TABLE OF CONTENTS
I. PRELIMINARIES
II. AUXILIARY RESULTS
III. THE MAIN RESULTS
I. PRELIMINARIES
I will answer the question within the framework of Hilbert's axioms, as described in Wikipedia.
We will assume implicitly the existence of a constant set, the set of points, and the following additional two axioms.
- A line consists of points.
- A plane consists of points.
For every pair of points $x$ and $y$, denote by $(xy)$ the set of points lying between $x$ and $y$, and denote by $[xy]$ the line segment between $x$ and $y$, i.e. $[xy]:=\{x,y\}\cup(xy)$.
For every pair of distinct points $x$ and $y$, denote by $\overline{xy}$ the line through $x$ and $y$, whose existence and uniqueness are guaranteed by axioms I1 and I2, respectively.
The following book (written in German) will be cited as [Grund].
Hilbert, David. Grundlagen der Geometrie, Mit Supplementen von Prof. Dr. Paul Bernays, 13th edition, Teubner, 1987
I use the German book as a reference, rather than the English translation by Edgar Jerome Townsend, which is readily available online, since Hilbert's axioms underwent refinements over the years (even after his death), which Townsend's translation, published 1950, does not reflect, as exemplified here.
II. AUXILIARY RESULTS
Lemma 1
Let $P$ be a plane.
- There exists some non-collinear $C\subseteq P$.
- Let $L\subseteq P$ be a line. Then $P\setminus L\neq\emptyset$.
Proof
- See here.
- By part 1.
$\square$
Lemma 2
Let $x$ be a point.
- $(xx)=\emptyset$.
- $[xx]=\{x\}$.
Proof
Suppose the opposite. Choose some $y\in(xx)$. Then $x$, $x$, and $y$ are pairwise distinct by axiom O1.
$[xx]=\{x,x\}\cup(xx)$ by the definition of $[xx]$, and $(xx)=\emptyset$ by part 1.
$\square$
Lemma 3
Let $x$, $y$, and $z$ be pairwise distinct and collinear points. Then exactly one of the following three conditions holds.
- $x\in(yz)$.
- $y\in(xz)$.
- $z\in(xy)$.
Proof
This post proves that at least one of the three conditions holds, while axiom O3 guarantees that at most one of them does.
$\square$
Lemma 4
Let $a$, $c$, and $d$ be points, let $b\in(ac)$, and suppose that $c\in(bd)$. Then $b,c\in(ad)$.
Proof
This is proved in [Grund] as the first part of Theorem 5's proof (on p. 7).
$\square$
Lemma 5
Let $a$ and $d$ be points, and let $c\in(ad)$.
- Let $b\in(ac)$. Then
- $c\in(bd)$.
- $b\in(ad)$.
- Let $b\in(cd)$. Then
- $c\in(ab)$.
- $b\in(ad)$.
Proof
This is proved in [Grund] as the second part of Theorem 5's proof (on p. 7).
This can be shown by an argument analogous to the one used to prove part 1.
$\square$
Lemma 6
Let $x$ and $z$ be points, let $y\in(xz)$, and let $p\in\overline{xz}\setminus\{x,y,z\}$. Then exactly one of the following conditions holds.
- $x\in(py)$, and $x,y\in(pz)$.
- $p\in(xy)$, $y\in(pz)$, and $p\in(xz)$.
- $p\in(yz)$, $y\in(xp)$, and $p\in(xz)$.
- $z\in(yp)$, and $y,z\in(xp)$.
Proof
For starters we note that $x$, $y$, $z$, and $p$ are pairwise distinct and collinear points. In fact, since $y\in(xz)$ it follows from axiom O1 that $x$, $y$, and $z$ are pairwise distinct and collinear points, in particular $x,y,z\in\overline{xz}$; and $p\in\overline{xz}\setminus\{x,y,z\}$ by the initial assumptions.
Now, firstly we show that at least one of the four conditions holds. The proof idea can be found in [Grund] in the final part of the Theorem 5's proof (on p. 8). By Lemma 3 either $x\in(zp)$ or $z\in(xp)$ or $p\in(xz)$, and likewise $x\in(yp)$ or $y\in(xp)$ or $p\in(xy)$. Hence one of the following five cases applies.
Suppose that $x\in(zp)$. In this case the first condition holds. In fact, $x\in(pz)$ by assumption; and since $x\in(pz)$ and $y\in(xz)$ (by the initial assumptions), part 2.1 of Lemma 5 implies that $x\in(py)$, whereas part 2.2 of Lemma 5 implies that $y\in(pz)$.
Suppose that $z\in(xp)$. In this case the fourth condition holds. In fact, $z\in(xp)$ by assumption; and since $z\in(xp)$ and $y\in(xz)$ (by the initial assumptions), part 1.1 of Lemma 5 implies that $z\in(yp)$, whereas part 1.2 of Lemma 5 implies that $y\in(xp)$.
Suppose that $p\in(xz)$ and $y\in(xp)$. In this case the third condition holds. In fact, $p\in(xz)$ and $y\in(xp)$ by assumption; and from this follows by part 1.1 of Lemma 5 that $p\in(yz)$.
Suppose that $p\in(xz)$ and $p\in(xy)$. In this case the second condition holds. In fact, $p\in(xy)$ and $p\in(xz)$ by assumption; and since $y\in(xz)$ (by the initial assumptions) and $p\in(xy)$, part 1.1 of Lemma 5 implies that $y\in(pz)$.
Suppose that $p\in(xz)$ and that $x\in(yp)$. This case cannot occur, a fact we will prove by deriving a contradiction. On the one hand $x\in(yz)$ by Lemma 4, since $x\in(yp)$ and $p\in(xz)$; but on the other hand $x\notin(yz)$ by Lemma 3, since $y\in(xz)$ by the initial assumptions.
Secondly we show that no two distinct conditions of the four listed in the statement of our lemma hold simultaneously. We consider each of the six possible combinations, and in each case derive a contradiction by Lemma 3.
Suppose that conditions 1 and 2 hold simultaneously. Then $x\in(py)$ by condition 1, and $p\in(xy)$ by condition 2.
Suppose that conditions 1 and 3 hold simultaneously. Then $x\in(py)$ by condition 1, and $y\in(xp)$ by condition 3.
Suppose that conditions 1 and 4 hold simultaneously. Then $x\in(py)$ by condition 1, and $y\in(xp)$ by condition 4.
Suppose that conditions 2 and 3 hold simultaneously. Then $p\in(xy)$ by condition 2, and $y\in(xp)$ by condition 3.
Suppose that conditions 2 and 4 hold simultaneously. Then $p\in(xy)$ by condition 2, and $y\in(xp)$ by condition 4.
Suppose that conditions 3 and 4 hold simultaneously. Then $p\in(yz)$ by condition 3, and $z\in(yp)$ by condition 4.
$\square$
Lemma 7
Let $x,z$ be points, and let $y\in(xz)$.
- $(xz)=\coprod\big\{(xy),\{y\},(yz)\big\}$, i.e. $(xz)$ is the disjoint union of $(xy)$, $\{y\}$, and $(yz)$. In particular, $(xy),(yz)\subseteq(xz)$.
- Let $p\in(xz)\setminus\{y\}$. Then exactly one of the following conditions holds: $p\in(xy)$ or $p\in(yz)$.
Proof
For starters we note that $x$, $y$, and $z$ are pairwise distinct and collinear points by axiom O1, since $y\in(xz)$.
Now, firstly we show that $(xy)$, $\{y\}$, and $(yz)$ are pairwise disjoint. In fact, $(xy)\cap\{y\}=\emptyset$, since otherwise we would have $y\in(yz)$, which would yield by axiom O1 the contradiction that $y$, $z$, and $y$ were pairwise distinct. Similarly, $(yz)\cap\{y\}=\emptyset$. As for $(xy)\cap(yz)$, suppose to the contrary that this intersection is non-empty. Choose $p\in(xy)\cap(yz)$. Since $y\in(xz)$ and $p\in(yz)$, part 2.1 of Lemma 5 implies that $y\in(xp)$. Since $y\in(xp)$ and $p\in(xy)$, part 1.1 of Lemma 5 implies that $y\in(pp)$. But $(pp)=\emptyset$ by part 1 of Lemma 2.
Secondly we show that $(xz)=(xy)\cup\{y\}\cup(yz)$.
Let $p\in(xz)$. If $p=y$, then $p\in\{y\}$. So assume that $p\neq y$. Since $p\in(xz)$, axiom O1 implies that $x$, $z$, and $p$ are pairwise distinct and collinear. So $p\in\overline{xz}\setminus\{x,y,z\}$. It follows from Lemma 6 that one of the following four conditions holds:
- $x\in(py)$, and $x,y\in(pz)$.
- $p\in(xy)$, $y\in(pz)$, and $p\in(xz)$.
- $p\in(yz)$, $y\in(xp)$, and $p\in(xz)$.
- $z\in(yp)$, and $y,z\in(xp)$.
The first of these conditions must be ruled out, since otherwise we would have $x\in(pz)$ and $p\in(xz)$ (by the selection of $p$), in contradiction to %Absurd. Likewise, the fourth of these conditions must be ruled out, since otherwise we would have $z\in(xp)$ and $p\in(xz)$. It follows from the remaining two conditions that $p\in(xy)$ or $p\in(yz)$.
Conversely, let $q\in(xy)\cup\{y\}\cup(yz)$. If $q\in\{y\}$, we have $q\in(xz)$ by the initial assumptions. If $q\in(xy)$, we have $y\in(xz)$ (by the initial assumptions) and $q\in(xy)$, which together imply that $q\in(xz)$, by part 1.2 of Lemma 5. Finally, if $q\in(yz)$, we have $y\in(xz)$ and $q\in(yz)$, which together imply that $q\in(xz)$, by part 2.2 of Lemma 5.
By part 1.
$\square$
We will strengthen Pasch's axiom (axiom O4) in two ways: by weakening the premises to allow the three points to be collinear, and by strengthening the conclusion to deduce that a line that intersects one side of a triangle intersect exactly one of the other two sides. We encapsulate these modifications in the following lemma.
Lemma 8 (Pasch)
Let $L$ be a line, and let $x$, $y$, and $z$ be such that there is a plane containing both $L$ and $\{x,y,z\}$, such that $L\cap\{x,y,z\}=\emptyset$, and such that $(xy)\cap L\neq\emptyset$. Then exactly one of the following conditions holds: $(zx)\cap L\neq\emptyset$, or $(zy)\cap L\neq\emptyset$.
Proof
Suppose that $x=y$. This case cannot occur, since on the one hand $(xy)\cap L\neq\emptyset$ by the initial assumptions, while on the other hand $(xy)\cap L=\emptyset$, since $(xy)=\emptyset$ by part 1 of Lemma 2.
Suppose that $x\neq y$, and that $z\in\{x,y\}$. Assume w.l.g. that $z=x$. Then $(zx)\cap L=\emptyset$ by part 1 of Lemma 2, whereas $(zy)\cap L\neq\emptyset$ by the initial assumptions.
Suppose that $x$, $y$, and $z$ are pairwise distinct and collinear. By Lemma 3 we need consider three possibilities.
Suppose that $y\in(xz)$. We will show that $(zx)\cap L\neq\emptyset$ and $(zy)\cap L=\emptyset$.
To see that $(zx)\cap L\neq\emptyset$, choose some $p\in(xy)\cap L$, a choice possible since $(xy)\cap L\neq\emptyset$ by the initial assumptions. Then since $y\in(xz)$ (by the original assumption of this case) and $p\in(xy)$, part 1.2 of Lemma 5 implies that $p\in(xz)$, hence $p\in(zx)\cap L$.
To see that $(zy)\cap L=\emptyset$, suppose the opposite. On the one hand $z\notin L$, since $L\cap\{x,y,z\}=\emptyset$ by the initial assumptions. On the other hand, choose $q\in(yz)\cap L$, a choice possible by the contradictory hypothesis. Since $y\in(xz)$ (by the original assumption of this case) and $p\in(xy)$, it follows from part 1.1 of Lemma 5 that $y\in(pz)$. This fact together with the fact that $q\in(yz)$ imply by part 2.1 of Lemma 5 that $y\in(pq)$. Therefore by axiom O1 $p$, $q$, and $y$ are pairwise distinct and collinear; in particular $y\in\overline{pq}$. Since $p,q\in L$, axiom I2 implies that $\overline{pq}=L$. Hence $y\in L$, contradicting the initial assumption that $L\cap\{x,y,z\}=\emptyset$.
Suppose that $x\in(yz)$. Then, by an argument analogous to the one used in the preceding bullet point, $(zx)\cap L=\emptyset$ and $(zy)\cap L\neq\emptyset$.
Suppose that $z\in(xy)$. Choose some $p\in(xy)\cap L$, a choice possible since $(xy)\cap L\neq\emptyset$ by the initial assumptions. Since $L\cap\{x,y,z\}=\emptyset$, $p\neq z$. Hence by part 2 of Lemma 7 exactly one of the following conditions holds: $p\in(xz)$ or $p\in(zy)$.
We show that $p\in(xz)$ iff $(zx)\cap L\neq\emptyset$. If $p\in(xz)$, then $p\in(xz)\cap L$, hence $(zx)\cap L\neq\emptyset$. Conversely, suppose that $(zx)\cap L\neq\emptyset$. Choose $q\in(zx)\cap L$. It suffices to show that $q=p$, or equivalently that $q\in\{p\}$. Observing that $L$ and $\overline{xy}$ are distinct lines (since $L\cap\{x,y,z\}=\emptyset$ but $x,y\in\overline{xy}$), and that $p\in L\cap\overline{xy}$ (since $p\in L$; and since $x$, $y$, and $p$ are collinear by axiom O1 because $p\in(xy)$), we conclude by axiom I2 that $L\cap\overline{xy}=\{p\}$. Now, $q\in L$, so it remains to show that $q\in\overline{xy}$. In fact, since $q\in(xz)$, axiom O1 guarantees that $x$, $z$, and $q$ are collinear; in particular $q\in\overline{xz}$. But $x$, $y$, and $z$ are collinear by assumption, hence by axiom I2 $\overline{xz}=\overline{xy}$.
To see that $p\in(zy)$ iff $(zy)\cap L\neq\emptyset$ use an argument analogous to the one used in the previous paragraph.
Suppose that $x$, $y$, and $z$ are pairwise distinct and not collinear. Then since $(xy)\cap L\neq\emptyset$ by the initial assumptions, Pasch's axiom implies that either $(zx)\cap L\neq\emptyset$ or $(zy)\cap L\neq\emptyset$. This post rules out the possibility that these two inequalities hold simultaneously.
$\square$
III. THE MAIN RESULTS
Lemma 9
Let $P$ be a plane, let $L$ be a line contained in $P$, and let $x_1,x_2\in P\setminus L$ be such that $(x_1x_2)\cap L\neq\emptyset$. Define
$$
\begin{align*}
H_1&:=\big\{x\in P:[xx_1]\cap L=\emptyset\big\},\\
H_2&:=\big\{x\in P:[xx_2]\cap L=\emptyset\big\}.
\end{align*}
$$
Then
- $x_1\in H_1$, and $x_2\in H_2$. In particular, $H_1,H_2\neq\emptyset$.
- $P=\coprod\{L,H_1,H_2\}$.
- Let $x$ and $y$ be such that either $x,y\in H_1$ or $x,y\in H_2$. Then $[xy]\cap L=\emptyset$.
- $H_1$ and $H_2$ are convex.
Proof
To show that, say, $x_1\in H_1$, we need to show that $[x_1x_1]\cap L=\emptyset$, i.e. that $\{x_1\}\cap L=\emptyset$ by part 2 of Lemma 2. This follows from the initial assumptions.
Firstly we show that $H_1\cap L=\emptyset$. Suppose the opposite. Choose $x\in H_1\cap L$. Then on the one hand $x\in[xx_1]\cap L$ since $x\in[xx_1]=\{x,x_1\}\cup(xx_1)$ and $x\in L$, but on the other hand $[xx_1]\cap L=\emptyset$ by the definition of $H_1$ and the fact that $x\in H_1$. The proposition $H_2\cap L=\emptyset$ can be proved analogously.
Secondly we show that $H_1\cap H_2=\emptyset$. Suppose the opposite. Choose some $z\in H_1\cap H_2$. Since $(x_1x_2)\cap L\neq\emptyset$ by the initial assumptions, Pasch's Lemma yields that either $(zx_1)\cap L\neq\emptyset$ or $(zx_2)\cap L\neq\emptyset$. However, $(zx_1)\cap L=\emptyset$ by the definition of $H_1$ and the fact that $z\in H_1$, and similarly $(zx_2)\cap L=\emptyset$.
Thirdly we show that $P=L\cup H_1\cup H_2$. Since $L,H_1,H_2\subseteq P$ by the initial assumptions, it suffices to show that $P\setminus(L\cup H_1)\subseteq H_2$. Let $z\in P\setminus(L\cup H_1)$. We need to show that $[zx_2]\cap L=\emptyset$. Since $z,x_2\notin L$, it suffices to show that $(zx_2)\cap L=\emptyset$. Since $(x_1x_2)\cap L\neq\emptyset$ by the initial assumptions, Pasch's Lemma implies that exactly one of $(zx_1)\cap L\neq\emptyset$ and $(zx_2)\cap L\neq\emptyset$ holds. It must be the first of these inequalities, since otherwise we would have $[zx_1]\cap L=\emptyset$, which would imply that $z\in H_1$, contradicting the selection of $z$.
Assume w.l.g. that $x,y\in H_1$. It suffices to show that $(xy)\cap L=\emptyset$, since by part 1 $\{x,y\}\cap L\subseteq H_1\cap L=\emptyset$. Suppose the opposite. Then on the one hand by Pasch's Lemma either $(x_1x)\cap L\neq\emptyset$, or $(x_1y)\cap L\neq\emptyset$. But on the other hand both $(xx_1)\cap L=\emptyset$, and $(yx_1)\cap L=\emptyset$ by the definition of $H_1$ and the fact that $x,y\in H_1$.
We show only that $H_1$ is convex; an analogous argument will show that $H_2$ is convex. Let $x,y\in H_1$. If $x=y$, then $[xy]=\{x\}\subseteq H_1$ by part 2 of Lemma 2. So suppose that $x\neq y$, and let $z\in[xy]$. By the definition of $H_1$ we need to show that $z\in P$, and that $[zx_1]\cap L=\emptyset$. If $z\in\{x,y\}$, this follows from the definition of $H_1$ and the fact that $x,y\in H_1$. So assume that $z\notin\{x,y\}$, which is to say $z\in(xy)$. It suffices to show that $z\in P\setminus L$, and that $(zx_1)\cap L=\emptyset$.
We show that $z\in P$. Since $z\in(xy)$, axiom O1 implies that $x$, $y$, and $z$ are pairwise distinct and collinear; in particular $z\in\overline{xy}$. Since $x,y\in H_1$, part 2 implies that $x,y\in P$. Hence by axiom I6 $\overline{xy}\subseteq P$.
To see that $z\notin L$, suppose the opposite. Then $z\in(xy)\cap L$, hence in particular $(xy)\cap L\neq\emptyset$. Then on the one hand Pasch's Lemma implies that $(x_1x)\cap L\neq\emptyset$ or $(x_1y)\cap L\neq\emptyset$. But on the other hand both $(xx_1)\cap L=\emptyset$ and $(yx_1)\cap L=\emptyset$ by the definition of $H_1$ and the fact that $x,y\in H_1$.
To see that $(zx_1)\cap L=\emptyset$, suppose the opposite. Then on the on hand Pasch's Lemma implies that $(xz)\cap L\neq\emptyset$ or $(xx_1)\cap L\neq\emptyset$. But on the other hand both $(xz)\cap L=\emptyset$ and $(xx_1)\cap L=\emptyset$: in fact, $(xz)\cap L\subseteq(xy)\cap L\subseteq[xy]\cap L=\emptyset$, where the first equality is by part 1 of Lemma 7, and the last one is by part 3 of our lemma since $x,y\in H_1$; and $(xx_1)\cap L=\emptyset$ by the definition of $H_1$ and the fact that $x\in H_1$.
$\square$
Lemma 10
Let $P$ be a plane, let $L$ be a line contained in $P$, let $H_1$ and $H_2$ be convex sets of points, and assume that $P=\coprod\{L,H_1,H_2\}$. Let $x_1\in H_1$. Then $H_1=\big\{x\in P:[xx_1]\cap L=\emptyset\big\}$.
Proof
Let $x\in H_1$. We need to show that $[xx_1]\cap L=\emptyset$. In fact, $[xx_1]\subseteq H_1$ since $x,x_1\in H_1$ and $H_1$ is convex; and $H_1\cap L=\emptyset$ by the initial assumptions.
Conversely, define $G_1:=\big\{x\in P:[xx_1]\cap L=\emptyset\big\}$, and let $x\in G_1$. Suppose to the contrary that $x\notin H_1$. Choose $x_0\in L$ (possible by axiom I3), and choose a point $x_2$ such that $x_0\in(x_1x_2)$ (possible by axiom O2).
$x_1$, $x_2$, and $x_0$ are pairwise distinct and collinear by axiom O1. Then by axiom I2 $x_1,x_2\in\overline{x_0x_1}=\overline{x_0x_2}$. This implies that $x_2\in P\setminus L$: in fact, $x_2\in\overline{x_0x_1}\subseteq P$ by axiom I6 since $x_0,x_1\in L\cup H_1\cup H_2=P$; and $x_2\notin L$, since otherwise we would have $x_1\in\overline{x_0x_2}=L$ by axiom I2, in contradiction to the fact, shown above, that $H_1\cap L=\emptyset$.
Define $G_2:=\big\{x\in P:[xx_2]\cap L=\emptyset\big\}$. Then $G_1\cap G_2=\emptyset$ by part 2 of @Adamant. Since also $x\in G_1$ (by assumption), to obtain a contradiction it suffices to show that $x\in G_2$. Since $H_2\subseteq G_2$ by an argument analogous to the one used in the opening paragraph of this proof, it suffices to show that $x\in H_2$. This is a consequence of the following observations:
- $x\in G_1\subseteq P=L\cup H_1\cup H_1$.
- $x\notin L$, since on the one hand $x\in[xx_1]$, and on the other hand $[xx_1]\cap L=\emptyset$ by the definition of $G_1$ and the fact that $x\in G_1$.
- $x\notin H_1$ by the contradictory hypothesis.
$\square$
Theorem 11
Let $P$ be a plane, and let $L$ be a line contained in $P$. Define $\mathcal{H}$ to consist of all $\mathbf{H}$ satisfying that there exist sets of points $H_1$ and $H_2$ such that the following four conditions hold:
- $H_1,H_2\neq\emptyset$.
- $P=\coprod\{L,H_1,H_2\}$.
- $H_1$ and $H_2$ are convex.
- $\mathbf{H}=\{H_1,H_2\}$.
Then $\mathcal{H}$ is a singleton. We call $\mathcal{H}$'s single member the set of open half planes of $P$ determined by $L$.
Proof
Choose $x_1\in P\setminus L$ (possible by part 2 of Lemma 1). As in the proof of Lemma 10, choose $x_0$ and $x_2$, such that
- $x_1,x_2\in P\setminus L$, and
- $x_0\in(x_1x_2)\cap L$,
and define
$$
\begin{align*}
G_1&:=\big\{x\in P:[xx_1]\cap L=\emptyset\big\},\\
G_2&:=\big\{x\in P:[xx_2]\cap L=\emptyset\big\},\\
\mathbf{G}&:=\{G_1,G_2\}.
\end{align*}
$$
Then $\mathbf{G}\in\mathcal{H}$ by parts 1, 2, and 4 of Lemma 9. Now, let $\mathbf{H}\in\mathcal{H}$. Choose $H_1$ and $H_2$ that satisfy the four conditions of our theorem. We need to show that $\{H_1,H_2\}=\{G_1,G_2\}$.
Firstly suppose that $x_1\in H_1$. Then $H_1=G_1$ by Lemma 10. By the same token $H_2=G_2$ provided $x_2\in H_2$. Suppose to the contrary that $x_2\notin H_2$. Since also $x_2\in P\setminus L$ and since $P=L\cup H_1\cup H_2$, we conclude that $x_2\in H_1$. But then $[x_2x_1]\cap L=\emptyset$ by Lemma 10, in contradiction to the fact that $x_0\in(x_1x_2)\cap L$.
Secondly suppose that $x_1\notin H_1$. Since also $x_1\in P\setminus L$ and $P=L\cup H_1\cup H_2$, we conclude that $x_1\in H_2$. Then by an argument analogous to the one used in the previous paragraph, $H_2=G_1$ and $H_1=G_2$.
$\square$
Theorem 12
Let $P$ be a plane, let $L$ be a line contained in $P$, let $H$ be an open half plane of $P$ determined by $L$, and let $y\in H$. Then $H=\{x\in P:[xy]\cap L=\emptyset\}$.
Proof
Define $\mathbf{H}$ to be the set of open half planes of $P$ determined by $L$. Choose $H'$ to be such that $\mathbf{H}=\{H,H'\}$. Then by parts 3 and 2 of Theorem 11, respectively, $H$ and $H'$ are convex, and $P=\coprod\{L,H,H'\}$. The desired conclusion now follows from Lemma 10.
$\square$
Lemma 13
Let $P$ be a plane, let $L$ be a line contained in $P$, let $H_1$ and $H_2$ be such that $\{H_1,H_2\}$ is the set of open half planes of $P$ determined by $L$, let $x_1\in H_1$, and let $x_2\in H_2$. Then $(x_1x_2)\cap L\neq\emptyset$.
Proof
Suppose the opposite. Then $[x_1x_2]\cap L\neq\emptyset$. Then $x_1\in H_1$ by assumption, and $x_1\in H_2$ by Theorem 12, in contradiction to the fact that $H_1\cap H_2=\emptyset$ by the second condition of Theorem 11.
$\square$
Theorem 14
Let $P$ be a plane, let $L$ be a line contained in $P$, let $H_1$ and $H_2$ be such that $\{H_1,H_2\}$ is the set of open half planes of $P$ determined by $L$, and let $x_2\in H_2$.
- $H_1=\big\{x\in P\setminus L:[xx_2]\cap L\neq\emptyset\big\}$.
- $H_1=\big\{x\in P:(xx_2)\cap L\neq\emptyset\big\}$.
Proof
Let $x_1\in H_1$. Then by Lemma 13 $[x_1x_2]\cap L\neq\emptyset$.
Conversely, let $x\in\{x\in P\setminus L:[xx_2]\cap L\neq\emptyset\big\}$. Suppose to the contrary that $x\notin H_1$. Then on the one hand $x\in H_2$, since $x\in P\setminus(L\cup H_1)$ and $P=L\cup H_1\cup H_2$ by part 2 of Theorem 11. But on the other hand $x\notin H_2$, for suppose the opposite. Then $[xx_2]\cap L=\emptyset$ by Theorem 12, in contradiction to the original selection of $x$.
Let $x\in H_1$. Then by Lemma 13 $(xx_2)\cap L\neq\emptyset$.
Conversely, let $x\in\big\{x\in P:(xx_2)\cap L\neq\emptyset\big\}$. Then $[xx_2]\cap L\neq\emptyset$. Then $x\in H_1$ by part 1 provided $x\notin L$. Suppose to the contrary that $x\in L$. Choose $y\in(xx_2)\cap L$, a choice possible since $(xx_2)\cap L\neq\emptyset$ by the original selection of $x$. By axiom O1 $x$, $x_2$, and $y$ are pairwise distinct and collinear; in particular, $x_2\in\overline{xy}$. By axiom I2 $\overline{xy}=L$, since $x\in L$ by the contradictory hypothesis, and $y\in L$ by the choice of $y$. Then $x_2\in L$. But $x_2\in H_2$ by the initial assumptions, and $H_2\cap L=\emptyset$ by part 2 of Theorem 11.
$\square$
