Euclid Book 13 Proposition 2
Statement
If the square of a line segment () is five times the square of a shorter segment (5) in it, then, when the double () of the shorter segment is cut in the golden ratio, the longer segment () is the remaining part of the original line segment.
2
CD
2
AD
AB
AC
Computational Explanation
GeometricScene{{A.,B.,C.,D.,F.,H.,K.,L.,P.},{a.,b.,c.}},Line[{A.,C.,B.}],Line[{C.,A.,D.}],EuclideanDistance[A.,C.]a.,EuclideanDistance[C.,B.]b.,EuclideanDistance[A.,D.]EuclideanDistance[A.,B.],GeometricAssertion[{Polygon[{D.,C.,F.,L.}],Polygon[{D.,A.,H.,K.}]},"Regular"],Area[Polygon[{D.,C.,F.,L.}]],Area[Polygon[{D.,A.,H.,K.}]],5,Line[{A.,P.,B.}],,P.C.,,
1
2
a.+b.
2
2
a.+
a.+b.
2
2
a.+b.
2
2
a.+
a.+b.
2
2
a.+b.
2
EuclideanDistance[A.,B.]
EuclideanDistance[A.,P.]
EuclideanDistance[A.,P.]
EuclideanDistance[P.,B.]
EuclideanDistance[A.,B.]
EuclideanDistance[A.,C.]
EuclideanDistance[A.,C.]
EuclideanDistance[C.,B.]
a.+b.
a.
a.
b.
P.C. |
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Explanations
For let the square on the straight line be five times the square on the segment of it, and let be double of ; I say that, when is cut in extreme and mean ratio, the greater segment is .
Let the squares, be described on , respectively, let the figure in be drawn, and let be drawn through.
Now, since the square on is five times the square on , is five times . Therefore the gnomon is quadruple of .
And, since is double of , therefore the square on is quadruple of the square on , that is, is quadruple of .
But the gnomon was also proved quadruple of ; therefore the gnomon is equal to .
And, since is double of , while is equal to , and to , therefore is also double of .
But, are also double of ; therefore is equal to , .
But the whole gnomon was also proved equal to the whole ; therefore the remainder is equal to .
And is the rectangle , , for is equal to ; and is the square on ; therefore the rectangle , is equal to the square on .
Therefore, as is to , so is to . But is greater than ; therefore is also greater than . Therefore, when the straight line is cut in extreme and mean ratio, is the greater segment.
AB
AC
CD
AC
CD
CB
Let the squares
AF
CG
AB
CD
AF
BE
Now, since the square on
BA
AC
AF
AH
MNO
AH
And, since
DC
CA
DC
CA
CG
AH
But the gnomon
MNO
AH
MNO
CG
And, since
DC
CA
DC
CK
AC
CH
KB
BH
But
LH
HB
HB
KB
LH
HB
But the whole gnomon
MNO
CG
HF
BG
And
BG
CD
DB
CD
DG
HF
CB
CD
DB
CB
Therefore, as
DC
CB
CB
BD
DC
CB
CB
BD
CD
CB