*

quently, the Sides that are about the equal Angles of the Triangles ABG, DEF, are reciprocal: But thole Triangles that have one Angle of the one equal to one Angle of the other, and the Sides about the equal Angles reciprocal, are ‡ equal. Therefore the Triangle I 15 of this ABG is equal to the Triangle DEF; and because BC is to EF, as E F is to BG; and if three Right Lines be proportional, the firft has a duplicate Pro-Def. 10.5. portion to the third, of what it has to the fecond; BC to B G fhall have a duplicate Proportion of that which BC has to EF; and as B C is to B G, fo is the Triangle ABC to the Triangle A BG; whence the Triangle ABC bears to the Triangle A B G, a duplicate Proportion to what BC doth to EF; but the Triangle ABC is equal to the Triangle D E F. Therefore the Triangle A B C, 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.

Coroll. 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 firft, to a fimilar and
fimilarly defcribed Triangle upon the second; be-
cause it has been proved, that as CB is to BG, fo is
the Triangle ABC to the Triangle ABG; that is, to
the Triangle D E F; which was to be demonftrated.

PROPOSITION XX.
THEOREM.

Similar Polygons are divided into fimilar Trian-
gles, equal in Number, and homologous to the
Wholes; and, Polygon to Polygon, is in the du-
plicate Proportion of that which one bomole-
gous Side bas to the other.

LE

ET ABCDE, F G HK L, be fimilar Polygons, and let the Side A B be homologous to the Side F G. I fay, the Polygons A B C D E, FGHK L, are divided into equal Numbers of fimilar Triangles, and homologous to the Wholes; and the Polygon ABCDE, to the Polygon F G H KL, is

in

* Def. 1 of bis.

in the duplicate Proportion of that which the Side A B has to the Side F G.

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

**

Then, because the Polygon A B C D E is fimilar to the Polygon FGH KL, the Angle B A E is equal to the Angle GFL; and B A is to A E, as GF is to FL. Now, fince A B E, FGL, are two Triangles having one Angle of the one equal to one Angle of the other, and the Sides about the equal Angles 16 of this. proportional; the Triangle A B E will be equiangular to the Triangle FG L, and 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 Angle F G H, because of the Similarity of the Polygons; therefore the remaining Angle EBC is equal to the remaining Angle L G H: (And fince by the Similarity of the Triangles ABE, FGL), as EB is to B A, fo is L G to GF; and fince, alfo, by the Similarity offthe Polygons, A B is to BC as F G, is to GH; it fhall be + by Equality of Proportion, as E B is to BC, fo is LG to GH; that is, the Sides about the equal Angles E B C, LG H, are proportional. Wherefore the Triangle E BC is equiangular to the Triangles LGH, and, confequently, alfo fimilar to it. For the fame Reason, the Triangle ECD is likewife fimilar to the Triangle LHK; therefore the fimilar Polygons ABCDE, F G H K L, are divided into equal Numbers of fimilar Triangles.

† 22. 5.

I fay, they are alfo homologous to the wholes; that is, that the Triangles are proportional, and the Antecedents are A BE, EBC, ECD; and their Confequents FGL, LGH, LHK: And the Polygon ABCDE, to the Polygon F G H KL, is in the duplicate Proportion of an homologous Side of the one, to an homologous Side of the other; that is, as A B to FG.

For, because the Triangle A B E is fimilar to the 19 of this. Triangle FGL, the Triangle A B E fhall be to the Triangle FGL, in the duplicate Proportion of BE to GL: For the fame Reafon, the Triangle B E C, to the Triangle GL H, is in a duplicate Proportion of BE to GL: Therefore the Triangle ABE is + to the Triangle FGL, as the Triangle BEC is to the Triangle GLH. Again, because the Triangle E B C,

+ 11.5.

is fimilar to the Triangle LG H, the Triangle EBC, to the Triangle LG H, fhall be in the duplicate Proportion of the Right Line C E, to the Right Line HL; and fo, likewife, the Triangle E CD to the Triangle LHK, fhall be in the duplicate Proportion of C E to HL. Therefore the Triangle BEC is to the Triangle LG H, as the Triangle CED is to the Triangle L HK. But it has been proved, that the Triangle EBC is to the Triangle LGH, as the Triangle A BE is to the Triangle FGL. Therefore, as the Triangle ABE is to the Triangle F G L, fo is the Triangle BIE C to the Triangle G HL; and fo is the Triangle ECD to the Triangle LH K. But as one of the Antecedents is to one of the Confequents, fo are tall the Antecedents to all the Confequents. 12. 5. Wherefore, as the Triangle A B E is to the Triangle FG L, fo is the Polygon A B C D E to the Polygon FGHKL: But the Triangle A B E, to the Triangle F GL, is in the duplicate Proportion of the 19 of this homologous Side A B to the homologous Side F G; for fimilar Triangles are in the duplicate Proportion of the homologous Sides. Wherefore, the Polygon ABCDE, to the Polygon F G H K L, is in the duplicate Proportion of the homologous Side A B to the homologous Side FG. Therefore, fimilar 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; which was to be demonftrated.

It may be demonstrated, after the fame manner, that fimilar quadrilateral Figures are to each other in the duplicaie Proportion of their homologous Sides; and this has been already proved in Triangles. Coroll. 1. Therefore, univerfally, fimilarly Right-lined A Figures are to one another in the duplicate Proportion of their homologous Sides; and if X be taken a third Proportional to A B and FG, then A B will have to X a duplicate Proportion of that which AB has to F G; 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. There

BGX