3.4: Half-planes

Proposition $\PageIndex{1}$

Assume $X, Y \not\in (PQ)$. Then the angles $PQX$ and $PQY$ have the same sign if and only if $[XY]$ does not intersect $(PQ)$.

Proof

The if-part follows from Lemma 3.3.2.

Assume $[XY]$ intersects $(PQ)$; suppose that $Z$ denotes the point of intersection. Without loss of generality, we can assume $Z \ne P$.

Note that $Z$ lies between $X$ and $Y$. According to Lemma 3.1.1, $\angle PZX$ and $\angle PZY$ have opposite signs. It proves the statement if $Z = Q$.

If $Z \ne Q$, then $\angle ZQX$ and $QZX$ have opposite signs by 3.7. The same way we get that $\angle ZQY$ and $\angle QZY$ have opposite signs.

If $Q$ lies between $Z$ and $P$, then by Lemma 3.1.1 two pairs of angles $\angle PQX$, $\angle ZQX$ and $\angle PQY$, $\angle ZQY$ have opposite signs. It follows that $\angle PQX$ and $\angle PQY$ have opposite signs as required.

In the remaining case $[QZ) = [QP)$ and therefore $\angle PQX = \angle ZQX$ and $\angle PQY = \angle ZQY$. Therefore again $\angle PQX$ and $\angle PQY$ have opposite signs as required.

Corollary $\PageIndex{1}$

The complement of a line $(PQ)$ in the plane can be presented in a unique way as a union of two disjoint subsets called half-planes such that

(a) Two points $X, Y \not\in (PQ)$ lie in the same half-plane if and only if the angles $PQX$ and $PQY$ have the same sign.

(b) Two points $X, Y \not\in (PQ)$ in the same half-plane if and only if $[XY]$ does not intersect $(PQ)$.

We say that $X$ and $Y$ lie on one side of $(PQ)$ if they lie in one of the half-planes of $(PQ)$ and we say that $P$ and $Q$ lie on the opposite sides of $l$ if they lie in the different half-planes of $l$.

Exercise $\PageIndex{1}$

Suppose that the angles $AOB$ and $A'OB'$ are vertical and $B \not\in (OA)$. Show that the line $(AB)$ does not intersect the segment $[A'B']$.

Hint

Note that $O$ and $A'$ lie on the same side of $(AB)$. Analogously $O$ and $B'$ lie on the same side of $(AB)$. Hence the result.

Consider the triangle $ABC$. The segments $[AB], [BC]$, and $[CA]$ are called sides of the triangle.

Theorem $\PageIndex{1}$ Pasch's theorem

Assume line $l$ does not pass thru any vertex of a triangle. Then it intersects either two or zero sides of the triangle.

Proof

Assume that the line $l$ intersects side $[AB]$ of the triangle $ABC$ and does not pass thru $A, B,$ and $C$.

By Corollary $\PageIndex{1}$, the vertexes $A$ and $B$ lie on opposite sides of $l$.

The vertex $C$ may lie on the same side with $A$ and on the opposite side with $B$ or the other way around. By Corollary $\PageIndex{1}$, in the first case, $l$ intersects side $[BC]$ and does not intersect $[AC]$; in the second case, $l$ intersects side $[AC]$ and does not intersect $[BC]$. Hence the statement follows.

Exercise $\PageIndex{2}$

Show that two points $X, Y \not\in (PQ)$ lie on the same side of $(PQ)$ if and only if the angles $PXQ$ and $PYQ$ have the same sign.

Hint

Apply Theorem 3.3.1 for $\triangle PQX$ and $\triangle PQY$ and then apply Corollary $\PageIndex{1}$a.

Exercise $\PageIndex{3}$

Let $\triangle ABC$ be a nondegenerate triangle, $A' \in [BC]$ and $B' \in [AC]$. Show that the segments $[AA']$ and $[BB']$ intersect.

Hint

We can assume that $A' \ne B, C$ and $B' \ne A, C$; otherwise the statement trivially holds.

Note that $(BB')$ does not intersect $[A'C]$. Applying Pasch's theorem (Theorem 3.4.1) for $\triangle AA'C$ and $(BB')$, we get that $(BB')$ intersets $[AA']$; denote the point of intersection by $M$.

The same way we get that $(AA')$ intersects $[BB']$; that is $M$ lies on $[AA']$ and $[BB']$.

Exercise $\PageIndex{4}$

Assume that the points $X$ and $Y$ lie on opposite sides of the line $(PQ)$. Show that the half-line $[PX)$ does not intersect $[QY)$.

Hint

Assume that $Z$ is the point of intersection.

Note that $Z \ne P$ and $Z \ne Q$. Therefore, $Z \in (PQ)$.

Show that $Z$ and $X$ lie on one side of $(PQ)$. Repeat the argument to show that $Z$ and $Y$ lie on one side of $(PQ)$. It follows that $X$ and $Y$ lie on the same side of $(PQ)$ - a contradiction.

Advanced Exercise $\PageIndex{1}$

Note that the follwing quantity

$$^{~} \measuredangle ABC = \begin{cases} \pi & \text{if } \measuredangle ABC = \pi \\ -\measuredangle ABC & \text{if } \measuredangle ABC < \pi \end{cases}$$

can serve as the angle measure; that is, the axioms hold if one exchanges $\measuredangle$ to $^{~} \measuredangle$ everywhere.

Show that $\measuredangle$ and $^{~} \measuredangle$ are the only possible angle measures on the plane.
Show that without Axiom IIIc, this is no longer true.