Euclid Book 13 Proposition 3
Statement
If a line segment () is cut in the golden ratio (), then the square of the sum of the shorter segment and the half of the longer segment () is five times the square of the half (5) of the longer segment.
AB
AB
AC
AC
CB
2
(BC+CD)
2
AD
Computational Explanation
GeometricScene{{A.,B.,C.,D.,E.,F.,H.,K.,L.,M.},{a.,b.}},Line[{A.,D.,C.,B.}],EuclideanDistance[A.,C.]a.,EuclideanDistance[C.,B.]b.,,D.Midpoint[Line[{A.,C.}]],Line[{{A.,H.,F.},{F.,L.,E.},{L.,K.,D.},{H.,K.,M.},{B.,M.,E.}}],GeometricAssertion[{Polygon[{A.,B.,E.,F.}],Style[Polygon[{H.,K.,L.,F.}],Pink]},"Regular"],Style[Polygon[{B.,D.,K.,M.}],Blue],Area[Polygon[{B.,D.,K.,M.}]]5Area[Polygon[{H.,K.,L.,F.}]],5,5
EuclideanDistance[A.,B.]
EuclideanDistance[A.,C.]
EuclideanDistance[A.,C.]
EuclideanDistance[C.,B.]
2
(EuclideanDistance[B.,C.]+EuclideanDistance[C.,D.])
2
EuclideanDistance[A.,D.]
2
b.+
a.
2
2
a.
2
Area[Polygon[{B.,D.,K.,M.}]]5Area[Polygon[{H.,K.,L.,F.}]] |
 |
Explanations
For let any straight line be cut in extreme and mean ratio at the point , let be the greater segment, and let be bisected at ; I say that the square on is five times the square on .
For let the square be described on , and let the figure be drawn double.
Since is double of , therefore the square on is quadruple of the square on , that is, is quadruple of .
And, since the rectangle, is equal to the square on , and is the rectangle , , therefore is equal to .
But is quadruple of ; therefore is also quadruple of .
Again, since is equal to , is also equal to .
Hence the square is also equal to the square .
Therefore is equal to , that is, to ; hence is also equal to .
But is equal to ; therefore is also equal to .
Let be added to each; therefore the gnomon is equal to .
But was proved quadruple of ; therefore the gnomon is also quadruple of the square .
Therefore the gnomon and the square are five times .
But the gnomon and the square are the square .
And is the square on , and the square on .
Therefore the square on is five times the square on .
AB
C
AC
AC
D
BD
DC
For let the square
AE
AB
Since
AC
DC
AC
DC
RS
FG
And, since the rectangle
AB
BC
AC
CE
AB
BC
CE
RS
But
RS
FG
CE
FG
Again, since
AD
DC
HK
KF
Hence the square
GF
HL
Therefore
GK
KL
MN
NE
MF
FE
But
MF
CG
CG
FE
Let
CN
OPQ
CE
But
CE
GF
OPQ
FG
Therefore the gnomon
OPQ
FG
FG
But the gnomon
OPQ
FG
DN
And
DN
DB
GF
DC
Therefore the square on
DB
DC