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In this lesson, some theorems about lines and angles will be explored and proven. The theorems will be applied using real-life examples.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Parallel lines
Perpendicular lines
Vertical angles
Supplementary angles
Rigid motions
Corresponding angles
Alternate interior angles
Perpendicular bisector

Explore

Investigating Properties of Vertical Angles

Before any theorems will be introduced, try to discover some properties of angles using the interactive applet. While exploring, think about how those properties could be proven. The applet allows for translations and rotations of the angles. Consider a pair of vertical angles.

Is it possible to map one angle onto another by performing a transformation?
What conjecture can be made about any two vertical angles?

Extra

How to Use the Applet

To translate an angle, click on the sides of the angle or in the region between the sides.
To rotate an angle about $P,$ move the corresponding slider.

Discussion

Congruence of Vertical Angles

Vertical angles can be mapped onto each other by using a rotation. Since rotations are rigid motions, the angle measures are preserved. This leads to the following theorem.

Rule

Vertical Angles Theorem

Vertical angles are always congruent.

Two intersecting lines that form two pairs of vertical angles

Based on the characteristics of the diagram, the following relations hold true.

$∠ 1 ≅ ∠ 3$
$∠ 2 ≅ ∠ 4$

Proof

Geometric Approach

Analyzing the diagram, it can be seen that $∠ 1$ and $∠ 2$ form a straight angle, so these are supplementary angles. Similarly, $∠ 2$ and $∠ 3$ are also supplementary angles.

Therefore, by the Angle Addition Postulate, the sum of $m ∠ 1$ and $m ∠ 2$ is $18 0^{\circ},$ and the sum of $m ∠ 2$ and $m ∠ 3$ is also $18 0^{\circ} .$ These facts can be used to express $m ∠ 2$ in terms of $m ∠ 1$ and in terms of $m ∠ 3 .$

Angle Addition Postulate	Isolate $m ∠ 2$
$m ∠ 1 + m ∠ 2$ $=$ $18 0^{\circ}$	$m ∠ 2$ $=$ $18 0^{\circ} - m ∠ 1$
$m ∠ 2 + m ∠ 3$ $=$ $18 0^{\circ}$	$m ∠ 2$ $=$ $18 0^{\circ} - m ∠ 3$

By the Transitive Property of Equality, the expressions representing

m ∠ 2

can be set equal to each other.

{m ∠ 2 = 18 0^{\circ} - m ∠ 1 m ∠ 2 = 18 0^{\circ} - m ∠ 3 ⇓ 18 0^{\circ} - m ∠ 1 = 18 0^{\circ} - m ∠ 3

Then the equation can be simplified.

18 0^{\circ} - m ∠ 1 = 18 0^{\circ} - m ∠ 3

SubEqn

$LHS - 18 0^{\circ} = RHS - 18 0^{\circ}$

- m ∠ 1 = - m ∠ 3

MultEqn

$LHS \cdot (- 1) = RHS \cdot (- 1)$

m ∠ 1 = m ∠ 3

By the definition of congruent angles, this means that the vertical angles

∠ 1

and

∠ 3

are congruent angles. The same process can be used to prove

∠ 2

and

∠ 4

congruent.

Two-Column Proof

The previous proof can be summarized in the following two-column table.

Statements	Reasons
$ℓ_{1}$ and $ℓ_{2}$ lines	Given
$∠ 1$ and $∠ 2$ supplementary	Definition of straight angle
$m ∠ 1 + m ∠ 2 = 18 0^{\circ}$	Definition of supplementary angles
$m ∠ 2 = 18 0^{\circ} - m ∠ 1$	Subtraction Property of Equality
$∠ 2$ and $∠ 3$ supplementary	Definition of straight angle
$m ∠ 2 + m ∠ 3 = 18 0^{\circ}$	Definition of supplementary angles
$m ∠ 2 = 18 0^{\circ} - m ∠ 3$	Subtraction Property of Equality
$18 0^{\circ} - m ∠ 1 = 18 0^{\circ} - m ∠ 3$	Transitive Property of Equality
$m ∠ 1 = m ∠ 3$	Subtraction and Multiplication Properties of Equality

Proof

Using Transformations

Consider the points $A,$ $B,$ $C,$ and $D$ on each ray that starts at the point of intersection $E$ of the two lines.

Suppose that points

A

and

B

are rotated

18 0^{\circ}

about point

E .

The points

A

and

B

are mapped onto the points

A^{'}

and

B^{'}

after the rotation. This means that

∠ A E B

is mapped onto

∠ A^{'} E B^{'} .

Since rotations are a rigid motion,

∠ A E B

and

∠ A^{'} E B^{'}

are congruent angles.

∠ A E B ≅ ∠ A^{'} E B^{'}

Since the point

A^{'}

lies on

E C

and point

B^{'}

lies on

E D,

∠ A^{'} E B^{'}

is congruent to

∠ C E D .

∠ A^{'} E B^{'} ≅ ∠ C E D

By applying the Transitive Property of Congruence, it can be confirmed that

∠ A E B

is congruent to

∠ C E D .

{∠ A E B ≅ ∠ A^{'} E B^{'} ∠ A^{'} E B^{'} ≅ ∠ C E D ⇓ ∠ A E B ≅ ∠ C E D

Example

Solving Problems Using Vertical Angles

In Flowerland Village, there is a crossroad between Tulip Street and Rose Street. There is a plan to continue the construction of Tulip Street toward the southwest. At the moment, the crossroad forms two angles, whose measures are expressed by

3 x + 14

and

6 x - 5,

respectively.

A crossroad before and after the construction

Find the measures of all four angles the crossroad will form after the construction of Tulip Street is finished.

Hint

Use the given expressions to form an equation for $x .$ Identify the relationship between $∠ 1$ and $∠ 3,$ as well as $∠ 2$ and $∠ 4$ by analyzing their positions.

Solution

The angles formed by the crossroad before the construction form a linear pair. Therefore, they are supplementary angles. Using this information, the following equation can be formed.

6 x - 5 + 3 x + 14 = 18 0^{\circ}

By solving it, the value of

x

can be determined.

6 x - 5 + 3 x + 14 = 18 0^{\circ}

▼

Solve for

x

AddSubTerms

Add and subtract terms

9 x + 9 = 18 0^{\circ}

SubEqn

$LHS - 9 = RHS - 9$

9 x = 17 1^{\circ}

DivEqn

$LHS / 9 = RHS / 9$

x = 1 9^{\circ}

Now that the value of

x

is known, the measures of

∠ 1

and

∠ 2

can be calculated.

$x = 19$
$m ∠ 1 = 3 x + 14$	$m ∠ 2 = 6 x - 5$
$m ∠ 1 = 3 (19) + 14$	$m ∠ 2 = 6 (19) - 5$
$m ∠ 1 = 57 + 14$	$m ∠ 2 = 114 - 5$
$m ∠ 1 = 7 1^{\circ}$	$m ∠ 2 = 10 9^{\circ}$

Next, by analyzing the position of

∠ 1

and

∠ 3,

as well as

∠ 2

and

∠ 4,

it can be noted that these are vertical angles. Therefore, by the Vertical Angles Theorem, they are two pairs of congruent angles.

m ∠ 1 = m ∠ 3 m ∠ 2 = m ∠ 4 \Rightarrow m ∠ 3 = 7 1^{\circ} m ∠ 4 = 10 9^{\circ}

In this way, it was obtained that

∠ 1

and

∠ 3

are each

7 1^{\circ},

and

∠ 2

and

∠ 4

are each

10 9^{\circ} .

Explore

Investigating Parallel Lines Cut by a Transversal

Consider two parallel lines cut by a transversal. The applet shows a pair of corresponding angles, $∠ A$ and $∠ B .$ Is it possible to translate one line so that one of these angles maps onto another?

What conjecture can be made about any two corresponding angles?

Discussion

Corresponding Angles Theorem and Its Converse

The observed relation between corresponding angles is presented and proven in the following theorem.

Rule

Corresponding Angles Theorem

If parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent.

Two parallel lines intersected by a transversal forming four pairs of corresponding angles

Based on the characteristics of the diagram, the following relations hold true.

If $ℓ_{1} ∥ ℓ_{2},$ then $∠ 1 ≅ ∠ 5,$ $∠ 2 ≅ ∠ 6,$ $∠ 3 ≅ ∠ 7,$ and $∠ 4 ≅ ∠ 8 .$

Proof

It is given that

ℓ_{1}

and

ℓ_{2}

are parallel lines. Therefore, by the definition of parallel lines, there is a translation that maps one line onto another.

It should be noted that when translating one line onto another, the pairs of corresponding angles overlap and seem to have the same measures. In fact, because a translation is a rigid motion, and a rigid motion preserves length and angle measures, the pairs of corresponding angles are congruent angles.

Note that the converse statement is also true.

Rule

Converse Corresponding Angles Theorem

If two lines and a transversal form corresponding angles that are congruent, then the lines are parallel.

Based on the characteristics of the diagram, the following relation holds true.

If $∠ 1 ≅ ∠ 5,$ $∠ 2 ≅ ∠ 6,$ $∠ 3 ≅ ∠ 7,$ or $∠ 4 ≅ ∠ 8,$ then $ℓ_{1} ∥ ℓ_{2} .$

Proof

This theorem can be proven by an indirect proof. Let $ℓ_{1}$ and $ℓ_{2}$ be two lines intersected by a transversal line $ℓ_{3}$ forming corresponding congruent angles $∠ 1$ and $∠ 2 .$

Since the goal is to prove that

ℓ_{1}

is parallel to

ℓ_{2}

, it will be temporarily assumed that

ℓ_{1}

and

ℓ_{2}

are not parallel.

Temporary Assumption ℓ_{1} ∦ ℓ_{2}

By the Parallel Postulate, there exists a line

n

parallel to

ℓ_{2}

that passes through the point of intersection between

ℓ_{1}

and

ℓ_{3} .

This line forms

∠ 3

and

∠ 4 .

By the Angle Addition Postulate,

m ∠ 1

is equal to the sum of

m ∠ 3

and

m ∠ 4 .

m ∠ 1 = m ∠ 3 + m ∠ 4

Since

n

and

ℓ_{2}

are parallel lines that are cut by a transversal, by the Corresponding Angles Theorem,

∠ 3

and

∠ 2

are congruent. By the definition of congruence, these angles have the same measure.

∠ 3 ≅ ∠ 2 ⇕ m ∠ 3 = m ∠ 2

By the Substitution Property of Equality,

m ∠ 2

can be substituted for

m ∠ 3

into the equation for

m ∠ 1 .

m ∠ 1 = m ∠ 3 + m ∠ 4 ↓ m ∠ 1 = m ∠ 2 + m ∠ 4

From the above equation and since

m ∠ 4

is a positive number, it can be concluded that

m ∠ 1

is greater than

m ∠ 2 .

m ∠ 1 > m ∠ 2

This contradicts the given fact that

∠ 1

and

∠ 2

are congruent. The contradiction came from assuming that

ℓ_{1}

and

ℓ_{2}

are not parallel lines. Therefore,

ℓ_{1}

and

ℓ_{2}

must be parallel lines.

Example

Solving Problems Using Corresponding Angles

In a Flowerland Village house, there are stairs with hand railings like shown in the diagram. The measures of $∠ 1$ and $∠ 2$ are expressed as $5 t - 2$ and $4 t + 12,$ respectively.

What are the measures of

∠ 1

and

∠ 2 ?

Hint

How do measures of $∠ 1$ and $∠ 2$ relate to each other? Use the given expressions to form an equation for $t .$

Solution

Analyzing the diagram, it can be noted that

∠ 1

and

∠ 2

are corresponding angles formed by two parallel lines and a transversal. Therefore, by the Corresponding Angles Theorem, these angles are congruent. Hence, the measures of these angles are the same.

∠ 1 m ∠ 1 ≅ ∠ 2 ⇓ = m ∠ 2

By substituting

m ∠ 1

with

5 t - 2

and

m ∠ 2

with

4 t + 12,

the equation for

t

can be formed.

m ∠ 1 = m ∠ 2

SubstituteII

$m ∠ 1 = 5 t - 2$ , $m ∠ 2 = 4 t + 12$

5 t - 2 = 4 t + 12

▼

Solve for

t

SubEqn

$LHS - 4 t = RHS - 4 t$

t - 2 = 12

AddEqn

$LHS + 2 = RHS + 2$

t = 14

Now that the value of

t

is known, the measure of each of the angles can be calculated.

m ∠ 1 = 5 t - 2

▼

Substitute

14

for

t

and evaluate

Substitute

$t = 14$

m ∠ 1 = 5 (14) - 2

Multiply

m ∠ 1 = 70 - 2

SubTerm

Subtract term

m ∠ 1 = 68

Since the angles are congruent, it can be concluded that they both measure to be

6 8^{\circ} .

m ∠ 1 = m ∠ 2 m ∠ 1 = 6 8^{\circ} \Rightarrow m ∠ 2 = 6 8^{\circ}

Discussion

Alternate Interior Angles Theorem and Its Converse

Like corresponding angles, alternate interior angles are also formed by two parallel lines cut by a transversal.

Rule

Alternate Interior Angles Theorem

If parallel lines are cut by a transversal, then alternate interior angles are congruent.

Two parallel lines cut by a transversal forming two pairs of congruent angles

Based on the characteristics of the diagram, the following relations hold true.

If $ℓ_{1} ∥ ℓ_{2},$ then $∠ 1 ≅ ∠ 2$ and $∠ 3 ≅ ∠ 4 .$

Proof

Geometric approach

To prove that alternate interior angles are congruent, it will be shown that $∠ 1$ and $∠ 2$ are congruent.

Notice that by definition

∠ 2

and

∠ 5

are vertical angles. By the Vertical Angles Theorem, they are therefore congruent angles.

∠ 2 ≅ ∠ 5

Furthermore, by definition

∠ 5

and

∠ 1

are corresponding angles. Hence, by the Corresponding Angles Theorem,

∠ 5

and

∠ 1

are also congruent angles.

∠ 5 ≅ ∠ 1

Applying the Transitive Property of Congruence,

∠ 2

and

∠ 1

can be concluded to be congruent angles as well.

{∠ 2 ≅ ∠ 5 ∠ 5 ≅ ∠ 1 \Rightarrow ∠ 2 ≅ ∠ 1

The same reasoning applies to the other pair of alternate interior angles. Therefore, when a pair of parallel lines is cut by a transversal, the pairs of alternate interior angles are congruent.

Two-Column Proof

The previous proof can be summarized in the following two-column table.

Statements	Reasons
$∠ 2$ and $∠ 5$ are vertical angles	Def. of vertical angles
$∠ 2 ≅ ∠ 5$	Vertical Angles Theorem
$∠ 5$ and $∠ 1$ are corresponding angles	Def. of corresponding angles
$∠ 5 ≅ ∠ 1$	Corresponding Angles Theorem
$∠ 2 ≅ ∠ 1$	Transitive Property of Congruence

Proof

Using Transformations

Apart from the points of intersection, consider two more points on each line.

Next,

A,

B,

C,

and

D

will be translated parallel to the transversal until the points

A,

C,

and

D

lie on

ℓ_{2} .

Then,

A,

B,

and

C

will be rotated

18 0^{\circ}

about

F .

It should be noted that since point

D

lies on the transversal, when translating it to

ℓ_{2}

the point will fall into the same position as

F .

Therefore,

D

will not be affected by the rotation around

F .

After this combination of rigid motions,

A,

B,

C,

and

D

are mapped onto

A^{'},

B^{'},

C^{'},

and

D^{'} .

This means that

∠ A D B

is mapped onto

∠ A^{'} D^{'} B^{'} .

Therefore,

∠ A D B

and

∠ A^{'} D^{'} B^{'}

are congruent angles.

∠ A D B ≅ ∠ A^{'} D^{'} B^{'}

Note that

D^{'}

and

F

share the same location. It can also be seen that

A^{'}

lies on

F G

and

B^{'}

lies on

F H .

Because of this,

∠ A^{'} D^{'} B^{'}

is congruent to

∠ G F H .

∠ A^{'} D^{'} B^{'} ≅ ∠ G F H

Applying the Transitive Property of Congruence,

∠ A D B

is congruent to

G F H .

{∠ A D B ≅ ∠ A^{'} D^{'} B^{'} ∠ A^{'} D^{'} B^{'} ≅ ∠ G F H ⇓ ∠ A D B ≅ ∠ G F H

The converse statement is also true.

Rule

Converse Alternate Interior Angles Theorem

If two lines and a transversal form alternate interior angles that are congruent, then the two lines are parallel.

Based on the characteristics of the diagram, the following relation holds true.

If $∠ 1 ≅ ∠ 2$ or $∠ 3 ≅ ∠ 4,$ then $ℓ_{1} ∥ ℓ_{2} .$

Proof

The proof will be based on the given diagram, but it holds true for any pair of lines cut by a transversal. Consider only one pair of congruent alternate interior angles and one more angle.

It needs to be proven that

ℓ_{1}

and

ℓ_{2}

are parallel lines. It is already given that

∠ 1

is congruent to

∠ 2 .

∠ 1 ≅ ∠ 2

The diagram shows that

∠ 2

and

∠ α

are vertical angles. By the Vertical Angles Theorem, these angles are congruent.

∠ 2 ≅ ∠ α

Notice the common angle of

∠ 2

in both relationships. By the Transitive Property of Congruence, since

∠ 1

is congruent to

∠ 2

and

∠ 2

is congruent to

∠ α,

then

∠ 1

is congruent

∠ α .

{∠ 1 ≅ ∠ 2 ∠ 2 ≅ ∠ α ⇓ ∠ 1 ≅ ∠ α

The diagram also shows that

∠ 1

and

∠ α

are corresponding angles. Given that relation, the Converse Corresponding Angles Theorem can be applied.

Converse Corresponding Angles Theorem
If two lines are cut by a transversal so that the corresponding angles are congruent, then the lines are parallel.

Since $∠ 1$ and $∠ α$ are corresponding congruent angles, then $ℓ_{1}$ and $ℓ_{2}$ are parallel lines. To summarize, all of the steps will be described in a two-column proof.

Statement	Reason
$∠ 1 ≅ ∠ 2$	Given
$∠ 2 ≅ ∠ α$	Vertical Angles Theorem
$∠ 1 ≅ ∠ α$	Transitive Property of Congruence
$ℓ_{1} ∥ ℓ_{2}$	Converse Corresponding Angles Theorem

Discussion

Alternate Exterior Angles Theorem and Its Converse

Similar properties can be discovered for alternate exterior angles.

Rule

Alternate Exterior Angles Theorem

If parallel lines are cut by a transversal, then the pairs of alternate exterior angles are congruent.

Based on the characteristics of the diagram, the following relations hold true.

If $ℓ_{1} ∥ ℓ_{2},$ then $∠ 1 ≅ ∠ 2$ and $∠ 3 ≅ ∠ 4 .$

Proof

Geometric Approach

In order to prove that alternate exterior angles are congruent, it will be shown that $∠ 1$ and $∠ 2$ are congruent.

Notice that by definition,

∠ 2

and

∠ 8

are corresponding angles. Therefore, by the Corresponding Angles Theorem, they are congruent angles.

∠ 2 ≅ ∠ 8

Furthermore, by definition,

∠ 8

and

∠ 1

are vertical angles. Therefore, by the Vertical Angles Theorem,

∠ 8

and

∠ 1

are congruent angles.

∠ 8 ≅ ∠ 1

Then, by applying the Transitive Property of Congruence,

∠ 2

and

∠ 1

can be concluded to be congruent angles as well.

{∠ 2 ≅ ∠ 8 ∠ 8 ≅ ∠ 1 \Rightarrow ∠ 2 ≅ ∠ 1

The same reasoning applies to the other pair of alternate exterior angles. Therefore, when a pair of parallel lines is cut by a transversal, the pairs of alternate exterior angles are congruent.

Two-Column Proof

The previous proof can be summarized in the following two-column table.

Statements	Reasons
$∠ 2$ and $∠ 8$ are corresponding angles	Def. of corresponding angles
$∠ 2 ≅ ∠ 8$	Corresponding Angles Theorem
$∠ 8$ and $∠ 1$ are vertical angles	Def. of vertical angles
$∠ 8 ≅ ∠ 1$	Vertical Angles Theorem
$∠ 2 ≅ ∠ 1$	Transitive Property of Congruence

Proof

Using Transformations

Consider the points of intersection as well as two more points on each line.

Next,

A,

B,

C,

and

D

will be translated in the direction of the transversal so that points

A,

C,

and

D

lie on

ℓ_{2} .

Then,

A,

B,

and

C

will be rotated

18 0^{\circ}

about

F .

After this combination of rigid motions,

A,

B,

C,

and

D

are mapped onto

A^{'},

B^{'},

C^{'},

and

D^{'} .

This means that

∠ A D B

is mapped onto

∠ A^{'} D^{'} B^{'} .

Therefore,

∠ A D B

and

∠ A^{'} D^{'} B^{'}

are congruent angles.

∠ A D B ≅ ∠ A^{'} D^{'} B^{'}

Since

D^{'}

and

F

share the same location,

A^{'}

lies on

F G

and

B^{'}

lies on

F H .

Because of this,

∠ A^{'} D^{'} B^{'}

is congruent to

∠ G F H .

∠ A^{'} D^{'} B^{'} ≅ ∠ G F H

Applying the Transitive Property of Congruence,

∠ A D B

is congruent to

∠ G F H .

{∠ A D B ≅ ∠ A^{'} D^{'} B^{'} ∠ A^{'} D^{'} B^{'} ≅ ∠ G F H ⇓ ∠ A D B ≅ ∠ G F H

It has been proved that one pair of alternate exterior angles is congruent. Further, since

C^{'}

lies on

E F,

it can also be proven that

∠ C^{'} D^{'} B^{'}

is congruent to

∠ E F H .

∠ C^{'} D^{'} B^{'} ≅ ∠ E F H

Applying the Transitive Property of Congruence again,

∠ C D B

is congruent to

∠ E F H .

{∠ C D B ≅ ∠ C^{'} D^{'} B^{'} ∠ C^{'} D^{'} B^{'} ≅ ∠ E F H ⇓ ∠ C D B ≅ ∠ E F H

To conclude, it has been obtained that both pairs of alternate exterior angles are congruent.

$∠ A D B ≅ ∠ G F H$ and $∠ C D B ≅ ∠ E F H$

Rule

Converse Alternate Exterior Angles Theorem

If two lines and a transversal form alternate exterior angles that are congruent, then the two lines are parallel.

Based on the properties of the diagram, the following relation holds true.

If $∠ 1 ≅ ∠ 2$ or $∠ 3 ≅ ∠ 4,$ then $ℓ_{1} ∥ ℓ_{2} .$

Proof

The proof will be based on the given diagram, but it holds true for any pair of lines cut by a transversal. Consider only one pair of congruent alternate exterior angles and one more angle.

It needs to be proven that

ℓ_{1}

and

ℓ_{2}

are parallel lines. It is already given that

∠ 1

is congruent to

∠ 2 .

∠ 1 ≅ ∠ 2

From the diagram, it can also be noted that

∠ 2

and

∠ α

are vertical angles. By the Vertical Angles Theorem, these angles are congruent.

∠ 2 ≅ ∠ α

By the Transitive Property of Congruence, because

∠ 1

is congruent to

∠ 2

and

∠ 2

is congruent to

∠ α,

∠ 1

is congruent to

∠ α .

{∠ 1 ≅ ∠ 2 ∠ 2 ≅ ∠ α ⇓ ∠ 1 ≅ ∠ α

Further,

∠ 1

and

∠ α

are corresponding angles. Hence, the Converse Corresponding Angles Theorem can be applied.

Converse Corresponding Angles Theorem
If two lines are cut by a transversal so that the corresponding angles are congruent, then the lines are parallel.

Since $∠ 1$ and $∠ α$ are corresponding congruent angles, $ℓ_{1}$ and $ℓ_{2}$ are parallel lines. Each step of the proof will now be summarized in a two-column proof.

Statement	Reason
$∠ 1 ≅ ∠ 2$	Given
$∠ 2 ≅ ∠ α$	Vertical Angles Theorem
$∠ 1 ≅ ∠ α$	Transitive Property of Congruence
$ℓ_{1} ∥ ℓ_{2}$	Converse Corresponding Angles Theorem

Example

Using Alternate Interior Angles to Solve Problems

In order to build Tulip Street on the south side of the Lilian river, which goes through Flowerland Village, there is a need to build a bridge. Devontay, an architect, proposed the following plan for the bridge.

It is known that the measure of $∠ 1$ is equal to $4 a + 11$ and the measure of $∠ 2$ is equal to $8 a - 53 .$ What are the measures of $∠ 1$ and $∠ 2 ?$

Hint

How do the measures of $∠ 1$ and $∠ 2$ relate to each other? Use the given expressions to form an equation for $a .$

Solution

By analyzing the diagram it can be noted that $∠ 1$ and $∠ 2$ are alternate interior angles.

Therefore, by the Alternate Interior Angles Theorem, these angles are congruent. Hence, the measures of these angles are the same.

∠ 1 m ∠ 1 ≅ ∠ 2 ⇓ = m ∠ 2

By substituting

m ∠ 1

with

4 a + 11

and

m ∠ 2

with

8 a - 53,

the equation for

a

can be formed.

m ∠ 1 = m ∠ 2

SubstituteII

$m ∠ 1 = 4 a + 11$ , $m ∠ 2 = 8 a - 53$

4 a + 11 = 8 a - 53

▼

Solve for

a

SubEqn

$LHS - 4 a = RHS - 4 a$

11 = 4 a - 53

AddEqn

$LHS + 53 = RHS + 53$

64 = 4 a

DivEqn

$LHS / 4 = RHS / 4$

16 = a

RearrangeEqn

Rearrange equation

a = 16

Knowing the value of

a,

the measure of each of these angles can be calculated.

m ∠ 1 = 4 a + 11

▼

Substitute

16

for

a

and evaluate

Substitute

$a = 16$

m ∠ 1 = 4 (16) + 11

Multiply

m ∠ 1 = 64 + 11

AddTerms

Add terms

m ∠ 1 = 75

Since the angles are congruent, it can be concluded that they both measure

7 5^{\circ} .

m ∠ 1 = m ∠ 2 m ∠ 1 = 7 5^{\circ} \Rightarrow m ∠ 2 = 7 5^{\circ}

Explore

Investigating Points on a Perpendicular Bisector

Up to now, some basic theorems about angles have been seen and proven through some rigid motions. Before the end of the lesson, one last theorem about segments will be learned. Consider a perpendicular bisector of a segment

\overline{A B} .

Move

C

and the endpoints of the segment and compare the distances

A C

and

B C .

What conjecture can be made about position of

C

in respect to the endpoints

A

and

B

of the segment? Does this conjecture also apply to other points on the perpendicular bisector

\overline{C M} ?

Discussion

Perpendicular Bisector Theorem and Its Converse

Rule

Perpendicular Bisector Theorem

Any point on a perpendicular bisector of a line segment is equidistant from the endpoints of the line segment.

Point C on the perpendicular bisector equidistant from endpoints A and B

Based on the characteristics of the diagram, $C M$ is the perpendicular bisector of $\overline{A B} .$ Therefore, $C$ is equidistant from $A$ and $B .$

$A C = B C$

Proof

Geometric Approach

Suppose $C M$ is the perpendicular bisector of $\overline{A B} .$ Then $M$ is the midpoint of $\overline{A B} .$

Consider a triangle with vertices $A,$ $M,$ and $C,$ and another triangle with vertices and $B,$ $M,$ and $C .$

Both $△ A C M$ and $△ B C M$ have a right angle and congruent legs $\overline{A M}$ and $\overline{B M} .$ Since all right angles are congruent, $∠ A M C ≅ ∠ B M C .$ Furthermore, by the Reflexive Property of Congruence, $\overline{C M}$ is congruent to itself.

By the Side-Angle-Side Congruence Theorem, the triangles are congruent. Therefore, since corresponding parts of congruent figures are congruent, their hypotenuses $\overline{A C}$ and $\overline{B C}$ are also congruent. By the definition of congruent segments, $\overline{A C}$ and $\overline{B C}$ have the same length. This means that $C$ is equidistant from $A$ and $B .$

Using this reasoning it can be proven that any point on a perpendicular bisector is equidistant from the endpoints of the segment.

Two-Column Proof

The proof can be summarized in the following two-column table.

Statements	Reasons
$\overline{A M} ≅ \overline{M B}$ $∠ A M C$ and $∠ B M C$ are right angles	Definition of a perpendicular bisector.
$∠ A M C ≅ ∠ B M C$	All right angles are congruent.
$\overline{C M} ≅ \overline{C M}$	Reflexive Property of Congruence.
$△ A C M ≅ △ B C M$	SAS Congruence Theorem.
$\overline{A C} ≅ \overline{B C}$	Corresponding parts of congruent figures are congruent.
$A C = B C$	Definition of congruent segments.

Proof

Using Transformations

Suppose $C M$ is the perpendicular bisector of $\overline{A B} .$

Using the given points $A,$ $B,$ $C,$ and $M$ as vertices, two triangles can be formed. The resulting triangles, $△ A C M$ and $△ B C M,$ can be proven to be congruent by identifying a congruence transformation that maps one triangle onto the other.

Since

\overline{A M}

and

\overline{B M}

are congruent, the distance between

A

and

M

is equal to the distance between

B

and

M .

Therefore,

A

is the image of

B

after a reflection across

C M .

Since

C

lies on

C M,

a reflection across

C M

maps

C

onto itself. The same is true for

M .

Reflection Across $C M$
Preimage	Image
$B$	$A$
$C$	$C$
$M$	$M$

The above table shows that the images of the vertices of

△ B C M

are the vertices of

△ A C M .

Therefore,

△ A C M

is the image of

△ B C M

after a reflection across

C M .

Since a reflection is a rigid motion, this proves that the triangles are congruent.

Corresponding parts of congruent figures are congruent, so

\overline{A C}

and

\overline{B C}

are congruent. By the definition of congruent segments,

A C

and

B C

are equal. This means that

C

is equidistant from

A

and

B .

A C = B C

The same reasoning can be applied to any point on a perpendicular bisector, showing that the point is equidistant from the endpoints of the segment.

Rule

Converse Perpendicular Bisector Theorem

If a point is equidistant from the endpoints of a line segment, then it lies on the perpendicular bisector of the segment.

Based on the characteristics of the diagram, the following relation holds true.

$A C = C B ⇓ C M ⊥ \overline{A B} and A M = M B$

Proof

Converse Perpendicular Bisector Theorem

Consider $\overline{A B}$ and a point $C$ equidistant from $A$ and $B .$

To prove that

C

lies on the perpendicular bisector of

\overline{A B},

it will be shown that the line perpendicular to

\overline{A B}

through

C

bisects

\overline{A B} .

M

is the point of intersection between the line and the segment, it must be proven that

A M = M B .

This line forms two right triangles that share a common leg

\overline{C M} .

Because all right angles are congruent,

∠ A M C

is congruent to

∠ B M C .

Also, by the Reflexive Property of Congruence,

\overline{C M}

is congruent to itself. Since

A C

is equal to

B C,

\overline{A C}

is congruent to

\overline{B C} .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ ∠ A M C ≅ ∠ B M C \overline{C M} ≅ \overline{C M} \overline{A C} ≅ \overline{B C}

By the Hypotenuse Leg Theorem,

△ A M C

and

△ B M C

are congruent triangles. Because corresponding parts of congruent figures are congruent,

\overline{A M}

is congruent to

\overline{B M} .

Additionally, it was already known that $C M$ and $\overline{A B}$ are perpendicular.

$C M ⊥ \overline{A B} and A M = M B$

By the definition of a perpendicular bisector, $C M$ is the perpendicular bisector of $\overline{A B} .$ Therefore, $C$ lies on the perpendicular bisector of $\overline{A B} .$

Closure

Solving Problems Using Perpendicular Bisectors

In Flowerland Village, there are two related families, Funnystongs and Cleverstongs, who live opposite each other. Mr. Funnystong and Mr. Cleverstong want to pave a road between the houses so that every point of the road is equidistant to their houses.

The locating of two houses opposite to each other

If the houses are

16

meters away, how far from the houses and along what line should the road be paved?

Answer

Distance: $8$ meters from each house.
Direction: Along the perpendicular bisector to the segment with endpoints at the houses.

Hint

What does the the Perpendicular Bisector Theorem state?

Solution

Recall what the Perpendicular Bisector Theorem states.

Any point on a perpendicular bisector is equidistant from the endpoints of the line segment.

With this theorem in mind, the position of the road can be determined. To do so, draw a segment whose endpoints are located at the houses.

Before drawing the perpendicular bisector of this segment, its midpoint should be found. Since the distance between the houses is $16$ meters, the perpendicular bisector will pass through a point that is $\frac{1 6}{2} = 8$ meters away from the houses.

Based on the theorem, it can be said that each point on the bisector is equidistant from the houses. Therefore, the road between the houses should be paved along the segment's perpendicular bisector.

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Catch-Up and Review

Extra

Proof

Two-Column Proof

Proof

Hint

Solution

Proof

Proof

Hint

Solution

Proof

Two-Column Proof

Proof

Proof

Proof

Two-Column Proof

Proof

Proof

Hint

Solution

Proof

Two-Column Proof

Proof

Proof

Answer

Hint

Solution