
# 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.

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 $$[AB]$$.