Decomposition of a plane

Question

Angular line is a union of two half-lines with the same origin in the plane. Open angle is the set of all points which are on the same side of the angular line (they are on the same side if there is a polygonal line that connects two points and doesn't intersect the angular line). Now, I'm trying to prove that every angular line $pOq$ is generating a decomposition of a plane $π$ into two open angles. Given angular line $pOq$, consisting of half-lines $p, q$ with the same origin $O$, pick a random point $P$ on $p$ and point $Q$ on $q$. Then, choose a random point $A$ on the segment $PQ$ (such that $P-A-Q$). Next, on the line $AO$, choose a point $B$ such that $A-O-B$. Now, $π_1$ is the angle with points that are on the same side of the $pOq$ as the point $A$ and $π_2$ is the set of points that are on the same side of $pOq$ as the point B. How to prove that $π_1\cup π_2\cup pOq = π$ ?

Answer 1

Well, I think you will need the theorem that given the line $l$, it divides the plane into exactly two halfplanes i.e.

Theorem Given the line $l$, the two-argument relation $\sim_l$ over the set $\pi\setminus l$ given by the formula $$A\sim_l B :\iff l\cap\overline{AB}=\emptyset$$ is an equivalence relation with exactly two equivalence classes. ($\overline{AB}$ is the open segment).

We say that a set $h\subset \pi$ is an (open) halfplane whenever there exists a line $l$ such that $h$ is one of the equivalence classes of $\sim_l$. The line $l$ is called the boundary of $h$ and it is uniquely determined by $h$ (I think uniqueness is not necessary here).

Proposition. Any halflpane is a convex set. An intersection of a family of convex sets is convex.

Lemma. Given a line $l$ and and an (open) halfline $r$ with origin $O\in l$, either $r\subset l$ or $r\subset h$ for exactly one of the halfplanes determined by $l$.

Assume we are given two halflines $p,q$ with the same origin $O$ such that they are not contained in one line. Let $l_p, l_q$ be the lines containing $p,q$ respectively. By lemma $q$ is contained in exactly one halplane with boundary $l_p$. Denote it by $h_{pq}$ and denote complementary halfplane by $h_{pq}^*$. Similarly define $h_{qp}$ and $h_{qp}^*$. Observe that $p^*\subset h_{qp}^*,q^*\subset h_{pq}^*$.

Now let's follow your construction. First notice that $A\in h_{pq}\cap h_{qp}$ and $B\in h_{pq}^*\cap h_{qp}^*$ (I leave out an easy proof for you). We must prove that $\pi_1\cup\pi_2\cup pOq=\pi$. So let $M\in \pi$ be an arbitrary point. Of course $$M\in\pi=l_p\cup h_{pq}\cup h_{pq}^*=\{O\}\cup p\cup p^*\cup h_{pq}\cup h_{pq}^*$$ and similarly $$M\in\pi=l_q\cup h_{qp}\cup h_{qp}^*=\{O\}\cup q\cup q^*\cup h_{qp}\cup h_{qp}^*$$ Combining these we get few cases:

1. $M\in h_{pq}\cap h_{qp}$. By convexity of $h_{pq}\cap h_{qp}$ we know that a segment $\overline{AM}$ lies entirely within this set and hence it does not intersect the lines $l_p,l_q$ and as a consequence it does not intersect $pOq$. So $M\in\pi_1$.

2. $M\in h_{pq}^*$. By convexity of $h_{pq}^*$ segment $\overline{BM}$ lies entirely within this set. Hence it does not intersect $l_p$ and consequently $p\cup\{O\}$. Suppose it intersects $q$ in a point $X$. Then $X\in \overline{BM}\subset h_{pq}^*$ and $X\in q\subset h_{pq}$, but these halfplanes are disjoint. A Contradiction. Hence $M\in \pi_2$.

3. $M\in h_{qp}^*$. A case symmetric to case 2.

4. $M\in l_p\setminus h_{qp}^*$. Since $p^*\subset h_{qp}^*$, $M\in p\cup\{O\}\subset pOq$.

5. $M\in l_q\setminus h_{pq}^*$. A case symmetric to case 4.