In geometry, Routh's theorem determines the ratio of areas between a given triangle and a triangle formed by the pairwise intersections of three cevians. The theorem states that if in triangle
A
B
C
{\displaystyle ABC}
points
D
{\displaystyle D}
,
E
{\displaystyle E}
, and
F
{\displaystyle F}
lie on segments
B
C
{\displaystyle BC}
,
C
A
{\displaystyle CA}
, and
A
B
{\displaystyle AB}
, then writing
C
D
B
D
=
x
{\displaystyle {\tfrac {CD}{BD}}=x}
,
A
E
C
E
=
y
{\displaystyle {\tfrac {AE}{CE}}=y}
, and
B
F
A
F
=
z
{\displaystyle {\tfrac {BF}{AF}}=z}
, the signed area of the triangle formed by the cevians
A
D
{\displaystyle AD}
,
B
E
{\displaystyle BE}
, and
C
F
{\displaystyle CF}
is
S
A
B
C
⋅
(
x
y
z
−
1
)
2
(
x
y
+
y
+
1
)
(
y
z
+
z
+
1
)
(
z
x
+
x
+
1
)
,
{\displaystyle S_{ABC}\cdot {\frac {(xyz-1)^{2}}{(xy+y+1)(yz+z+1)(zx+x+1)}},}
where
S
A
B
C
{\displaystyle S_{ABC}}
is the area of the triangle
A
B
C
{\displaystyle ABC}
.
This theorem was given by Edward John Routh on page 82 of his Treatise on Analytical Statics with Numerous Examples in 1896. The particular case
x
=
y
=
z
=
2
{\displaystyle x=y=z=2}
has become popularized as the one-seventh area triangle. The
x
=
y
=
z
=
1
{\displaystyle x=y=z=1}
case implies that the three medians are concurrent (through the centroid).
Proof
Suppose that the area of triangle
A
B
C
{\displaystyle ABC}
is 1. For triangle
A
B
D
{\displaystyle ABD}
and line
F
R
C
{\displaystyle FRC}
, Menelaus's theorem implies
A
F
F
B
×
B
C
C
D
×
D
R
R
A
=
−
1
{\displaystyle {\frac {AF}{FB}}\times {\frac {BC}{CD}}\times {\frac {DR}{RA}}=-1}