INTERIOR ANGLES ON THE SAME SIDE OF THE TRANSVERSAL THEOREM

Theorem :

If two parallel lines are cut by a transversal, then interior angles on the same side of the transversal are supplementary.

Given : m||n, p is transversal.

Prove : ∠4 and ∠5 are supplementary and ∠3 and ∠6 are supplementary.

Statement

m||n, p is transversal. 

Reason

Given

∠1 & ∠4 - linear pair

∠2 & ∠3 - linear pair


Definition of linear pair

∠1 & ∠4 - Supplementary

m∠1 + m∠4  =  180°

∠2 & ∠3 - Supplementary

m∠2 + m∠3  =  180°



Supplementary Postulate

∠1  ∠5 and ∠2  ∠6

Corresponding Angles Theorem

∠4 & ∠5 - Supplementary

∠3 & ∠6 - Supplementary


Substitution Property

Consecutive Interior Angles :

If two lines are cut by a transversal, the pair of angles on the same side of the transversal and inside the two lines are called consecutive interior angles.

Interior Angles on the Same Side of the Transversal Theorem – Converse

When two lines are cut by a transversal, if interior angles on the same side of the transversal add up to to 180°, then the two lines are parallel.

In the figure above, lines m and n are parallel. Because, a pair of angles on the same side of the transversal p and inside the two lines m and n are supplementary.

110° + 70°  =  180°

Solved Problems

Problem 1 :

In the figure shown below, m∠3 = 105°. Find the measure of ∠6.

Solution :

In the figure above, ∠3 and ∠6 are on the same side of the transversal p and inside inside the parallel lines m and n

m∠3 + m∠6  =  180°

Substitute m∠3 = 105°. 

105° + m∠6  =  180°

Subtract 105° from each side. 

m∠6  =  75°

Problem 2 :

In the figure shown below, m∠3 = 102°. Find the measures ∠6, ∠12 and ∠13.

Solution :

In the figure above, lines m and n are parallel, p and q are parallel.

∠3 and ∠6 are on the same side of the transversal p and inside the parallel lines m and n.

m∠3 + m∠6  =  180°

Substitute m∠3 = 102°. 

102° + m∠6  =  180°

Subtract 102° from each side. 

m∠6  =  78°

∠3 and ∠12 are on the same side of the transversal m and inside the parallel lines p and q.

m∠3 + m∠12  =  180°

Substitute m∠3 = 102°. 

102° + m∠12  =  180°

Subtract 102° from each side. 

m∠12  =  78°

∠12 and ∠13 are on the same side of the transversal q and inside the parallel lines m and n.

m∠12 + m∠13  =  180°

Substitute m∠12 = 78°. 

78° + m∠13  =  180°

Subtract 78° from each side. 

m∠13  =  102°

Therefore, 

m∠6  =  78°

m∠12  =  78°

m∠13  =  102°

Problem 3 :

In the figure shown below, lines m and n are parallel and p is transversal. Find the value of x.

Solution :

In the figure above 5x° and (3x + 28)° are on the same side of the transversal p and inside the parallel lines m and n.

5x° + (3x + 28)°  =  180°

5x + 3x + 28  =  180

8x + 28  =  180

Subtract 28 from each side. 

8x  =  152

Divide each side by 8.

x  =  19

Problem 4 :

Using a 3rd parallel Line – Auxiliary Line, find the value of x.

Solution :

In the figure above, a° and 50° are corresponding angles and they are equal.

a°  =  50°

b° and 100° are on the same side of the transversal and inside the parallel lines.

b° + 100°  =  180°

Subtract 100° from each side. 

b°  =  80°

In the figure above, 

x  =  a + b

=  50 + 80

=  130

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