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Congruence, Proof, and Constructions

Theorems About Triangles

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In order to operate with triangles, the relationships between the angles and sides of a triangle need to be investigated. Therefore, in this lesson, some theorems about triangles will be built upon several theorems about lines and angles.

Catch-Up and Review

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

Challenge

Investigating Triangles and Their Properties

To strengthen roof trusses, usually triangular shaped structures are used. In the diagram, AC=CF, AB=BD, and DE=EF. The beams BC, DF, CE, and AD are built in a way that BCDF and CEAD.

Knowing that CE is 43 inches long and DF is 68 inches long, what would be the lengths of AD and BC?

Explore

Investigating a Triangle's Interior Angles

Consider ABC, where D and E are the midpoints of AB and AC, respectively. Perform two rotations on ABC — one about point D and the other about point E.
Considering the images and preimage, investigate the sum of the interior angles of ABC.

Discussion

Properties of a Triangle's Interior Angles

Considering the previous exploration, the sum of interior angles of a triangle can be derived.

Rule

Interior Angles Theorem

The sum of the interior angles of a triangle is
The triangle ABC with movable vertices A, B, and C and the angle measures written

Based on this diagram, the following relation holds true.

This theorem is also known as the Triangle Angle Sum Theorem.

Proof

Consider a triangle with vertices A, B, and C, and the parallel line to BC through A. Let ∠1 and ∠2 be the angles outside ABC formed by this line and the sides AB and AC.

Triangle ABC with the line parallel to BC

By the Alternate Interior Angles Theorem, B is congruent to ∠1 and C is congruent to ∠2.

Two pairs of alternate interior angles
By the definition of congruent angles, ∠1 and B have the same measure. For the same reason, ∠2 and C also have the same measure.
Furthermore, in the diagram it can be seen that BAC, ∠1, and ∠2 form a straight angle. Therefore, by the Angle Addition Postulate their measures add to
By the Substitution Property of Equality, the sum of the measures of BAC, B, and C is equal to
Finally, in ABC, BAC can be named A.

Example

Solving Problems Using the Interior Angles Theorem

Dylan is designing a wooden sofa made of oak wood for his local park. The sides of the sofa will have identical dimensions in the shape of a triangle. He already has decided on the angle measures of the top corner and bottom-right corner of each side.

To cut the sides of the sofa out of the board using a table saw, which can cut at angles, Dylan needs to find the measure of the third angle. Dylan's hands are full — help him find the measure of the third angle.

Hint

Solution

Since the sides have a triangular shape and the measures of two angles are known, the Interior Angles Theorem can be used to find the missing angle measure. Let x be the measure of the missing angle.

Solving this equation for x, the measure of the missing angle can be found.

Explore

Investigating a Triangle's Exterior Angles

Once again, consider ABC, where D and E are the midpoints of AB and AC, respectively. This time, begin by rotating ABC about D. Then, rotate the resulting figure about E.
Considering the images and preimage, what can be concluded about exterior angle ACF and its remote interior angles A and B?

Discussion

Properties of a Triangle's Exterior Angles

The previous exploration shows that there is a clear relation between an exterior angle of a triangle and its remote interior angles.

Rule

Triangle Exterior Angle Theorem

The measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent interior angles.
Triangle with an exterior angle marked
Based on the diagram above, the following relation holds true.

mPCA=mA+mB

Proof

Using Properties of Angles

Consider a triangle with vertices A, B, and C, with the exterior angle corresponding to C.

Triangle with an exterior angle marked
The diagram shows that C and PCA form a linear pair, so the sum of their measures is Additionally, by the Triangle Angle Sum Theorem, the sum of the angle measures of ABC is
Now mC can be isolated in Equation (I).
Next, the expression of mC can be substituted into Equation (II).
Solve for mPCA
mA+mBmPCA=0
mA+mB=mPCA
mPCA=mA+mB
It has been proven that the measure of PCA is equal to the sum of the measures of A and B. Therefore, it can be said that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent interior angles.

Proof

Using Transformations

Consider ABC, where D and E are the midpoints of AB and AC, respectively. Let PCA be the exterior angle of ABC.

Triangle with an exterior angle marked
Now, ABC can be rotated over D. Since a rotation is a rigid motion, the image of ABC after the rotation is congruent to ABC. Corresponding parts of congruent figures are congruent, so the measures of the angles and the lengths of the sides remain unchanged.
Triangle interior angles included side rotation
Since a rotation is equivalent to a reflection, is parallel to BC and is parallel to AC. Therefore, is a parallelogram and is congruent to Now the parallelogram will be rotated over E.
Triangle exterior angle parallelogram rotation
By the Parallelogram Opposite Angles Theorem, PCA is congruent to Congruent angles have the same measure by the definition.
Since is equal to the sum of mA and mB and because of the Transitive Property of Equality, mPCA is equal to the sum of mA and mB.
This completes the proof.

Example

Solving Problems Using the Triangle Exterior Angles Theorem

Dylan is almost ready to cut the sides of the sofa. Before doing so, he wants to be sure that people sitting on his sofa can lean back freely and feel comfortable. Therefore, he needs to find the measure of the angle exterior to the third angle.

Note that if the angle measure is less than the sofa is inclined backwards. Dylan is a bit busy with handling the wood. Help him find the measure of the exterior angle.

Solution

Recall that by the Triangle Exterior Angle Theorem, the measure of a triangle's exterior angle is equal to the sum of the measures of its two remote interior angles. Therefore, the measure of the exterior angle x can be expressed.

Since the measure of the exterior angle is less than the people sitting on this sofa can lean back and feel comfortable. Thanks for helping Dylan.

Explore

Investigating Properties of Isosceles Triangles

Given that ABC is a right triangle, reflect it across AB.
Examine the triangle formed by the preimage and image. Compare the base angles of the triangle.

Discussion

Properties of Isosceles Triangles

Reflecting a right triangle about either of its legs forms an isosceles triangle. Note that a reflection is a rigid motion, so the side lengths and the interior angles of the right triangle are preserved.

Rule

Isosceles Triangle Theorem

If two sides of a triangle are congruent, then the angles opposite them are congruent.
An isosceles triangle.
Based on this diagram, the following relation holds true.

ABAC BC

The Isosceles Triangle Theorem is also known as the Base Angles Theorem.

Proof

Geometric Approach

Consider a triangle ABC with two congruent sides, or an isosceles triangle.

An isosceles triangle ABC.
In this triangle, let P be the point of intersection of BC and the angle bisector of A.
An isosceles triangle ABC with an angle bisector AP.
From the diagram, the following facts about BAP and CAP can be observed.
Statement Reason
Definition of an angle bisector
Given
Reflexive Property of Congruence
Therefore, BAP and CAP have two pairs of corresponding congruent sides and one pair of congruent included angles. By the Side-Angle-Side Congruence Theorem, BAP and CAP are congruent triangles.
Corresponding parts of congruent figures are congruent. Therefore, B and C are congruent.
It has been proven that if two sides of a triangle are congruent, then the angles opposite them are congruent.

Proof

Using Transformations

Consider an isosceles triangle ABC.

An isosceles triangle ABC

A line passing through A and the midpoint of BC will be drawn. Let P be the midpoint.

An isosceles triangle ABC with a line through A and midpoint P of the base BC

Since BP and PC are congruent, the distance between B and P is equal to the distance between C and P. Therefore, B is the image of C after a reflection across Also, because A lies on a reflection across maps A onto itself. The same is true for P.

Reflection Across
Preimage Image
C B
A A
P P
The table shows that the images of the vertices of CAP are the vertices of BAP. It can be concluded that BAP is the image of CAP after a reflection across Since a reflection is a rigid motion, this proves that the triangles are congruent.
A reflection across AP that maps triangle CAP onto BAP
Corresponding parts of congruent figures are congruent, so B and C are congruent.

Example

Solving Problems Using the Isosceles Triangle Theorem

Dylan notices that he needs a support beam to support the seat. The bottoms of each side panel are 3 feet long. Therefore, if he places the support beam from the corner with the larger angle measure to the opposite side in a position where the endpoint of the support beam is 3 feet away from the bottom-right corner, then it will fit just right.

In this case, what should be the measure of the angle between the support beam and the bottom of the side panel?

Hint

Consider the Base Angles Theorem.

Solution

Placing the support beam as shown forms an isosceles triangle.

Recall that according to the Base Angles Theorem, base angles of an isosceles triangle are congruent. It can be seen that the measure of the vertex angle is Assuming that the measure of a base angle of the triangle is x, an equation can be written by the Interior Angles Theorem.
By solving this equation, the measure of the angle between the support beam and the bottom of the side can be found.

Explore

Investigating Properties of a Triangle's Midsegment

In the following applet, investigate the rigid motions by moving the slider.
What is the resulting figure formed by the preimage and images? What is the relationship between and

Discussion

Properties of a Triangle's Midsegment

As it is seen in the previous exploration, using the rigid motions, the Triangle Midsegment Theorem can be proven.

Rule

Triangle Midsegment Theorem

The line segment that connects the midpoints of two sides of a triangle — also known as a midsegment — is parallel to the third side of the triangle and half its length.
Triangle ABC with the midsegment DE
If DE is a midsegment of ABC, then the following statement holds true.

DEBC and

Proof

Using Coordinates

This theorem can be proven by placing the triangle on a coordinate plane. For simplicity, vertex B will be placed at the origin and vertex C on the x-axis.

Triangle ABC on a coordinate plane
Since B lies on the origin, its coordinates are (0,0). Point C is on the x-axis, meaning its y-coordinate is 0. The remaining coordinates are unknown and can be named a, b, and c.
If DE is the midsegment from BA to CA, then by the definition of a midpoint, D and E are the midpoints of BA and CA, respectively.
ABC with midsegment DE

To prove this theorem, it must be proven that DE is parallel to BC and that DE is half of BC.

BCDE

If the slopes of these two segments are equal, then they are parallel. The y-coordinate of both B(0,0) and C(a,0) is 0. Therefore, BC is a horizontal segment. Next, the coordinates of D and E will be found using the Midpoint Formula.

Segment Endpoints Substitute Simplify
BA B(0,0) and
CA C(a,0) and
The y-coordinate of both and is Therefore, DE is also a horizontal segment. Since all horizontal segments are parallel, it can be said that BC and DE are parallel.

Since both BC and DE are horizontal, their lengths are given by the difference of the x-coordinates of their endpoints.

Segment Endpoints Length Simplify
BC B(0,0) and C(a,0) BC=a0 BC=a
DE and
Since is half of a, it can be stated that the midsegment DE is half the length of BC.
Therefore, a midsegment of two sides of a triangle is parallel to the third side of the triangle and half its length.

Proof

Using Transformations
This proof will be developed based on the given diagram, but it is valid for any triangle.
Movable triangle with a midsegment
To prove this theorem, it must be proven that DE is parallel to BC and that DE is equal to half of BC. Each statement will be proven one at a time.

BCDE

This part can be proven by using rigid motions. First, translate ADE along DB so that D is mapped onto B. Since D is the midpoint of AB, A is mapped onto D.
Translation of the triangle in a triangle with a midsegment
Next, it must be proven that the image of E — which is marked as — lies on BC. This proof will be done using indirect reasoning. In this method, it is temporarily assumed that the negation of the statement is true.
1
Assume That Does Not Lie on BC
expand_more

Assume that the image of E after the translation along DB, does not lie on BC. This means that lies either above or below BC. The proof will be developed for only one case but is valid for both.

Point E' lies above the base BC of the triangle ABC

Based on the assumption, let F denote the point of intersection of and EC.

Point of intersection of the ray BE' and the segment EC

Next, it will be proven that ADE is congruent to

2
Show That
expand_more

It is given that AD=DB because D is the midpoint of AB. Additionally, since ADE is translated along DB, it can be concluded that Then, by the Transitive Property of Equality,

Image of the triangle ABC with the segments of equal lengths marked

Recall that Additionally, DE is the common side of ADE and All pieces of information can now be summarized.

  • DE is the common side of ADE and
Using all the information, ADE is congruent to by the Side-Side-Side Congruence Theorem.
Because corresponding angles of congruent triangles are congruent, ADE is congruent to
Angle ADE is congruent to angle DEE'
Since translations preserve angles, DE is parallel to BF. By the Alternate Interior Angles Theorem, is congruent to
Therefore, by the Transitive Property of Congruence,
Angle ADE is congruent to angle EE'F
Since DE is parallel to BF and the image of E is translated in the same direction as the image of D, is congruent to Additionally, is congruent to DAE.
One more time, by the Transitive Property of Congruence,
Angle E'EF is congruent to angle DAE

Summarize the obtained information about the triangles ADE and

Therefore, by the Angle-Side-Angle Congruence Theorem, ADE is congruent to

3
Contradiction
expand_more
It has been proven that Because corresponding sides of congruent triangles are congruent, it follows that EF is equal to AE.
This means that EF is equal to EC, because E is the midpoint of AC. However, since F lies between E and C, it cannot be true that EF=EC. Therefore, the temporary assumption leads to a contradiction.
Assumption of the Theorem Indirect Assumption
AE=EF=EC EF<EC
EF=EC and EF<EC

Therefore, this contradiction verifies that the image of E must lie on BC.

It has been proven that lies on BC. Because translations preserve angles, ADE is congruent to
Angle ADE is congruent to DBE'

By the Converse Corresponding Angles Theorem, BC is parallel to DE.

BCDE

It has been previously obtained that BC and DE are parallel. Now, another rigid motion to will be applied.
1
Rotate Around the Midpoint of
expand_more
Rotate counterclockwise about the midpoint of so that is mapped onto D. It can be noted that the triangle is rotated Since rotations preserve angles and lengths, this rotation maps B onto E. Therefore, BD is mapped onto
The Midsegment Theorem proof first rotation
As a result of the rotation, it can be concluded that and or are congruent triangles.
2
Rotate Around the Midpoint of
expand_more
Rotate counterclockwise about the midpoint of so that E is mapped onto It can be noted that the triangle is again rotated Since rotations preserve angles and lengths, this rotation maps D onto C. Therefore, and DE are mapped onto CE and respectively.
The Midsegment Theorem proof second rotation
As a result of the rotation, it can be concluded that and are congruent triangles.
By the Segment Addition Postulate, the length of BC can be calculated by adding the lengths of smaller segments.
Because corresponding sides of congruent triangles are congruent, DE is congruent to and DE is congruent to Therefore, and By the Substitution Property of Equality, BC can be expressed in terms of DE.
Finally, by the Division Property of Equality, the second statement of the theorem is obtained.

Example

Solving Problems Using the Triangle Midsegment Theorem

Finally, Dylan is ready to place the seat. He plans to place it just above the support beam such that it will be parallel to the bottom. Therefore, the corners of the seat will be at the midpoints of the sides.

How can he find the width of the seat knowing that the bottom of the side is 3 feet long.

Hint

The seat will be aligned with the midsegment of the triangular side.

Solution

Since the corners of the seat are at the midpoints of the triangular side, it will be aligned with the midsegment of the triangular side. Therefore, by the Triangle Midsegment Theorem, the width of the seat will be half the length of the bottom of the side.
The width of the seat is 1.5 feet.

Discussion

Solving Problems Using a Triangle's Properties

In this lesson, the investigated theorems about triangles have been proven using a variety of methods. Furthermore, with the help of these theorems, the challenge provided at the beginning of the lesson can be solved. Recall the diagram.

Consider the given information about the beams of the roof.

From here, what are the lengths of AD and BC?

Solution

By the definition of a midsegment, both BC and CE are midsegments of ADF. By the Triangle Midsegment Theorem, CE is half of AD, and BC is half of DF.

Knowing that CE is 43 inches and DF is 68 inches, these values can be substitute into these equations to find AD and BC.

Therefore, the length of AD is 86 inches and the length of BC is 34 inches.

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