SAT Parallel Lines

To successfully solve problems about parallel lines on the SAT, you need to know:

• angles on the SAT are always measured in degrees
• the definition of complementary and supplementary angles
• the definition of vertical angles
• the definition of an angle bisector
• the angle addition postulate
• the Properties of Parallel Lines:

Consider the two parallel lines $m$ and $n,$ and the transversal, $p.$

Interior angles are angles located on the inside of the parallel lines: $\angle 3, \angle 4, \angle 5,\ \text {and}\ \angle 6.$

Exterior angles are angles located on the outside of the parallel lines: $\angle 1, \angle 2, \angle 7,\ \text{and}\ \angle 8.$

Corresponding angles are two angles located in the same position relative to the parallel lines. Corresponding angles are congruent.

$$\angle 1 \cong \angle 5, \qquad \angle 2 \cong \angle 6, \qquad \angle 3 \cong \angle 7, \qquad \angle 4 \cong \angle 8$$

Alternate-interior angles are two nonadjacent interior angles located on opposite sides of the transversal. Alternate-interior angles are congruent.

$$\angle 3 \cong \angle 5 \qquad \text{and} \qquad \angle 4 \cong \angle 6$$

Alternate-exterior angles are two nonadjacent exterior angles located on opposite sides of the transversal. Alternate-exterior angles are congruent.

$$\angle 1 \cong \angle 7 \qquad \text{and} \qquad \angle 2 \cong \angle 8$$

Same-side interior angles are two nonadjacent interior angles located on the same side of the transversal. Same-side interior angles are supplementary.

$$m\angle 3 + m\angle 6 = 180^\circ \qquad \text{and} \qquad m\angle 4 + m\angle 5 = 180^\circ$$

Same-side exterior angles are two nonadjacent exterior angles located on the same side of the transversal.

$$m\angle 1 + m\angle 8 = 180^\circ \qquad\text{and} \qquad m\angle 2 + m\angle 7 = 180^\circ$$

In the figure above, $m \parallel n$ and $p \parallel q.$ If the measure of $\angle 1$ is $x^\circ,$ how many of the other angles also measure $x^\circ?$

(A) $\ \ 1$
(B) $\ \ 3$
(C) $\ \ 7$
(D) $\ \ 8$
(E) $\ \ 15$

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Correct Answer: C

Solution:

Tip: Know the Properties of Parallel Lines.
Tip: Vertical angles are congruent.
There are many ways to go about solving this problem, depending on which cluster of angles you start with and which properties of parallel lines you choose to use. We show one way.

$$\begin{array}{r c l l} \angle 1 &\cong& \angle 4 &\quad \text{vertical angles}\\ \angle 1 &\cong& \angle 5 &\quad \text{corresponding angles}\\ \angle 5 &\cong& \angle 8 &\quad \text{vertical angles}\\ \angle 5 &\cong& \angle 13 &\quad \text{corresponding angles}\\ \angle 13 &\cong& \angle 15 &\quad \text{vertical angles}\\ \angle 13 &\cong& \angle 9 &\quad \text{corresponding angles}\\ \angle 9 &\cong& \angle 12 &\quad \text{vertical angles}\\ \end{array}$$

There are seven angles congruent to $\angle 1$: $\angle 4, \angle 5, \angle 8, \angle 9, \angle 12, \angle 13,\ \text{and}\ \angle 15,$ and therefore seven other angles measure $x^\circ.$ In the diagram below, $\angle 1$ is colored in green, the angles congruent to it are colored in gray.

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Incorrect Choices:

(A)
If you think only $\angle 4$ is congruent to $\angle 1,$ you will get this wrong answer.

(B)
If you only consider 2 clusters of angles, not all 4, you will get this wrong answer.

(D)
If you count $\angle 1$ among the angles congruent to $\angle 1,$ you will get this wrong answer.

(E)
If you think all angles are congruent to $\angle 1,$ you will get this wrong answer. Note, if all the angles were $90^\circ,$ this choice would be correct, but the picture is drawn to scale, and the angles clearly aren't all equal to $90^\circ.$

Here is a counter-example: if $m \angle 1 = 30^\circ,$ $m \angle 1 + m \angle 2 = 180^\circ,$ then $m\angle 2 = 150^\circ.$ But $30 \neq 150,$ and so this choice is wrong.

In the figure above, $m \parallel n$ and $\overline{CD}$ bisects $\angle BCE.$ Which of the following is the value of $x?$

(A) $\ \ 30$
(B) $\ \ 40$
(C) $\ \ 50$
(D) $\ \ 60$
(E) $\ \ 70$

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Correct Answer: C

Solution 1:

Tip: Know the Properties of Parallel Lines.
Tip: The angles in a triangle sum to $180^\circ.$
Tip: The angle bisector divides an angle in half.
This solution applies the Properties of Parallel Lines and it uses the fact that the angles in a triangle add up to $180^\circ.$

The angle in the diagram that measures $70^\circ$ and $\angle DCE$ are corresponding angles. Therefore they are congruent and

$$m \angle DCE = 70^\circ.$$

$\overline{CD}$ is an angle bisector, and therefore

$$m\angle BCE = 2\cdot m \angle DCE = 2 \cdot 70^\circ = 140^\circ.$$

$\angle BCA$ and $\angle BCE$ are supplementary angles. Therefore they add up to $180^\circ,$ and

$$m \angle BCA = 180^\circ - 140^\circ = 40^\circ.$$

The angles in a triangle add up to $180^\circ,$ and so

$$\begin{aligned} m \angle BAC + m \angle ABC + m\angle BCA &= 180^\circ\\ 90^\circ + x^\circ + 40^\circ &=180^\circ\\ x^\circ + 130^\circ &=180^\circ\\ x^\circ &=50^\circ \end{aligned}$$

Solution 2:

Tip: Know the Properties of Parallel Lines.
Tip: The angle bisector divides an angle in half.
This solution uses the Properties of Parallel Lines. It begins in the same way that Solution 1 does, but it ends differently.

The angle in the diagram that measures $70^\circ$ and $\angle DCE$ are corresponding angles. Therefore they are congruent and

$$m \angle DCE = 70^\circ.$$

$\overline{CD}$ is an angle bisector, and therefore

$$m\angle BCE = 2\cdot m \angle DCE = 2 \cdot 70^\circ = 140^\circ.$$

$\angle BCA$ and $\angle BCE$ are supplementary angles. Therefore they add up to $180^\circ,$ and

$$m \angle BCA = 180^\circ - 140^\circ = 40^\circ.$$

$\angle BCA$ and $\angle CBD$ are alternate interior angles. Therefore they are congruent and

$$m\angle CBD = m\angle BCA = 40^\circ.$$

$\angle BAC$ and $\angle ABD$ are same-side interior angles. Therefore, they are supplementary and their measures add up to $180^\circ.$ Since $m\angle BAC = 90^\circ,$ then

$$m\angle ABD = 90^\circ.$$

$m\angle ABD = m\angle ABC + m\angle CBD.$ Therefore,

$$m\angle ABC = m\angle ABD - m\angle CBD = 90^\circ - 40^\circ = 50^\circ = x^\circ.$$

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Incorrect Choices:

(A)
This answer is just meant to confuse you.

(B)
This is $m\angle BCA$ and $m\angle CBD,$ not $m\angle ABC.$

(D)
$m\angle ABC$ looks like it may be $60^\circ.$ If you guess, you might get this wrong answer.

(E)
Tip: Just because a number appears in the question doesn’t mean it is the answer.

If you thought these examples difficult and you need to review the material, these links will help:

• Perpendicular Lines
• Parallel Lines
• Angles and Lines Problem Solving
• Types of Angles
• Finding Angle Measures
• Complementary and Supplementary Angles
• Vertical Angles
• SAT Lines and Angles

• Know the Properties of Parallel Lines.
• Angles on a line sum to $180^\circ.$
• $\angle A$ and $\angle B$ are complementary if $m\angle A + m\angle B=90^\circ.$
• $\angle A$ and $\angle B$ are supplementary if $m\angle A + m\angle B=180^\circ.$
• Vertical angles are congruent.
• The angle bisector divides an angle in half.
• Angles in a triangle sum to $180^\circ.$
• The two acute angles in a right triangle are complementary.
• An exterior angle in a triangle equals the sum of the two nonadjacent interior angles.
• If a diagram is drawn to scale, trust it.
• SAT General Tips