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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.**

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 BC∥DF and CE∥AD.

Knowing that CE is 43 inches long and DF is 68 inches long, what would be the lengths of AD and BC?
Consider △ABC, where D and E are the midpoints of AB and AC, respectively. Perform two $180_{∘}$ 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.

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

The sum of the interior angles of a triangle is $180_{∘}.$
### Proof

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 $180_{∘}.$
By the Substitution Property of Equality, the sum of the measures of ∠BAC, ∠B, and ∠C is equal to $180_{∘}.$
Finally, in △ABC, ∠BAC can be named ∠A.

Based on this diagram, the following relation holds true.

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

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.

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

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.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.

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Use the Interior Angles Theorem.

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.

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?

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

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 Properties of Angles
Next, the expression of m∠C can be substituted into Equation (II).
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
Since $m∠AB_{′′}P$ is equal to the sum of m∠A and m∠B and because of the Transitive Property of Equality, m∠PCA is equal to the sum of m∠A and m∠B.
This completes the proof.

Based on the diagram above, the following relation holds true.

m∠PCA=m∠A+m∠B

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

The diagram shows that ∠C and ∠PCA form a linear pair, so the sum of their measures is $180_{∘}.$ Additionally, by the Triangle Angle Sum Theorem, the sum of the angle measures of △ABC is $180_{∘}.$${m∠C+m∠PCA=180_{∘}m∠A+m∠B+m∠C=180_{∘} (I)(II) $

Now m∠C can be isolated in Equation (I).
$m∠A+m∠B+m∠C=180_{∘}$

Substitute

$m∠C=180_{∘}−m∠PCA$

$m∠A+m∠B+(180_{∘}−m∠PCA)=180_{∘}$

Solve for m∠PCA

RemovePar

Remove parentheses

$m∠A+m∠B+180_{∘}−m∠PCA=180_{∘}$

SubEqn

$LHS−180_{∘}=RHS−180_{∘}$

m∠A+m∠B−m∠PCA=0

AddEqn

LHS+m∠PCA=RHS+m∠PCA

m∠A+m∠B=m∠PCA

RearrangeEqn

Rearrange equation

m∠PCA=m∠A+m∠B

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

Now, △ABC can be rotated $180_{∘}$ 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.
Since a $180_{∘}-$rotation is equivalent to a reflection, $C_{′}A$ is parallel to BC and $C_{′}B$ is parallel to AC. Therefore, $C_{′}ACB$ is a parallelogram and $∠C_{′}AC$ is congruent to $∠CBC_{′}.$ Now the parallelogram $C_{′}ACB$ will be rotated $180_{∘}$ over E.

By the Parallelogram Opposite Angles Theorem, ∠PCA is congruent to $∠AB_{′′}P.$ Congruent angles have the same measure by the definition.

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 $90_{∘},$ the sofa is inclined backwards. Dylan is a bit busy with handling the wood. Help him find the measure of the exterior angle.

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Use the Triangle Exterior Angle Theorem.

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 $90_{∘},$ the people sitting on this sofa can lean back and feel comfortable. Thanks for helping Dylan.

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.

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.

If two sides of a triangle are congruent, then the angles opposite them are congruent.
The Isosceles Triangle Theorem is also known as the **Base Angles Theorem**.
### Proof

Geometric Approach

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

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

Based on this diagram, the following relation holds true.

AB≅AC ∠B≅∠C

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

In this triangle, let P be the point of intersection of BC and the angle bisector of ∠A. From the diagram, the following facts about △BAP and △CAP can be observed.

Statement | Reason |
---|---|

$∠BAP≅∠CAP$ | Definition of an angle bisector |

$BA≅CA$ | Given |

$AP≅AP$ | Reflexive Property of Congruence |

Consider an isosceles triangle △ABC.

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

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 $AP.$ Also, because A lies on $AP,$ a reflection across $AP$ maps A onto itself. The same is true for P.

Reflection Across $AP$ | |
---|---|

Preimage | Image |

C | B |

A | A |

P | P |

Corresponding parts of congruent figures are congruent, so ∠B and ∠C are congruent.

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?

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Consider the Base Angles Theorem.

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 $40_{∘}.$ Assuming that the measure of a base angle of the triangle is x, an equation can be written by the Interior Angles Theorem.
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 $BA_{′′}$ and $AC_{′′′}?$

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

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.
### Proof

Using Coordinates
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.
### BC∥DE

The y-coordinate of both $D(2b ,2c )$ and $E(2a+b ,2c )$ is $2c .$ Therefore, DE is also a horizontal segment. Since all horizontal segments are parallel, it can be said that BC and DE are parallel.
### $DE=21 BC$

Since $21 a$ is half of a, it can be stated that the midsegment DE is half the length of BC.
### Proof

Using Transformations

If DE is a midsegment of △ABC, then the following statement holds true.

DE∥BC and

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.

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

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.

$M(2x_{1}+x_{2} ,2y_{1}+y_{2} )$ | |||
---|---|---|---|

Segment | Endpoints | Substitute | Simplify |

BA | B(0,0) and $A(b,c)$ | $D(20+b ,20+c )$ | $D(2b ,2c )$ |

CA | C(a,0) and $A(b,c)$ | $E(2a+b ,20+c )$ | $E(2a+b ,2c )$ |

$BC∥DE✓ $

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=a−0 | BC=a |

DE | $D(2b ,2c )$ and $E(2a+b ,2c )$ | $DE=2a+b −2b $ | $DE=21 a$ |

$DE=21 BC✓ $

Therefore, a midsegment of two sides of a triangle is parallel to the third side of the triangle and half its length.
This proof will be developed based on the given diagram, but it is valid for any triangle. ### BC∥DE

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.
*expand_more*
*expand_more*
Because corresponding angles of congruent triangles are congruent, ∠ADE is congruent to $∠DEE_{′}.$
Since translations preserve angles, DE is parallel to BF. By the Alternate Interior Angles Theorem, $∠DEE_{′}$ is congruent to $∠EE_{′}F.$
*expand_more*
It has been proven that $E_{′}$ lies on BC. Because translations preserve angles, ∠ADE is congruent to $∠DBE_{′}.$
### $DE=21 BC$

It has been previously obtained that BC and DE are parallel. Now, another rigid motion to $△BDE_{′}$ will be applied.
*expand_more*
*expand_more*
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 $BE_{′}$ and DE is congruent to $CE_{′}.$ Therefore, $DE=BE_{′}$ and $DE=CE_{′}.$ 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.

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.

Next, it must be proven that the image of E — which is marked as $E_{′}$ — 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 $E_{′}$ Does Not Lie on BC

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

Based on the assumption, let F denote the point of intersection of $BE_{′}$ and EC.

Next, it will be proven that △ADE is congruent to $△EE_{′}F.$

2

Show That $△ADE≅△EE_{′}F$

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 $DB=EE_{′}.$ Then, by the Transitive Property of Equality, $AD=EE_{′}.$

Recall that $AE=DE_{′}.$ Additionally, DE is the common side of △ADE and $△E_{′}ED.$ All pieces of information can now be summarized.

- $AD=EE_{′}$
- $AE=DE_{′}$
- DE is the common side of △ADE and $△E_{′}ED.$

$∠ADE≅∠DEE_{′}and∠DEE_{′}≅∠EE_{′}F $

Therefore, by the Transitive Property of Congruence, $∠ADE≅∠EE_{′}F.$
Since DE is parallel to BF and the image of E is translated in the same direction as the image of D, $∠BDE_{′}$ is congruent to $∠E_{′}EF.$ Additionally, $∠BDE_{′}$ is congruent to ∠DAE. $∠E_{′}EF≅∠BDE_{′}and∠BDE_{′}≅∠DAE $

One more time, by the Transitive Property of Congruence, $∠E_{′}EF≅∠DAE.$
Summarize the obtained information about the triangles △ADE and $△EE_{′}F.$

- $AD=EE_{′}$
- $∠ADE≅∠EE_{′}F$
- $∠DAE≅∠E_{′}EF$

Therefore, by the Angle-Side-Angle Congruence Theorem, △ADE is congruent to $△EE_{′}F.$

3

Contradiction

It has been proven that $△ADE≅△EE_{′}F.$ 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.

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

BC∥DE

1

Rotate $△BDE_{′}$ Around the Midpoint of $DE_{′}$

Rotate $△BDE_{′}$ counterclockwise about the midpoint of $DE_{′}$ so that $E_{′}$ is mapped onto D. It can be noted that the triangle is rotated $180_{∘}.$ Since rotations preserve angles and lengths, this rotation maps B onto E. Therefore, BD is mapped onto $EE_{′}.$

As a result of the rotation, it can be concluded that $△BDE_{′}$ and $△B_{′′}D_{′′}E_{′′},$ or $△EE_{′}D,$ are congruent triangles.

2

Rotate $△DED_{′′}$ Around the Midpoint of $D_{′′}E$

Rotate $△DED_{′′}$ counterclockwise about the midpoint of $D_{′′}E$ so that E is mapped onto $D_{′′}.$ It can be noted that the triangle is again rotated $180_{∘}.$ Since rotations preserve angles and lengths, this rotation maps D onto C. Therefore, $DD_{′′}$ and DE are mapped onto CE and $CE_{′},$ respectively.

As a result of the rotation, it can be concluded that $△DED_{′′}$ and $△EE_{′}C_{′′}$ are congruent triangles.

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.

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

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.

$23 =1.5ft $

The width of the seat is 1.5 feet.
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.

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Use the Triangle Midsegment Theorem.

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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