
# 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.