Euclid Book 13 Proposition 4
Statement
If a line segment () is cut in the golden ratio (), then the sum of the squares of the whole and of the shorter segment (+) is triple the square of the longer segment (3).
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2
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Computational Explanation
GeometricScene{{A.,B.,C.,D.,E.,F.,G.,H.,K.},{a.,b.}},Line[{A.,C.,B.}],EuclideanDistance[A.,C.]a.,EuclideanDistance[C.,B.]b.,,Line[{{A.,H.,D.},{D.,G.,E.},{E.,K.,B.},{H.,F.,K.},{C.,F.,G.}}],GeometricAssertion[{Polygon[{A.,B.,E.,D.}],Style[Polygon[{C.,B.,K.,F.}],Pink]},"Regular"],Style[Polygon[{D.,H.,F.,G.}],Blue],Area[Polygon[{A.,B.,E.,D.}]]+Area[Polygon[{C.,B.,K.,F.}]]3Area[Polygon[{D.,H.,F.,G.}]],+3,+3
EuclideanDistance[A.,B.]
EuclideanDistance[A.,C.]
EuclideanDistance[A.,C.]
EuclideanDistance[C.,B.]
2
EuclideanDistance[A.,B.]
2
EuclideanDistance[B.,C.]
2
EuclideanDistance[A.,C.]
2
(a.+b.)
2
b.
2
a.
Area[Polygon[{A.,B.,E.,D.}]]+Area[Polygon[{C.,B.,K.,F.}]]3Area[Polygon[{D.,H.,F.,G.}]] |
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Explanations
Let be a straight line, let it be cut in extreme and mean ratio at , and let be the greater segment; I say that the squares on , are triple of the square on .
For let the square be described on , and let the figure be drawn.
Since then has been cut in extreme and mean ratio at , and is the greater segment, therefore the rectangle , is equal to the square on .
And is the rectangle , , and the square on ; therefore is equal to .
And, since is equal to , let be added to each; therefore the whole is equal to the whole ; therefore , are double of .
But, are the gnomon and the square ; therefore the gnomon and the square are double of .
But, further, was also proved equal to ; therefore the gnomon and the squares , are triple of the square .
And the gnomon and the squares , are the whole square and , which are the squares on , , while is the square on .
Therefore the squares on, are triple of the square on .
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CA
For let the square
ADEB
AB
Since then
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AC
And
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HG
AC
AK
HG
And, since
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CK
AK
CE
AK
CE
AK
But
AK
CE
LMN
CK
LMN
CK
AK
But, further,
AK
HG
LMN
CK
HG
HG
And the gnomon
LMN
CK
HG
AE
CK
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HG
AC
Therefore the squares on
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AC