7.3: Transversal Property

If the line $t$ intersects each line $\ell$ and $m$ at one point, then we say that $t$ is a transversal to $\ell$ and $m$. For example, on the diagram, line ($CB$) is a transversal to ($AB$) and ($CD$).

Theorem $\PageIndex{1}$: Transversal Property

$(AB) \parallel (CD)$ if and only if

$$2 \cdot (\measuredangle ABC + \measuredangle BCD) \equiv 0. \nonumber$$

Equivalently

$\measuredangle ABC + \measuredangle BCD \equiv 0$ or $\measuredangle ABC + \measuredangle BCD \equiv \pi.$

Moreover, if $(AB) \ne (CD)$, then in the first case, $A$ and $D$ lie on opposite sides of $(BC)$; in the second case, $A$ and $D$ lie on the same sides of $(BC)$.

Proof

"only-if" part. Denote by $O$ the midpoint of $[BC]$.

Assume $(AB) \parallel (CD)$. According to Theorem 7.2.1, $(CD)$ is a reflection of $(AB)$ across $O$.

Let $A'$ be the reflection of $A$ across $O$. Then $A' \in (CD)$ and by Proposition 7.2.1 we have that

$$\measuredangle ABO = \measuredangle A'CO.$$

Note that

$$\measuredangle ABO \equiv \measuredangle ABC, \ \ \ \ \measuredangle A'CO \equiv \measuredangle BCA'.$$

Since $A', C$ and $D$ lie on one line, Exercise 2.4.2 implies that

$$2 \cdot \measuredangle BCD \equiv 2 \cdot \measuredangle BCA'.$$

Finally note that 7.3.2, 7.3.3 and 7.3.4 imply 7.3.1.

"If"-part. By Theorem 7.2.1 there is a unique line $(CD)$ thru $C$ that is parallel to $(AB)$. From the "only-if" part we know that 7.3.1 holds.

On the other hand, there is a unique line $(CD)$ such that 7.3.1 holds. Indeed, suppose there are two such lines $(CD)$ and $(CD')$, then

$2 \cdot (\measuredangle ABC + \measuredangle BCD) \equiv 2 \cdot (\measuredangle ABC + \measuredangle BCD') \equiv 0$.

Therefore $2 \cdot \measuredangle BCD \equiv 2 \cdot BCD'$ and by Exercise 2.4.2, $D' \in (CD)$, or equivalently the line $(CD)$ coincides with $(CD')$.

Therefore if 7.3.1 holds, then $(CD) \parallel (AB)$.

Finally, if $(AB) \ne (CD)$ and $A$ and $D$ lie on the opposite sides of $(BC)$, then $\angle ABC$ and $\angle BCD$ have opposite signs. Therefore

$-\pi < \measuredangle ABC + \measuredangle BCD < \pi.$

Applying 7.3.1, we get $\measuredangle ABC + \measuredangle BCD = 0$.

Similarly if $A$ and $D$ lie on the same side of $(BC)$, then $\angle ABC$ and $\angle BCD$ have the same sign. Therefore

$0 < |\measuredangle ABC + \measuredangle BCD| < 2\cdot \pi$

and 7.3.1 implies that $\measurdangle ABC + \measuredangle BCD \equiv \pi$.

Exercise $\PageIndex{1}$

Let $\triangle ABC$ be a nondegenerate triangle, and $P$ lies between $A$ and $B$. Suppose that a line $\ell$ passes thru $P$ and is parallel to $(AC)$. Show that $\ell$ crosses the side $[BC]$ at another point, say $Q$, and

$\triangle ABC \sim \triangle PBQ.$

In particular,

$\dfrac{PB}{AB} = \dfrac{QB}{CB}.$

Hint

Since $\ell \parallel (AC)$, it cannot cross $[AC]$. By Pasch's theorem (Theorem 3.4.1), $\ell$ has to cross another side of $\triangle ABC$. Therefore $\ell$ cross $[BC]$; denote the point of intersection by $Q$.

Use the transversal property (Theorem $\PageIndex{1}$) to show that $\measuredangle BAC = \measuredangle BPQ$. The same argument shows that $\measuredangel ACB = \measuredangle PQB$; it remains to apply the AA similarity condition.

Exercise $\PageIndex{2}$

Trisect a given segment with a ruler and a compass.

Answer

Assume we need to trisect segment $[AB]$. Construct a line $\ell \ne (AB)$ with four points $A, C_1, C_2, C_3$ such that $C_1$ and $C_2$ trisect $[AC_3]$. Draw the line $(BC_3)$ and draw parallel lines thru $C_1$ and $C_2$. The points of intersections of these two lines with $(AB)$ trisect the segment $[AB]$.