Euclid Book 13 Proposition 1
Statement
If a line segment () is cut in the golden ratio (), then the square of the longer segment added to the half of the whole () is five times the square of the half (5).
AB
AB
AC
AC
CB
2
(AC+AD)
2
AD
Computational Explanation
GeometricScene{{A.,B.,C.,D.,F.,H.,K.,L.},{a.,b.}},Line[{A.,C.,B.}],EuclideanDistance[A.,C.]a.,EuclideanDistance[C.,B.]b.,,Line[{C.,A.,D.}],EuclideanDistance[A.,D.]EuclideanDistance[A.,B.],GeometricAssertion[{Polygon[{D.,C.,F.,L.}],Polygon[{D.,A.,H.,K.}]},"Regular"],Area[Polygon[{D.,C.,F.,L.}]]5Area[Polygon[{D.,A.,H.,K.}]],5,5
EuclideanDistance[A.,B.]
EuclideanDistance[A.,C.]
EuclideanDistance[A.,C.]
EuclideanDistance[C.,B.]
1
2
a.+b.
2
2
(EuclideanDistance[A.,C.]+EuclideanDistance[A.,D.])
2
EuclideanDistance[A.,D.]
2
a.+
a.+b.
2
2
a.+b.
2
Area[Polygon[{D.,C.,F.,L.}]]5Area[Polygon[{D.,A.,H.,K.}]] |
 |
Explanations
For let the straight line be cut in extreme and mean ratio at the point , and let be the greater segment; let the straight line be produced in a straight line with , and let be made half of ; I say that the square on is five times the square on .
For let the squares, be described on , , and let the figure in be drawn; let be carried through to .
Now, since has been cut in extreme and mean ratio at , therefore the rectangle , is equal to the square on .
And is the rectangle , , and the square on ; therefore is equal to .
And, since is double of , while is equal to , and to , therefore is also double of .
But, as is to , so is to ;therefore is double of . But , are also double of .
Therefore is equal to , .
But was also proved equal to ; therefore the whole square is equal to the gnomon .
And, since is double of , the square on is quadruple of the square on , that is, is quadruple of .
But is equal to the gnomon ; therefore the gnomon is also quadruple of ; therefore the whole is five times .
And is the square on , and the square on ; therefore the square on is five times the square on .
AB
C
AC
AD
CA
AD
AB
CD
AD
For let the squares
AE
DF
AB
DC
DF
FC
G
Now, since
AB
C
AB
BC
AC
And
CE
AB
BC
FH
AC
CE
FH
And, since
BA
AD
BA
KA
AD
AH
KA
AH
But, as
KA
AH
CK
CH
CK
CH
LH
HC
CH
Therefore
KC
LH
HC
But
CE
HF
AE
MNO
And, since
BA
AD
BA
AD
AE
DH
But
AE
MNO
MNO
AP
DF
AP
And
DF
DC
AP
DA
CD
DA