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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 BC∥DF and CE∥AD.
Explore
Investigating a Triangle's Interior Angles
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.
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 measures of the interior angles of a triangle is 180∘. 
Based on this diagram, the following relation holds true.
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.
m∠A+m∠B+m∠C=180∘
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.
By the Alternate Interior Angles Theorem, ∠B is congruent to ∠1 and ∠C is congruent to ∠2.
∠B≅∠1 ⇕ m∠B=m∠1 ∠C≅∠2 ⇕ m∠C=m∠2
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∘.
m∠BAC+m∠1+m∠2=180∘
By the Substitution Property of Equality, the sum of the measures of ∠BAC, ∠B, and ∠C is equal to 180∘.
m∠BAC+m∠1+m∠2=180∘⇓m∠BAC+m∠B+m∠C=180∘
Finally, in △ABC, ∠BAC can be named ∠A. m∠BAC+m∠B+m∠C=180∘⇓m∠A+m∠B+m∠C=180∘
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.
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Hint
Use the Interior Angles Theorem.
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.
35∘+40∘+x=180∘
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, or remote interior angles.
Based on the diagram above, the following relation holds true.
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∘.
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.
Now, △ABC can be rotated 180∘ about 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∘ about E. 
By the Parallelogram Opposite Angles Theorem, ∠PCA is congruent to ∠AB′′P. Congruent angles have the same measure by the definition.
m∠PCA=m∠A+m∠B
Proof
Using Properties of AnglesConsider a triangle with vertices A, B, and C, and one of the exterior angles corresponding to ∠C.
{m∠C+m∠PCA=180∘m∠A+m∠B+m∠C=180∘(I)(II)
Now m∠C can be isolated in Equation (I).
m∠C+m∠PCA=180∘⇕m∠C=180∘−m∠PCA
Next, the expression of m∠C can be substituted into Equation (II).
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
Proof
Using TransformationsConsider △ABC, where D and E are the midpoints of AB and AC, respectively. Let ∠PCA be one exterior angle of △ABC.
∠PCAm∠PCA≅∠AB′′P⇕=m∠AB′′P
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.
{m∠PCA=m∠AB′′Pm∠AB′′P=m∠A+m∠B⇓m∠PCA=m∠A+m∠B
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.
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Hint
Use the Triangle Exterior Angle Theorem.
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.
x=35∘+40∘⇓x=75∘
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.
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. 
Based on this diagram, the following relation holds true.
The Isosceles Triangle Theorem is also known as the Base Angles Theorem.
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.
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.
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.
Corresponding parts of congruent figures are congruent, so ∠B and ∠C are congruent.
AB≅AC ⇒ ∠B≅∠C
Proof
Geometric ApproachConsider a triangle ABC with two congruent sides, or an isosceles triangle.
| Statement | Reason |
|---|---|
| ∠BAP ≅ ∠CAP | Definition of an angle bisector |
| BA ≅ CA | Given |
| AP ≅ AP | Reflexive Property of Congruence |
△BAP≅△CAP
Corresponding parts of congruent figures are congruent. Therefore, ∠B and ∠C are congruent. ∠B≅∠C
It has been proven that if two sides of a triangle are congruent, then the angles opposite them are congruent.
Proof
Using TransformationsConsider 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 |
∠B≅∠C
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?
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Hint
Consider the Base Angles Theorem.
Solution
Placing the support beam as shown forms an isosceles triangle.
x+x+40∘=180∘
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 BA′′ and AC′′′?
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. 
If DE is a midsegment of △ABC, then the following statement holds true.
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.
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.
Since 21a is half of a, it can be stated that the midsegment DE is half the length of BC.
DE∥BC and DE=21BC
Proof
Using CoordinatesThis 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.
B(0,0)C(a,0)A(b,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. To prove this theorem, it must be proven that DE is parallel to BC and that DE is half of BC.
BC∥DE
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(2x1+x2,2y1+y2) | |||
|---|---|---|---|
| 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 ✓
DE=21BC
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=21a |
DE=21BC ✓
Therefore, a midsegment of two sides of a triangle is parallel to the third side of the triangle and half its length.
Proof
Using TransformationsThis proof will be developed based on the given diagram, but it is valid for any triangle. 
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.
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.1
Assume That E′ Does Not Lie on BC
expand_more 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
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 DB=EE