(
Mathematics - Affine Geometry)
The theorem
For an arbitrary
triangle ABC and an arbitrary point
Q,
let
A' be the
midpoint of the side opposite of
A,
B' the midpoint of the side opposite of
B,
C' the
midpoint of the side opposite of
C,
a the line passing
A' parallel to
lAQ,
b the
line passing
B' parallel to
lBQ,
c the line passing
C' parallel to
lCQ.
- The lines a, b, c are concurrent at a point P.
- The centroid G of triangle ABC lies on lPQ and
2(P - G) = G - Q.
Diagram
P C
| \ a /\
b| \ / \
| /\ \
| / \ \
| / \ \
B'./ \. A'
/ \
/ \
/ \
/ \
/ \
/ C' \
A-----------------.-----------------B
\ |
\ |
\ |
\ |
\ |
\ |
\ |
\ |
\ |
\ |
\ |
\|
Q
(lines lCQ and c omitted to avoid clutter)
Preliminary observations
One of the preliminary results needed to prove this
theorem is that a
central dilatation is a
collineation (a
bijection that maps lines into lines). A tedious grinding through
algebra shows this.
Another observation needed is the location of the
centroid G of a
triangle ABC.
G = (1/3)(A + B + C)
(see: centroid)
G = (1/3)A + (2/3)(1/2)(B + C)
This shows that
G lies on the
line that passes through
A (
see: equation for a line in Affine Geometry) and
A' = (1/2)(
B +
C), and also that the distance from
G to
A is 1/2 the distance from
G to
A'.
Proof
The
central dilatation
δG,-1/2 maps
triangle ABC into
triangle A'B'C' due to the location of the
centroid as mentioned.
The
central dilatation also maps
lines
lAQ,
lBQ, and
lCQ into
a,
b, and
c since they are parallel.
Hence
δG,-1/2(
Q) must lie on all three lines
a,
b, and
c, which is the
concurrent point
P.
Since
δG,-1/2(
Q) =
P, it follows that (
P -
G) = (-1/2)(
Q -
G).
Q.E.D.
Result
This theorem shows that the
centroid, the
orthocenter, and the
circumcenter of a
triangle are
collinear and lies on the
Euler line when the line exists because each
altitude is
parallel to the
perpendicular bisector of the same side of a
triangle.