Parallel Lines

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Proofs.

This page develops important results regarding parallel lines and transversals. Read carefully, and complete the proofs.

Parallel postulate (uniqueness of parallels): Given a line and a point not on the line, there is exactly one line through the given point parallel to the given line.

A $180^\circ$ rotation about a point on a line takes the line to itself.

Proof: Suppose point $P$ is on line $k$ . The point cuts the line into two opposite $\answer [format=string]{rays}$ . A $180^\circ$ rotation about $P$ swaps the two opposite rays, thereby mapping the line onto itself. $\blacksquare$

A $180^\circ$ rotation about a point not on a line takes the line to a parallel line.

Proof: Let $O$ be a point not on line $l$ . Let $P$ be an arbitrary point on $R(l)$ , the rotated image of $l$ . To show that $R(l)$ is parallel to $l$ , it is sufficient to show that $P$ cannot lie also on $l$ .

Because $P$ is on $R(l)$ , there is a point $Q$ on $l$ such that $P = R(Q)$ . The rotated image of $\overrightarrow {OQ}$ is , and because $\angle QOP$ is $180^\circ$ , it follows that $Q$ , $O$ , and $P$ are $\answer [format=string]{collinear}$ . Call that line $k$ . We know line $k$ is distinct from $l$ because point $\answer {O}$ is on $k$ but not on $l$ . Now, if $P$ were on $l$ , then points $P$ and $Q$ would be on two distinct lines, $k$ and $l$ , contradicting the assumption that on two points there is a unique line. The theorem is proved. $\blacksquare$

If two parallel lines are cut by a transversal, alternate interior angles and corresponding angles are congruent.

Proof

Given that parallel lines $m$ and $n$ are cut by transversal $k$ , prove that alternate interior angles are congruent.

Let $B$ and $C$ be the intersections of transversal $k$ with lines $m$ and $n$ , respectively. Let $P$ be the midpoint of $\overline {BC}$ .

(a): Rotate $180^\circ$ about $P$ , which takes $k$ to itself $m$ $n$ .
(b): The rotation maps $B$ to $\answer {C}$ because $PB = PC$ and the rotation preserves distances.
(c): Because $P$ is not on $m$ , the rotation maps $m$ to a parallel line through $C$ , which must be $k$ $m$ $n$ by the uniqueness of parallels.
(d): Thus, the rotation maps $\angle 2$ to $\angle 1$ $\angle 2$ $\angle 3$ $\angle 4$ . These alternate interior angles must be congruent because the rotation preserves angle measures.

$\blacksquare$

Note: The congruence of corresponding angles now follows from the congruence of vertical angles. But the next problem is another approach that uses a translation.

Proof

Given that parallel lines $m$ and $n$ are cut by transversal $k$ , prove that corresponding angles are congruent.

Let $B$ and $C$ be the intersections of transversal $k$ with lines $m$ and $n$ , respectively.

(a): Translate to the right along line $k$ by distance $BC$ , which takes $k$ to itself $m$ $n$ .
(b): The translation maps $B$ to $\answer {C}$ , and it maps $m$ to because the translation maintains parallels, and there is a unique parallel to $m$ through $C$ .
(c): The translation maps $\angle 1$ to $\angle 1$ $\angle 2$ $\angle 3$ $\angle 4$ . These corresponding angles must be congruent because the translation preserves angle measures.

$\blacksquare$

If two lines are cut by a transversal so that alternate interior angles are congruent, then the lines are parallel.

This theorem is the $\answer [format=string]{converse}$ of the previous theorem about alternate interior angles.

Proof

Given that $m$ and $n$ are cut by transversal $k$ with alternate interior angles congruent, prove that lines $m$ and $n$ are parallel.

Let $B$ and $C$ be the intersections of transversal $k$ with lines $m$ and $n$ , respectively. Let $P$ be the midpoint of $\overline {BC}$ .

(a): Rotate $180^\circ$ about $P$ , which takes $k$ to , and which swaps $B$ and $\answer {C}$ because distances are preserved.
(b): Because $\angle 2 \cong \angle 3$ and because a side of $\angle 2$ (i.e., $\overrightarrow {BP}$ ) is mapped to a side of $\angle 3$ (i.e., ), it must be that the other side of $\angle 2$ (which lies on $m$ ) is mapped to the other side of $\angle 3$ (which lies on line $\answer {n}$ ). Thus, $n$ is the image of $m$ .
(c): Because $P$ is not on $m$ , the $180^\circ$ rotation maps $m$ to a parallel line through $C$ . Thus, $n$ must be parallel to $m$ .

$\blacksquare$