Solving Problems with Angles in Parallel Lines

Solving problems with angles in parallel lines is like solving a murder mystery.  One clue leads on to the next and the next until the murderer is found.  However, it doesn’t end there.  The detectives need to explain their reasoning in court using the relevant laws and procedures should the murderer plead not guilty.  If their reasoning is flawed or not sufficiently precise the accused could be acquitted.

A similar process takes place when students are tasked with solving problems with angles in parallel lines.  The better students understand and can apply the various angle properties the more likely they are to find the value of the first angle which would lead on to the next angle and so on until the problem is solved.  However, it is not enough to solve the problem.  It is becoming more common nowadays for students to have to explain their reasoning.   If their explanation is not sufficiently detailed or precise communication marks are lost.

Investigations in to angles in parallel lines

When I teach solving problems with angles in parallel lines I encourage students to explain their reasoning from the outset.  I do this through mini investigations.  First, I ask students to draw three sets of parallel lines with a transversal intersecting each pair.

Solving Problems with Angles in Parallel Lines

For the alternate angle property students highlight a Z shaped path to identify and measure the internal angles.  They then investigate and describe the relationship between the two angles.  For corresponding angles we draw an F shape path and for interior we use a C shaped path.

The challenge here is not necessarily measuring the angles correctly or even identifying the relationship.  The alternate and corresponding angle properties are often easily found.  The interior property is less easy but still not particularly difficult.

The challenge for the students lies in describing the relationship on paper.  To help with this I encourage students to talk with each other and come up with a description they agree is both concise and mathematically accurate.  After a short class discussion we come up with these descriptions.


Solving Problems with Angles in Parallel Lines

Corresponding angles HBC and BEF appear in a F shape and are equal.


Solving Problems with Angles in Parallel Lines

Alternate angles DEB and EBC appear in a Z shape and are equal.


Solving Problems with Angles in Parallel Lines

Interior angles EBC and FEB appear in a C shape and add up to 180°.


Solving Problems with Angles in Parallel Lines

At this point students begin to use the angle properties to calculate unknown angles.  By writing down the angle property they have applied, students justify their reasoning.  As learning progresses students have to apply multiple angle properties within a single problem.  To add further challenge I ask some students to solve problems using more than one approach.


Solving Problems with Angles in Parallel Lines

76° and i are corresponding therefore equal.

h and i are alternate therefore equal.

h and j are interior therefore have a sum of 180°


Solving Problems with Angles in Parallel Lines

76° and h are vertically opposite therefore equal.

h and j are interior therefore have a sum of 180°

i and j lie along a straight line therefore have a sum of 180°


Using Proof to Cement Understanding

Towards the end of the lesson students are challenged to prove the alternate, corresponding and interior angle properties.  The purpose of this is to consolidate the need for accurate notation when referring to an angle and to help students see how the various properties are connected.  In my opinion being able to prove corresponding, alternate and interior properties is key to solving problems with angles in parallel lines.


Corresponding Angles

HBC + EBC = 180° (angles on a straight line)

EBC = 180 – x

EBC + BEF = 180° (interior angles add to 180°)

180 – x + y = 180

y = x


Alternate Angles

EBC + BEF = 180° (interior angles add to 180°)

BEF = 180 – x

BEF + DEB = 180° (angles on a straight line)

180 – x + y = 180

x = y


Interior Angles

EBC = DEB (alternate angles)

DEB + BEF = 180° (angles on a straight line)

x + y = 180°


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