Euclid Book 6 Proposition 20
Statement
Similar polygons may be divided into similar triangles by their diagonals, and the corresponding triangles have the same ratio to one another as the polygons have.
Computational Explanation
GeometricScene{{A.,B.,C.,D.,E.,F.,G.,H.,I.,J.},{}},{GeometricAssertion[{Polygon[{A.,B.,C.,D.,E.}],Polygon[{F.,G.,H.,I.,J.}]},"Similar","Convex"],Line[{{C.,A.},{D.,A.},{H.,F.},{I.,F.}}]},GeometricAssertion[{Triangle[{A.,B.,C.}],Triangle[{F.,G.,H.}]},"Similar"],GeometricAssertion[{Triangle[{A.,C.,D.}],Triangle[{F.,H.,I.}]},"Similar"],GeometricAssertion[{Triangle[{A.,D.,E.}],Triangle[{F.,I.,J.}]},"Similar"],
Area[Triangle[{A.,B.,C.}]]
Area[Triangle[{F.,G.,H.}]]
Area[Triangle[{A.,C.,D.}]]
Area[Triangle[{F.,H.,I.}]]
Area[Triangle[{A.,D.,E.}]]
Area[Triangle[{F.,I.,J.}]]
Area[Polygon[{A.,B.,C.,D.,E.}]]
Area[Polygon[{F.,G.,H.,I.,J.}]]
GeometricAssertion[{Triangle[{A.,B.,C.}],Triangle[{F.,G.,H.}]},Similar] |
 |
Explanations
Let , be similar polygons, and let correspond to ; I say that the polygons , are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon has to the polygon a ratio duplicate of that which has to .
Let, , , be joined.
Now, since the polygon is similar to the polygon , the angle is equal to the angle ; and, as is to , so is to .
Since then, are two triangles having one angle equal to one angle and the sides about the equal angles proportional, therefore the triangle is equiangular with the triangle ;so that it is also similar; therefore the angle is equal to the angle .
But the whole angle is also equal to the whole angle because of the similarity of the polygons; therefore the remaining angle is equal to the angle .
And, since, because of the similarity of the triangles,, as is to , so is to , and moreover also, because of the similarity of the polygons, as is to , so is to , therefore, ex aequali, as is to , so is to ;that is, the sides about the equal angles , are proportional; therefore the triangle is equiangular with the triangle ,so that the triangle is also similar to the triangle .
For the same reason the triangle is also similar to the triangle .
Therefore the similar polygons, have been divided into similar triangles, and into triangles equal in multitude.
I say that they are also in the same ratio as the wholes, that is, in such manner that the triangles are proportional, and, , are antecedents, while , , are their consequents, and that the polygon has to the polygon a ratio duplicate of that which the corresponding side has to the corresponding side, that is to . For let , be joined.
Then since, because of the similarity of the polygons,the angle is equal to the angle , and, as is to , so is to , the triangle is equiangular with the triangle ;therefore the angle is equal to the angle , and the angle to the angle .
And, since the angle is equal to the angle , and the angle is also equal to the angle , therefore the remaining angle is also equal to the remaining angle ;therefore the triangle is equiangular with the triangle .
Similarly we can prove that the triangle is also equiangular with the triangle . Therefore, proportionally, as is to , so is to ,and, as is to , so is to ; so that, in addition, ex aequali, as is to , so is to .
But, as is to , so is the triangle to , and to ; for they are to one another as their bases.
Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; therefore, as the triangle is to , so is to .
But, as is to , so is to ; therefore also, as is to , so is the triangle to the triangle .
For the same reason also, as is to , so is the triangle to the triangle .
And, as is to , so is to ; therefore also, as the triangle is to the triangle , so is the triangle to the triangle ; and, alternately, as the triangle is to the triangle ,so is the triangle to the triangle .
Similarly we can prove, if, be joined, that, as the triangle is to the triangle , so also is the triangle to the triangle .
And since, as the triangle is to the triangle ,so is to , and further to , therefore also, as one of the antecedents is to one of the consequents so are all the antecedents to all the consequents; therefore, as the triangle is to the triangle , so is the polygon to the polygon .
But the triangle has to the triangle a ratio duplicate of that which the corresponding side has to the corresponding side ; for similar triangles are in the duplicate ratio of the corresponding sides.
Therefore the polygon also has to the polygon a ratio duplicate of that which the corresponding side has to the corresponding side .
ABCDE
FGHKL
AB
FG
ABCDE
FGHKL
ABCDE
FGHKL
AB
FG
Let
BE
EC
GL
LH
Now, since the polygon
ABCDE
FGHKL
BAE
GFL
BA
AE
GF
FL
Since then
ABE
FGL
ABE
FGL
ABE
FGL
But the whole angle
ABC
FGH
EBC
LGH
And, since, because of the similarity of the triangles
ABE
FGL
EB
BA
LG
GF
AB
BC
FG
GH
EB
BC
LG
GH
EBC
LGH
EBC
LGH
EBC
LGH
For the same reason the triangle
ECD
LHK
Therefore the similar polygons
ABCDE
FGHKL
I say that they are also in the same ratio as the wholes, that is, in such manner that the triangles are proportional, and
ABE
EBC
ECD
FGL
LGH
LHK
ABCDE
FGHKL
AB
FG
AC
FH
Then since, because of the similarity of the polygons,the angle
ABC
FGH
AB
BC
FG
GH
ABC
FGH
BAC
GFH
BCA
GHF
And, since the angle
BAM
GFN
ABM
FGN
AMB
FNG
ABM
FGN
Similarly we can prove that the triangle
BMC
GNH
AM
MB
FN
NG
BM
MC
GN
NH
AM
MC
FN
NH
But, as
AM
MC
ABM
MBC
AME
EMC
Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; therefore, as the triangle
AMB
BMC
ABE
CBE
But, as
AMB
BMC
AM
MC
AM
MC
ABE
EBC
For the same reason also, as
FN
NH
FGL
GLH
And, as
AM
MC
FN
NH
ABE
BEC
FGL
GLH
ABE
FGL
BEC
GLH
Similarly we can prove, if
BD
GK
BEC
LGH
ECD
LHK
And since, as the triangle
ABE
FGL
EBC
LGH
ECD
LHK
ABE
FGL
ABCDE
FGHKL
But the triangle
ABE
FGL
AB
FG
Therefore the polygon
ABCDE
FGHKL
AB
FG