Parallel lines: the converse theorems

In eighth grade you learned what happens when a line cuts two parallel lines. In the figure the line $\displaystyle EF$ cuts the parallel lines $\displaystyle AB$ and $\displaystyle CD$:

• Corresponding angles are equal: $\displaystyle {\alpha}={\beta}$. They sit in the same place at the two crossings.
• Alternate angles are equal: $\displaystyle {\gamma}={\beta}$. Both are between the lines, on opposite sides of the transversal.
• Co-interior angles add up to $\displaystyle 180^{\circ}$: $\displaystyle {\gamma+\delta}={180^{\circ}}$. Both are between the lines, on the same side of the transversal.

Today the other way round: from the angles, you show that the lines are parallel.

Theorem: a line cuts two lines. If one of these three conditions holds, the lines are parallel:

• two corresponding angles are equal;
• two alternate angles are equal;
• two co-interior angles add up to $\displaystyle 180^{\circ}$.

The proof below supposes that the alternate angles $\displaystyle \gamma$ and $\displaystyle \beta$ are equal but the lines are not parallel, meeting at a point $\displaystyle S$ as in the figure – and reaches a contradiction.

• Triangle $\displaystyle PQS$ is formed. The angle $\displaystyle \gamma$ is an exterior angle of it, and $\displaystyle \beta$ is an interior angle not adjacent to it.
• An exterior angle equals the sum of the two interior angles not adjacent to it, so it is larger than each of them: $\displaystyle \gamma>\beta$.
• That contradicts $\displaystyle {\gamma}={\beta}$. So the lines do not meet – they are parallel.

The other two conditions lead to the same equal alternate angles (in the figure at the top of the lesson):

• Equal corresponding angles, $\displaystyle {\alpha}={\beta}$: also $\displaystyle {\alpha}={\gamma}$, since they are vertical angles. So $\displaystyle {\gamma}={\beta}$.
• Co-interior angles, $\displaystyle {\gamma+\delta}={180^{\circ}}$: also $\displaystyle {\beta+\delta}={180^{\circ}}$, since they are adjacent angles. So $\displaystyle {\gamma}={\beta}$.

In the figure at the top of the lesson, without knowing beforehand that the lines are parallel:

• If $\displaystyle {\angle EPB}={72^{\circ}}$ and $\displaystyle {\angle PQD}={72^{\circ}}$: these are equal corresponding angles, so $\displaystyle AB\parallel CD$.
• If $\displaystyle {\angle BPQ}={115^{\circ}}$ and $\displaystyle {\angle PQD}={65^{\circ}}$: these are co-interior angles, and $\displaystyle {115^{\circ}+65^{\circ}}={180^{\circ}}$, so $\displaystyle AB\parallel CD$.
• If $\displaystyle {\angle APQ}={48^{\circ}}$ and $\displaystyle {\angle PQD}={50^{\circ}}$: these are alternate angles, and they are not equal. If the lines were parallel, they would be equal (the eighth-grade theorem). So $\displaystyle AB$ and $\displaystyle CD$ are not parallel.

The segments $\displaystyle AD$ and $\displaystyle BC$ cross at the point $\displaystyle O$, the midpoint of each of them.
Given: $\displaystyle {AO}={DO}$ and $\displaystyle {BO}={CO}$.
Prove: $\displaystyle AB\parallel CD$.

The proof below shows that triangles $\displaystyle AOB$ and $\displaystyle DOC$ are congruent; the congruence gives two equal alternate angles:

In the figure the line $\displaystyle EF$ cuts $\displaystyle AB$ and $\displaystyle CD$.
In each part: is $\displaystyle AB\parallel CD$? Give the reason.

$\displaystyle {\angle BPQ}={70^{\circ}}$ and $\displaystyle {\angle PQD}={110^{\circ}}$.

$\displaystyle {\angle APQ}={64^{\circ}}$ and $\displaystyle {\angle PQD}={64^{\circ}}$.

$\displaystyle {\angle EPB}={75^{\circ}}$ and $\displaystyle {\angle CQF}={85^{\circ}}$.

$\displaystyle {\angle APQ}={108^{\circ}}$ and $\displaystyle {\angle PQD}={72^{\circ}}$.

In triangle $\displaystyle ABC$, $\displaystyle {AB}={AC}$.
The side $\displaystyle BA$ is extended beyond $\displaystyle A$ to the point $\displaystyle E$, and $\displaystyle AD$ bisects the angle $\displaystyle EAC$.
Prove: $\displaystyle AD\parallel BC$.

In triangle $\displaystyle ABC$, $\displaystyle {AB}={AC}$.
The point $\displaystyle D$ is on $\displaystyle AB$, the point $\displaystyle E$ on $\displaystyle AC$, and $\displaystyle {AD}={AE}$.
Prove: $\displaystyle DE\parallel BC$.