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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.
Considering the previous exploration, the sum of interior angles of a triangle can be derived.
m∠A+m∠B+m∠C=180∘
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.
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.
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.
The previous exploration shows that there is a clear relation between an exterior angle of a triangle and its remote interior angles.
m∠PCA=m∠A+m∠B
Consider a triangle with vertices A, B, and C, and one of the exterior angles corresponding to ∠C.
m∠C=180∘−m∠PCA
Remove parentheses
LHS−180∘=RHS−180∘
LHS+m∠PCA=RHS+m∠PCA
Rearrange equation
Consider △ABC, where D and E are the midpoints of AB and AC, respectively. Let ∠PCA be one exterior angle of △ABC.
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.
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.
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.
AB≅AC ⇒ ∠B≅∠C
Consider 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 |
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 |
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?
Consider the Base Angles Theorem.
Placing the support beam as shown forms an isosceles triangle.
As it is seen in the previous exploration, using the rigid motions, the Triangle Midsegment Theorem can be proven.
DE∥BC and DE=21BC
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.
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(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) |
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 |
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.
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