What I want to ask is possibly a rather classical result. I remember I read a book which says that if $f$ is a bijection (a.k.a. collineation of an affine space) of $\mathbb{R}^2$ sending collinear points to collinear points, then $f$ will preserve the ratio between points, that is, $$\overrightarrow{AB}=k\overrightarrow{BC}\qquad\Rightarrow\qquad\overrightarrow{f(A)f(B)}=k\overrightarrow{f(B)f(C)}$$ which (should?) be an affine map by the end.
I am sorry I cannot recall the book. So I am wondering if anyone knows a slick proof of this? I could only manage to show that $f$ preserves the middle points :(
Notation : $\Delta xyz$ is a 2-dimensional triangle. And $[xy]$ is a line segment between $x,\ y$
Def : $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ is collinear if for $x,\ y,\ z$ that are distinct points in a line, then $f(x),\ f(y),\ f(z)$ are in a line.
EXE : If $f$ is collinear and bijective, then $f$ sends a line into a line.
EXE : If $f$ is collinear and bijective, then so is $f^{-1}$
Proof : For a line $l$, assume that $f(l)\subset L$ where $L$ is a line.
Hence $f^{-1}(L)$ contains $l$. Assume that $x\in f^{-1}(L)-l$.
Consider another line $l'$ through $x$ s.t. $l'$ is not parallel to $l$.
Hence $f(l')$ is in $L$. This implies that if $l''$ is a parallel line wrt $l$ and $l''$ passes through $x$, then $$ f(X)\subset L,\ X:=\mathbb{R}^2-(l'' -\{x\}) $$
For $y\in l''$, there is $a,\ b\in X$ s.t. $a,\ b,\ y$ are collinear. Hence $f(\mathbb{R}^2)$ is in $L$, which is a contradiction.
Hence $f:l\rightarrow L$ is a bijection.
Coro : In the above proof, $l,\ l'$ are parallel lines iff their images are also parallel.
Coro : If $x_i$ is vertexes in parallelogram, then so is $f(x_i)$. In further, here $f$ preserves order of $x_i$.
Coro : Hence if $z$ is a mid point in $[xy]$, then $f(z)$ is still a mid point in $[f(x)f(y)]$. This implies that $$ f(tx+(1-t)y)=tf(x)+(1-t)f(y),\ x\neq y,\ t\in \mathbb{R} $$