to F C.

For because the Triangle ABE is fimilar to the Triangle FGL, the Triangle ABE, fhall be * to the 19 of Triangle FGL, in the duplicate Proportion of BE this. to GL: For the fame Reafon, the Triangle BEC, to the Triangle GLH, is * in a duplicate Proportion of BE to GL; Therefore the Triangle ABE is † to † 11.5. the Triangle FGL, as the Triangle BEC is to the Triangle GLH. Again, because the Triangle EBC is fimilar to the Triangle LGH; the Triangle EBC to the Triangle LGH, fhall be in the duplicate Proportion of the Right Line CE to the Right Line HL; and fo likewife the Triangle ECD to the Tri

angle

12.5.

A

F

BG X

angle LHK, fhall be in the duplicate Proportion of CE to HL. Therefore the Triangle BEC is to the Triangle LGH, as the Triangle CED is to the Triangle LHK. But it has been proved, that the Triangle E BC is to the Triangle LG H, as the Triangle ABE is to the Triangle FGL: Therefore as the Triangle ABE is to the Triangle FGL, fo is the Triangle BEC to the Triangle GHL; and fo is the Triangle ECD to the Triangle LHK. But as one of the Antecedents is to one of the Confequents, fo are all the Antecedents to all the Confequents. Wherefore as the Triangle ABE is to the Triangle FGL, fo is the Polygon ABCDE to the Polygon FGHKL: But the Triangle ABE to the Triangle FGL, is in the duplicate Proportion of the homologous Side AB to the homologous Side FG; for fimilar Triangles are in the duplicate Proportion of the homologous Sides. Wherefore the Polygon ABCDE, to the Polygon FGHK L, is in the duplicate Proportion of the homologous Side AB to the homologous Side FG. Therefore fimilar Polygons are divided into fimilar Triangles, equal in Number, and homologous to the Whales; and Polygon to Polygon, is in the duplicate Proportion of that which one homologous Side bas to the other; which was to be demonftrated.

It may be demonftrated after the fame Manner that fimilar quadrilateral Figures are to each other in the duplicate Proportion of their homologous Sides; and this has been already proved in Triangles.

Coroll. 1. Therefore univerfally fimilar Right-lin'd
Figures, are to one another in the duplicate Pro-
portion of their homologous Sides; and if X be
taken a third Proportional to AB and FG, then AB
will have to Xa duplicate Proportion of that which
AB has to FG; and a Polygon to a Polygon, and
a quadrilateral Figure to a quadrilateral Figure, will
be in the duplicate Proportion of that which one
homologous Side has to the other; that is, AB to
FG; but this has been proved in Triangles.
2. Therefore univerfally it is manifeft, if three Right-
Lines be proportional, as the first is to the third, fo
is a Figure defcribed upon the firft, to a fimilar and

fimilarly