Angles between parallel lines

When two lines cross, they make four angles around the crossing point $\displaystyle O$:

• $\displaystyle \alpha$ and $\displaystyle \beta$ are adjacent angles: they share an arm, and their other two arms make one straight line. They add up to $\displaystyle 180^{\circ}$.
• $\displaystyle \alpha$ and $\displaystyle \gamma$ are vertical angles: the arms of one are the extensions of the arms of the other. They are equal.
• Why vertical angles are equal: $\displaystyle \alpha$ and $\displaystyle \gamma$ are both adjacent to $\displaystyle \beta$, so each of them is $\displaystyle 180^{\circ}-\beta$.

In the figure the line $\displaystyle EF$ cuts the lines $\displaystyle AB$ and $\displaystyle CD$ at the points $\displaystyle P$ and $\displaystyle Q$.
There are four angles at each crossing – eight angles in all. Pairs of angles, one at $\displaystyle P$ and one at $\displaystyle Q$, have names:

• Corresponding angles – in the same place at the two crossings: $\displaystyle \alpha$ and $\displaystyle \beta$ (both above their line and to the right of the transversal).
• Alternate angles – both between the two lines, on opposite sides of the transversal: $\displaystyle \gamma$ and $\displaystyle \beta$.
• Co-interior angles – both between the two lines, on the same side of the transversal: $\displaystyle \gamma$ and $\displaystyle \delta$.

When a line cuts two parallel lines:

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

Parallel lines have the same direction, and all three parts of the theorem follow from that.

• Corresponding: we slide the line $\displaystyle AB$ along the transversal, without turning it, until it passes through $\displaystyle Q$. Its direction has not changed, so it lies exactly on $\displaystyle CD$. The angle $\displaystyle \alpha$ moves with it and lands on $\displaystyle \beta$: $\displaystyle {\alpha}={\beta}$.
• Alternate: $\displaystyle {\gamma}={\alpha}$, since they are vertical angles, and $\displaystyle {\alpha}={\beta}$. So $\displaystyle {\gamma}={\beta}$.
• Co-interior: $\displaystyle {\beta+\delta}={180^{\circ}}$, since they are adjacent angles, and $\displaystyle {\gamma}={\beta}$. So $\displaystyle {\gamma+\delta}={180^{\circ}}$.

In the figure above, $\displaystyle AB\parallel CD$ and $\displaystyle {\alpha}={112^{\circ}}$.
We find all eight angles.

• At $\displaystyle P$: $\displaystyle {\gamma}={112^{\circ}}$, since it is vertical to $\displaystyle \alpha$. The other two angles are adjacent to $\displaystyle \alpha$, so each is $\displaystyle {180^{\circ}-112^{\circ}}={68^{\circ}}$.
• At $\displaystyle Q$: $\displaystyle {\beta}={112^{\circ}}$, since it corresponds to $\displaystyle \alpha$. From here, as at $\displaystyle P$: the angle vertical to $\displaystyle \beta$ is $\displaystyle 112^{\circ}$, and $\displaystyle \delta$ and the angle vertical to it are $\displaystyle 68^{\circ}$.

At each crossing there are only two sizes of angle, and they add up to $\displaystyle 180^{\circ}$.

In the figure above: are the angles in each pair corresponding, alternate or co-interior?

$\displaystyle \angle APQ$ and $\displaystyle \angle PQD$

$\displaystyle \angle BPQ$ and $\displaystyle \angle PQD$

$\displaystyle \angle APQ$ and $\displaystyle \angle CQF$

$\displaystyle \angle BPQ$ and $\displaystyle \angle PQC$

In the figure, $\displaystyle AB\parallel CD$, and the line $\displaystyle EF$ cuts them at the points $\displaystyle P$ and $\displaystyle Q$.
$\displaystyle {\angle EPA}={68^{\circ}}$.
Find the angles, and give the reasons.

$\displaystyle \angle BPQ$

$\displaystyle \angle PQC$

$\displaystyle \angle DQF$

$\displaystyle \angle PQD$

A line cuts two parallel lines.
Two corresponding angles are $\displaystyle 5x-8$ and $\displaystyle 4x+16$ degrees.
Write an equation and solve it.
Find the sizes of the angles.

A line cuts two parallel lines.
Two co-interior angles are $\displaystyle x$ and $\displaystyle 2x-24$ degrees.
Write an equation and solve it.
Find the sizes of the angles.

In the figure, $\displaystyle AB\parallel CD$.
$\displaystyle {\angle BME}={34^{\circ}}$ and $\displaystyle {\angle END}={38^{\circ}}$.
Find $\displaystyle \angle MEN$, and give the reasons.
Hint: draw a line through $\displaystyle E$ parallel to $\displaystyle AB$.