(

*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

*l*_{AQ},

*b* the

line passing

*B'* parallel to

*l*_{BQ},

*c* the line passing

*C'* parallel to

*l*_{CQ}.

- The lines
*a*, *b*, *c* are concurrent at a point *P*.
- The centroid
*G* of triangle *ABC* lies on *l*_{PQ} and
2(*P* - *G*) = *G* - *Q*.

**Diagram**
P C
| \ a /\
b| \ / \
| /\ \
| / \ \
| / \ \
B'./ \. A'
/ \
/ \
/ \
/ \
/ \
/ C' \
A-----------------.-----------------B
\ |
\ |
\ |
\ |
\ |
\ |
\ |
\ |
\ |
\ |
\ |
\|
Q

(lines *l*_{CQ} 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
*δ*_{G,-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

*l*_{AQ},

*l*_{BQ}, and

*l*_{CQ} into

*a*,

*b*, and

*c* since they are parallel.
Hence

*δ*_{G,-1/2}(

*Q*) must lie on all three lines

*a*,

*b*, and

*c*, which is the

concurrent point

*P*.
Since

*δ*_{G,-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.