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Euclid Book 13
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Euclid Book 13 Proposition 5
Statement
I
f
a
l
i
n
e
s
e
g
m
e
n
t
(
A
B
)
i
s
c
u
t
i
n
t
h
e
g
o
l
d
e
n
r
a
t
i
o
(
A
B
A
C
A
C
C
B
)
,
a
n
d
a
s
e
g
m
e
n
t
(
A
D
)
e
q
u
a
l
t
o
t
h
e
l
o
n
g
e
r
s
e
g
m
e
n
t
(
A
C
)
i
s
a
d
d
e
d
t
o
i
t
,
t
h
e
n
t
h
e
w
h
o
l
e
l
i
n
e
s
e
g
m
e
n
t
(
D
B
)
f
o
r
m
s
t
h
e
g
o
l
d
e
n
r
a
t
i
o
w
i
t
h
t
h
e
o
r
i
g
i
n
a
l
(
D
B
A
B
A
B
A
D
)
.
Computational Explanation
G
e
o
m
e
t
r
i
c
S
c
e
n
e
{
{
A
.
,
B
.
,
C
.
,
D
.
}
,
{
a
.
,
b
.
}
}
,
L
i
n
e
[
{
A
.
,
C
.
,
B
.
}
]
,
E
u
c
l
i
d
e
a
n
D
i
s
t
a
n
c
e
[
A
.
,
C
.
]
a
.
,
E
u
c
l
i
d
e
a
n
D
i
s
t
a
n
c
e
[
C
.
,
B
.
]
b
.
,
E
u
c
l
i
d
e
a
n
D
i
s
t
a
n
c
e
[
A
.
,
B
.
]
E
u
c
l
i
d
e
a
n
D
i
s
t
a
n
c
e
[
A
.
,
C
.
]
E
u
c
l
i
d
e
a
n
D
i
s
t
a
n
c
e
[
A
.
,
C
.
]
E
u
c
l
i
d
e
a
n
D
i
s
t
a
n
c
e
[
C
.
,
B
.
]
,
L
i
n
e
[
{
D
.
,
A
.
,
C
.
}
]
,
E
u
c
l
i
d
e
a
n
D
i
s
t
a
n
c
e
[
D
.
,
A
.
]
E
u
c
l
i
d
e
a
n
D
i
s
t
a
n
c
e
[
A
.
,
C
.
]
,
E
u
c
l
i
d
e
a
n
D
i
s
t
a
n
c
e
[
D
.
,
B
.
]
E
u
c
l
i
d
e
a
n
D
i
s
t
a
n
c
e
[
A
.
,
B
.
]
E
u
c
l
i
d
e
a
n
D
i
s
t
a
n
c
e
[
A
.
,
B
.
]
E
u
c
l
i
d
e
a
n
D
i
s
t
a
n
c
e
[
A
.
,
D
.
]
,
2
a
.
+
b
.
a
.
+
b
.
a
.
+
b
.
a
.
E
u
c
l
i
d
e
a
n
D
i
s
t
a
n
c
e
[
D
.
,
B
.
]
E
u
c
l
i
d
e
a
n
D
i
s
t
a
n
c
e
[
A
.
,
B
.
]
E
u
c
l
i
d
e
a
n
D
i
s
t
a
n
c
e
[
A
.
,
B
.
]
E
u
c
l
i
d
e
a
n
D
i
s
t
a
n
c
e
[
A
.
,
D
.
]
Explanations
For let the straight line
A
B
be cut in extreme and mean ratio at the point
C
, let
A
C
be the greater segment, and let
A
D
be equal to
A
C
.
I say that the straight line
D
B
has been cut in extreme and mean ratio at
A
, and the original straight line
A
B
is the greater segment.
For let the square
A
E
be described on
A
B
, and let the figure be drawn.
Since
A
B
has been cut in extreme and mean ratio at
C
, therefore the rectangle
A
B
,
B
C
is equal to the square on
A
C
.
[
V
I
.
D
e
f
.
3
]
[
V
I
.
1
7
]
And
C
E
is the rectangle
A
B
,
B
C
, and
C
H
the square on
A
C
; therefore
C
E
is equal to
H
C
. But
H
E
is equal to
C
E
, and
D
H
is equal to
H
C
; therefore
D
H
is also equal to
H
E
. Therefore the whole
D
K
is equal to the whole
A
E
.
And
D
K
is the rectangle
B
D
,
D
A
, for
A
D
is equal to
D
L
; and
A
E
is the square on
A
B
; therefore the rectangle
B
D
,
D
A
is equal to the square on
A
B
.
Therefore, as
D
B
is to
B
A
, so is
B
A
to
A
D
.
[
V
I
.
1
7
]
And
D
B
is greater than
B
A
; therefore
B
A
is also greater than
A
D
.
[
V
.
1
4
]
Therefore
D
B
has been cut in extreme and mean ratio at
A
, and
A
B
is the greater segment.
Classes
Euclid's Elements
Theorems
Geometric Algebra
EuclidBook13
Related Theorems
EuclidBook13Proposition1
EuclidBook13Proposition2
EuclidBook13Proposition3
EuclidBook13Proposition4