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.
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°
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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