3.3 Angles Associated with Parallel Lines
As we have seen in the previous section, given certain relationships between alternate interior angles or corresponding angles or same-side interior angles, we can prove that lines are parallel. Let us now see if the converses of these theorems hold true.
- Theorem 3.7: P
$\rightarrow$ CA
If two parallel lines are cut by a transversal, then corresponding angles are congruent.
Given: $m$ and $n$
are parallel with transversal $t$. Prove: $\angle 1\cong \angle 2$ |
Proof (indirect): Given that $m$ and $n$ are parallel lines and that $\angle 1$ and $\angle 2$ are corresponding angles, we want to prove that $\angle 1$ is congruent to $\angle 2$. Assume that $\angle 1$ is not congruent to $\angle 2$. Then there must be a line through $P$ such that $\angle 3$ is congruent to $\angle 2$. This makes $\ell \parallel m$ by CA $\rightarrow$ P. We are now faced with the fact that we were given that $n\parallel m$ and that now $\ell \parallel m$. This is a contradiction of the Parallel Postulate which says that through an external point (our point $P$), there is exactly one line parallel to a given line (the line $m$). This contradiction means that our assumption (that $\angle 1 \ncong \angle 2$) must be wrong. Hence, if two lines are parallel, a pair of corresponding angles must be congruent.
Using theorem 3.7, it should be easy to prove the following theorems.
- Theorem 3.8: P
$\rightarrow$ AI
If two parallel lines are cut by a transversal, then alternate interior angles are congruent.
Given: In
quadrilateral $ABCD$, $\overline{AB}\parallel\overline{CD}$ and
$\overline{AD}\parallel\overline{BC}$. Prove: $\overline{AB}\cong \overline{CD}$ and $\overline{AD}\cong\overline{BC}$. |
Solution: Draw diagonal $\overline{AC}$, so that we have $\triangle ABC$ and $\triangle ADC$. We now use P $\rightarrow$ AI twice. Since $\overline{AB}\parallel\overline{CD}$, we know that $\angle 2\cong\angle 3$. Since $\overline{AD}\parallel\overline{BC}$, we know that $\angle 1\cong\angle 4$. (It is important here to keep track of which angles are alternate interior angles from which parallel lines.) Since $\overline{AC}\cong\overline{AC}$, we have $\triangle ABC\cong\triangle CDA$ by ASA. From the congruent triangles, we then have $\overline{AB}\cong\overline{CD}$ and $\overline{AD}\cong\overline{BC}$. |
We have one other result concerning the angles that are formed when parallel lines are cut by a transversal. In this case, the angles are not congruent, but rather supplementary.
- Theorem 3.9:
If two parallel lines are cut by a transversal, then same-side interior
angles are supplementary.
The proof of this is left for the exercises.