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quentiy, 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 Sudes about the equal
Angles reciprocal, are I equal. Therefore the Triangle I 15 of this
ABG is equal to the Triangle DEF; and because
B C is to EF, as E F is to BG; and if three Right
Lines be proportional, the first has * a duplicate Pro-Def.10.5.
portion to the third, of what it has to the second; BC
to B G Tall have a duplicate Proportion of that which
BC has to EF; and as B C is to B G, fo is the Tri-
angle A B C to the Triangle A B G; whence the
Triangle A B C bears to the Triangle A B G, a dupli-
cate Proportion to what B C doth to EF, but the
Triangle ABC is equal to the Triangle DEF.
Therefore the Triangle A B C, to the Triangle DEF,
shall be in the duplicate Proportion of that which the
Side B C has to the Side E F. Wherefore, fimilar
Triangles are in the duplicate Proportion of their homolo-
gous Sides ; which was to be demonstrated.
Coroll. From hence it is manifeft, if three Right Lines

be proportional; then, as the firft is to the third, so
is a Triangle made upon the first, to a similar and
fimilarly described Triangle upon the second ; be-
cause it has been proved, that as C B is to B G, fois
the Triangle ABC to the Triangle ABG ; that is, to
the Triangle D E F; which was to be demonstrated.
PROPOSITION XX.

THEOREM.
Similar Polygons are divided into similar Trian-

gles, equal in Number, and bomologous to the
Wholes; and, Polygon to Polygon, is in the dy-
plicate Proportion of that which one bomolo.

gous Side bas to the otber. LET ABCDEFGHEL, be similar Poly.

gons, and let the Side A B be homologous to the Side FG. I say, the Polygons ABCDE, FGHKL, are divided into equal Numbers of similar Triangles, and homologous to the Wholes; and the Polygon A B C D E, to the Polygon FGHKL, is

!

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

För let B E, E C, GL, L H, be joined.

Then, because the Polygon ABCDE is fimilar to the Polygon FGHK L, the Angle B A E is equal to the Angle G FL; and B A is to A E, as GF is to FL. Now, fince A BE, FGL, are two Triangles having one Angle of the one equal to one. An

gle of the other, and the sides about the equal Angles 16 of this proportional; the Triangle A B E will be & equian

gular to the Triangle F G L, and also fimilar to it.

Therefore the Angle A B E is equal to the Angle : Defir of FGL ; but the whole Angle A B C is * equal to the

whole Angle F GH, because of the Similarity of the Polygons; therefore the remaining Angle É BC is equal to the remaining Angle LGH: (And since by the Similarity of the Triangles ABE, FGL), as E B is to B A, fo is L G to GF; and fince, also, by

the Similarity offthe Polygons, A B is to B Ç as FG, 22.5 is to GH; it shall be + by Equality of Proportion,

as E B is to B C, fo is LG to GH; that is, the sides about the equal Angles E BC, LGH, are proportional. Wherefore the Triangle E B C is equiangular to the Triangles LGH, and, consequently, allo similar to it. For the same Reason, the Triangle ECD is likewise fimilar to the Triangle LHK; therefore the similar Polygons ABCDE, FGHKL, are divided into equal Numbers of similar Triangles.

I say, they are also homologous to the wholes; that is, that the Triangles ate proporcional, and the Antecedents are A BE, EBC, ECD; and their Consequents F GL, LGH, LHK: And the Polygon ABCDE, to the Polygon F GHKL, is in the duplicate Proportion of an homologous Side of the one, to an homologous Side of the other; that is, as AB to F G.

For, because the Triangle A B E is similar to the * 19 of tbis. Triangle F GL, the Triangle A B E shall be * to the

Triangle F GL, in the duplicate Proportion of B E to
GL: For the same Reason, the Triangle B E C, to

- the Triangle G LH, is * in a duplicate Proportion of + 11.5 BE to GL: Therefore the Triangle A BE is + to

the Triangle FGL, as the Triangle BEC is to the Triangle G LH. Again, because the Triangle E B C,

is

is fimilar to the Triangle LGH, the Triangle EBC, to the Triangle L GH, shall be in the duplicate Proportion of the Right Line C E, to the Right Line HL; and so, likewise, the Triangle ECD to the Triangle LHK, shall be in the duplicate Proportion of CE to HL. Therefore the Triangle B E C is to the Triangle L GH, as the Triangle CE D is to the Triangle LHK. But it has been proved, that the Triangle E B C is to the Triangle LGH, as the Triangle A BE is to the Triangle F GL. Therefore, as the Triangle A B E is to the Triangle F GL, lo is the Triangle BIE C to the Triangle Ğ HL; and to is the Triangle E C D to the Triangle LHK. But as one of the Antecedents is to one of the Consequents, fo are all the Antecedents to all the Confequents. I 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 TriangleF GL, * is in the duplicate Proportion of the 19 of this. homologous Side A B to the homologous Side FG; for (imilar Triangles are in the duplicate Proportion of the homologous Sides. Wherefore, the Polygon ABC D E, to the Polygon F GHKL, '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 demonstrated.

It may be demonstrated, after the same manner,
that similar 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, universally, similarly Right-lined A

Figures are to one another in the duplicate Propor-
tion of their homologous Sides; and if X be taken
à third Proportional to A B and FG, then A B will
bave to X a 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

BGX
homologous Side has to the other; that is, A B to
FG; but this has been proved in Triangles.

2. There

F

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