Angles in Parallel Lines and Transversals Problems

Problems related to angles made by parallel lines and transversals are presented below along with detailed solutions. Review angles in parallel lines and transversals for the underlying concepts.

Problem 1

In the figure below, \( AA' \parallel CC' \). The measure of \( \angle A'AB \) is \( 135^\circ \) and the measure of \( \angle C'CB \) is \( 147^\circ \). Find \( \angle ABC \).

Problem 2

In the figure below, lines \( A'A'' \parallel C'C'' \). \( AB \) is the bisector of \( \angle CAA'' \) and \( BC \) is the bisector of \( \angle ACC'' \). Show that \( \angle ABC = 90^\circ \).

Problem 3

In the figure below, lines \( BC \parallel DD' \). The measure of angle \( x \) is \( 127^\circ \) and the measure of angle \( y \) is \( 115^\circ \). Find all interior angles of \( \triangle ADD' \).

Solutions

Solution to Problem 1

Construct \( BB' \) parallel to \( AA' \) and \( CC' \).

Then \( \angle ABC = \angle ABB' + \angle CBB' \).

Since \( AA' \parallel BB' \), \( \angle ABB' \) and \( \angle w' \) are alternate interior angles, so \( \angle ABB' = \angle w' \).

Since \( CC' \parallel BB' \), \( \angle CBB' \) and \( \angle z' \) are alternate interior angles, so \( \angle CBB' = \angle z' \).

Angles \( w \) and \( w' \) are supplementary: \[ \angle w' = 180^\circ - \angle w = 180^\circ - 135^\circ = 45^\circ \]

Angles \( z \) and \( z' \) are supplementary: \[ \angle z' = 180^\circ - \angle z = 180^\circ - 147^\circ = 33^\circ \]

Therefore: \[ \angle ABC = \angle w' + \angle z' = 45^\circ + 33^\circ = 78^\circ \]

Solution to Problem 2

Since \( A'A'' \parallel C'C'' \), \( \angle A'AC \) and \( \angle ACC'' \) are alternate interior angles, so: \[ \angle A'AC = \angle ACC'' \]

Angles \( \angle A'AC \) and \( \angle A''AC \) are supplementary: \[ \angle A''AC = 180^\circ - \angle A'AC = 180^\circ - \angle ACC'' \]

Rearranging: \[ \angle A''AC + \angle ACC'' = 180^\circ \]

Since \( AB \) and \( CB \) are angle bisectors, in \( \triangle ABC \): \[ \angle ABC = 180^\circ - \frac{\angle A''AC + \angle ACC''}{2} = 180^\circ - \frac{180^\circ}{2} = 90^\circ \]

Solution to Problem 3

Angles \( x \) and \( \angle ABC \) are supplementary: \[ \angle ABC = 180^\circ - x = 180^\circ - 127^\circ = 53^\circ \]

Angles \( y \) and \( \angle ACB \) are supplementary: \[ \angle ACB = 180^\circ - y = 180^\circ - 115^\circ = 65^\circ \]

Since \( BC \parallel DD' \), \( \angle ADD' \) and \( \angle ABC \) are corresponding angles: \[ \angle ADD' = \angle ABC = 53^\circ \]

Similarly, \( \angle AD'D \) and \( \angle ACB \) are corresponding angles: \[ \angle AD'D = \angle ACB = 65^\circ \]

In \( \triangle ADD' \), the sum of angles is \( 180^\circ \): \[ \angle DAD' = 180^\circ - (\angle ADD' + \angle AD'D) = 180^\circ - (53^\circ + 65^\circ) = 62^\circ \]

Thus the interior angles of \( \triangle ADD' \) are \( 62^\circ \), \( 53^\circ \), and \( 65^\circ \).

More References

• Parallel Lines and Angles
• Intercept Theorem and Problems with Solutions
• Geometry Tutorials and Problems
• Triangle Problems