Consider a (real) affine plane $ \mathbb A$ with associated (real) vector space $V$. We can take $\mathbb A=V=\mathbb R^2$ for example, so getting the usual affine plane of elementary Euclidean Affine Geometry. We orient $\mathbb A$ by considering an orientation 2-form on $V$
$$ \omega : V \times V \to \mathbb R, $$
i.e. a non-null alternating bilinear 2-form on $V$.
Let $\;\theta = ABC\hat{}$ be an angle in $\mathbb A$ with vertex $B$, and let $\;a=A-B,\;c=C-B\;$ be the vectors that determine the position of $A$ and $C$, respectively, with respect to $B$. Angle $ABC\hat{}$ is also denoted $(B;a,c)\hat{}$.
We assume that $\theta$ is not a flat or a null angle.
A point $D\in\mathbb A$ is said to be $\it internal$ to $(B;a,c)\hat{}$ if $D$ belongs to the so-called $\it (oriented)\; open\; angular\;sector$
$$ S^°(B;a,c) := \Big\{P\in\mathbb A \;: \;\text{pos}\Big(\omega(a,P-B),\;\omega(P-B,c),\;\omega(c,a)\Big)\geq 2\Big\}, $$
where $\;\text{pos}(\alpha,\beta,\gamma)\;$ is the number of strictly positive values in round brackets.
Since the orientation is not fixed by the problem, we can freely choose it. We can therefore assume throughout the sequel that $\omega(a,c)>0$ (and so $\omega(c,a)<0$ for the alternating property). [Incidentally, this means we are choosing the orientation that makes the angle $\theta$ convex]. Moreover, putting $b=C-A= (C-B)+(B-A)=c-a$, and still using the alternating properties, it follows $\omega(a,b)=\omega(a,a+b)=\omega(a,c)>0$.
Take $D$ internal to $\theta$. Putting $\;d=D-B\;$, necessarely we have:
$$ \omega(a,d)>0 \hskip 4ex \omega(d,c)>0 \hskip4ex \omega(c,a)<0. $$
From these properties, we can deduce the other one:
$$ \omega(d,b) = \omega(-b,d) = \omega(a-c,d) = \omega(a,d) + \omega(d,c) > 0. $$
It is easy to prove that $D\neq B$ and the stright line $BD$ isn't parallel to $AC$. Indeed $B=D$ implies $\omega(a,d)=\omega(a,0)=0$, contradiction; and say $BD$ parallel to $AC$ is equivalent to say $b=\alpha d$ for some $\alpha\in\mathbb R$, from which $\omega(d,a)=\omega(d,c-\alpha d)=\omega(d,c)$, and this is absurd in that the first and the last member have opposite signs. Therefore the lines
$$ B+\langle d \rangle \text{$\;\;$ and$\;\;$} A+\langle b\rangle $$
intersect in a point $E$, i.e. there exists $\lambda,\mu\in\mathbb R$ such that
$$ E = B+\lambda d = A + \mu b, $$
or, equivalently:
$$ a = A-B = \lambda d - \mu b.$$
From that, it follows:
- $\;\;\omega(a,d) =\omega(\lambda d-\mu b,d) =\omega(-\mu b,d) = \mu\omega(d,b) \implies \mu>0.$
- $\;\;\omega(a,b) = \omega(\lambda d - \mu b,b) =\lambda\omega(d,b) \implies \lambda>0.$
- $\;\;\lambda\omega(d,c) = \omega(\lambda d,a+b) = \omega(a+\mu b,a+b) = \omega(a,b)+\omega(\mu b,a) =$ $\hskip4ex=\omega(a,b)-\omega(a,\mu b)=\omega(a,b-\mu b) = (1-\mu)\omega(a,b) \implies \mu<1.$
The double inequality $0<\mu<1$ means that $E$ is internal to the segment $AC$, by definition.
Conversely (finally!), if $E$ is internal to the segment $AC$, i.e. $0<\mu<1$, we have:
$$ \lambda\omega(a,d)=\omega(a,\lambda d)=\omega(a,a+\mu b)=\mu\omega(a,b) $$
$$ \lambda\omega (d,c)= \text{[see 3. above]}=(1-\mu)\omega(a,b) $$
so it all depends on the sign of $\lambda$. If $\lambda>0$, we have $\omega(a,d)>0$ and $\omega(d,c)>0$, i.e. $D\in S^°(B;a,c)$; whereas if $\lambda <0$, we have $\omega(a,d)<0$ and $\omega(d,c)<0$, i.e. $D\in S^°(B;-a,-c)$, the vertically opposite angular sector of $S^°(B;a,c)$, as it is immediate to verify.
