Triangle Theorems: Insights into the Triangle Midsegment Theorem

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, $A C = C F,$ $A B = B D,$ and $D E = E F .$ The beams $\overline{B C},$ $\overline{D F},$ $\overline{C E},$ and $\overline{A D}$ are built in a way that $\overline{B C} ∥ \overline{D F}$ and $\overline{C E} ∥ \overline{A D} .$

Knowing that

\overline{C E}

43

inches long and

\overline{D F}

68

inches long, what would be the lengths of

\overline{A D}

and

\overline{B C} ?

Explore

Investigating a Triangle's Interior Angles

Consider

△ A B C,

where

D

and

E

are the midpoints of

\overline{A B}

and

\overline{A C},

respectively. Perform two

18 0^{\circ}

rotations on

△ A B C

— one about point

D

and the other about point

E .

Considering the images and preimage, investigate the sum of the interior angles of

△ A B C .

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

18 0^{\circ} .

The triangle ABC with movable vertices A, B, and C and the angle measures written

Based on this diagram, the following relation holds true.

$m ∠ A + m ∠ B + m ∠ C = 18 0^{\circ}$

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 $\overline{B C}$ through $A .$ Let $∠ 1$ and $∠ 2$ be the angles outside $△ A B C$ formed by this line and the sides $\overline{A B}$ and $\overline{A C} .$

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.

∠ B ≅ ∠ 1 ⇕ m ∠ B = m ∠ 1 ∠ C ≅ ∠ 2 ⇕ m ∠ C = m ∠ 2

Furthermore, in the diagram it can be seen that

∠ B A C,

∠ 1,

and

∠ 2

form a straight angle. Therefore, by the Angle Addition Postulate their measures add to

18 0^{\circ} .

m ∠ B A C + m ∠ 1 + m ∠ 2 = 18 0^{\circ}

By the Substitution Property of Equality, the sum of the measures of

∠ B A C,

∠ B,

and

∠ C

is equal to

18 0^{\circ} .

m ∠ B A C + m ∠ 1 + m ∠ 2 = 18 0^{\circ} ⇓ m ∠ B A C + m ∠ B + m ∠ C = 18 0^{\circ}

Finally, in

△ A B C,

∠ B A C

can be named

∠ A .

m ∠ B A C + m ∠ B + m ∠ C = 18 0^{\circ} ⇓ m ∠ A + m ∠ B + m ∠ C = 18 0^{\circ}

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

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.

3 5^{\circ} + 4 0^{\circ} + x = 18 0^{\circ}

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

3 5^{\circ} + 4 0^{\circ} + x = 18 0^{\circ}

AddTerms

Add terms

7 5^{\circ} + x = 18 0^{\circ}

SubEqn

$LHS - 7 5^{\circ} = RHS - 7 5^{\circ}$

x = 10 5^{\circ}

Explore

Investigating a Triangle's Exterior Angles

Once again, consider

△ A B C,

where

D

and

E

are the midpoints of

\overline{A B}

and

\overline{A C},

respectively. This time, begin by rotating

△ A B C

about

D .

Then, rotate the resulting figure about

E .

Considering the images and preimage, what can be concluded about exterior angle

∠ A C F

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.

$m ∠ P C A = m ∠ A + m ∠ B$

Proof

Using Properties of Angles

Consider a triangle with vertices $A,$ $B,$ and $C,$ and one of the exterior angles corresponding to $∠ C .$

The diagram shows that

∠ C

and

∠ P C A

form a linear pair, so the sum of their measures is

18 0^{\circ} .

Additionally, by the Triangle Angle Sum Theorem, the sum of the angle measures of

△ A B C

18 0^{\circ} .

{m ∠ C + m ∠ P C A = 18 0^{\circ} m ∠ A + m ∠ B + m ∠ C = 18 0^{\circ} (I) (II)

Now

m ∠ C

can be isolated in Equation (I).

m ∠ C + m ∠ P C A = 18 0^{\circ} ⇕ m ∠ C = 18 0^{\circ} - m ∠ P C A

Next, the expression of

m ∠ C

can be substituted into Equation (II).

m ∠ A + m ∠ B + m ∠ C = 18 0^{\circ}

Substitute

$m ∠ C = 18 0^{\circ} - m ∠ P C A$

m ∠ A + m ∠ B + (18 0^{\circ} - m ∠ P C A) = 18 0^{\circ}

▼

Solve for

m ∠ P C A

RemovePar

Remove parentheses

m ∠ A + m ∠ B + 18 0^{\circ} - m ∠ P C A = 18 0^{\circ}

SubEqn

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

m ∠ A + m ∠ B - m ∠ P C A = 0

AddEqn

$LHS + m ∠ P C A = RHS + m ∠ P C A$

m ∠ A + m ∠ B = m ∠ P C A

RearrangeEqn

Rearrange equation

m ∠ P C A = m ∠ A + m ∠ B

It has been proven that the measure of

∠ P C A

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 $△ A B C,$ where $D$ and $E$ are the midpoints of $\overline{A B}$ and $\overline{A C},$ respectively. Let $∠ P C A$ be one exterior angle of $△ A B C .$

Now,

△ A B C

can be rotated

18 0^{\circ}

about

D .

Since a rotation is a rigid motion, the image of

△ A B C

after the rotation is congruent to

△ A B C .

Corresponding parts of congruent figures are congruent, so the measures of the angles and the lengths of the sides remain unchanged.

Since a

18 0^{\circ} -

rotation is equivalent to a reflection,

\overline{C^{'} A}

is parallel to

\overline{B C}

and

\overline{C^{'} B}

is parallel to

\overline{A C} .

Therefore,

C^{'} A C B

is a parallelogram and

∠ C^{'} A C

is congruent to

∠ C B C^{'} .

Now the parallelogram

C^{'} A C B

will be rotated

18 0^{\circ}

about

E .

By the Parallelogram Opposite Angles Theorem,

∠ P C A

is congruent to

∠ A B^{''} P .

Congruent angles have the same measure by the definition.

∠ P C A m ∠ P C A ≅ ∠ A B^{''} P ⇕ = m ∠ A B^{''} P

Since

m ∠ A B^{''} P

is equal to the sum of

m ∠ A

and

m ∠ B

and because of the Transitive Property of Equality,

m ∠ P C A

is equal to the sum of

m ∠ A

and

m ∠ B .

{m ∠ P C A = m ∠ A B^{''} P m ∠ A B^{''} P = m ∠ A + m ∠ B ⇓ m ∠ P C A = 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.

Note that if the angle measure is less than

9 0^{\circ},

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

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 = 3 5^{\circ} + 4 0^{\circ} ⇓ x = 7 5^{\circ}

Since the measure of the exterior angle is less than $9 0^{\circ},$ the people sitting on this sofa can lean back and feel comfortable. Thanks for helping Dylan.

Explore

Investigating Properties of Isosceles Triangles

Given that

△ A B C

is a right triangle, reflect it across

\overline{A B} .

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.

$\overline{A B} ≅ \overline{A C}$ $\Rightarrow$ $∠ B ≅ ∠ C$

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

Proof

Geometric Approach

Consider a triangle $A B C$ with two congruent sides, or an isosceles triangle.

In this triangle, let

P

be the point of intersection of

\overline{B C}

and the angle bisector of

∠ A .

From the diagram, the following facts about

△ B A P

and

△ C A P

can be observed.

Statement	Reason
$∠ B A P ≅ ∠ C A P$	Definition of an angle bisector
$\overline{B A} ≅ \overline{C A}$	Given
$\overline{A P} ≅ \overline{A P}$	Reflexive Property of Congruence

Therefore,

△ B A P

and

△ C A P

have two pairs of corresponding congruent sides and one pair of congruent included angles. By the Side-Angle-Side Congruence Theorem,

△ B A P

and

△ C A P

are congruent triangles.

△ B A P ≅ △ C A P

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 Transformations

Consider an isosceles triangle $△ A B C .$

A line passing through $A$ and the midpoint of $\overline{B C}$ will be drawn. Let $P$ be the midpoint.

Since $\overline{B P}$ and $\overline{P C}$ 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 $A P .$ Also, because $A$ lies on $A P,$ a reflection across $A P$ maps $A$ onto itself. The same is true for $P .$

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

The table shows that the images of the vertices of

△ C A P

are the vertices of

△ B A P .

It can be concluded that

△ B A P

is the image of

△ C A P

after a reflection across

A P .

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.

∠ 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?

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

4 0^{\circ} .

Assuming that the measure of a base angle of the triangle is

x,

an equation can be written by the Interior Angles Theorem.

x + x + 4 0^{\circ} = 18 0^{\circ}

By solving this equation, the measure of the angle between the support beam and the bottom of the side can be found.

x + x + 4 0^{\circ} = 18 0^{\circ}

AddTerms

Add terms

2 x + 4 0^{\circ} = 18 0^{\circ}

SubEqn

$LHS - 4 0^{\circ} = RHS - 4 0^{\circ}$

2 x = 14 0^{\circ}

DivEqn

$LHS / 2 = RHS / 2$

x = 7 0^{\circ}

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

\overline{B A^{''}}

and

\overline{A C^{'''}} ?

Discussion

Midsegment of a Triangle

A midsegment of a triangle is a segment that connects the midpoints of two of the sides of a triangle. The midsegment of the triangle is parallel to the third side of the triangle.

Since a triangle has three sides, there are three midsegments between each pair of sides.

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.

\overline{D E}

is a midsegment of

△ A B C,

then the following statement holds true.

$\overline{D E} ∥ \overline{B C}$ and $D E = \frac{1}{2} B C$

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.

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 .

B (0, 0) C (a, 0) A (b, c)

\overline{D E}

is the midsegment from

\overline{B A}

\overline{C A},

then by the definition of a midpoint,

D

and

E

are the midpoints of

\overline{B A}

and

\overline{C A},

respectively.

To prove this theorem, it must be proven that $\overline{D E}$ is parallel to $\overline{B C}$ and that $D E$ is half of $B C .$

$\overline{B C} ∥ \overline{D E}$

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, $\overline{B C}$ is a horizontal segment. Next, the coordinates of $D$ and $E$ will be found using the Midpoint Formula.

$M (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
Segment	Endpoints	Substitute	Simplify
$\overline{B A}$	$B (0, 0)$ and $A (b, c)$	$D (\frac{0 + b}{2}, \frac{0 + c}{2})$	$D (\frac{b}{2}, \frac{c}{2})$
$\overline{C A}$	$C (a, 0)$ and $A (b, c)$	$E (\frac{a + b}{2}, \frac{0 + c}{2})$	$E (\frac{a + b}{2}, \frac{c}{2})$

The

y -

coordinate of both

D (\frac{b}{2}, \frac{c}{2})

and

E (\frac{a + b}{2}, \frac{c}{2})

\frac{c}{2} .

Therefore,

\overline{D E}

is also a horizontal segment. Since all horizontal segments are parallel, it can be said that

\overline{B C}

and

\overline{D E}

are parallel.

\overline{B C} ∥ \overline{D E} ✓

$D E = \frac{1}{2} B C$

Since both $\overline{B C}$ and $\overline{D E}$ are horizontal, their lengths are given by the difference of the $x -$ coordinates of their endpoints.

Segment	Endpoints	Length	Simplify
$\overline{B C}$	$B (0, 0)$ and $C (a, 0)$	$B C = a - 0$	$B C = a$
$\overline{D E}$	$D (\frac{b}{2}, \frac{c}{2})$ and $E (\frac{a + b}{2}, \frac{c}{2})$	$D E = \frac{a + b}{2} - \frac{b}{2}$	$D E = \frac{1}{2} a$

Since

\frac{1}{2} a

is half of

a,

it can be stated that the midsegment

\overline{D E}

is half the length of

\overline{B C} .

D E = \frac{1}{2} B C ✓

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.

To prove this theorem, it must be proven that

\overline{D E}

is parallel to

\overline{B C}

and that

D E

is equal to half of

B C .

Each statement will be proven one at a time.

$\overline{B C} ∥ \overline{D E}$

This part can be proven by using rigid motions. First, translate

△ A D E

along

\overline{D B}

so that

D

is mapped onto

B .

Since

D

is the midpoint of

\overline{A B},

A

is mapped onto

D .

Next, it must be proven that the image of

E

— which is marked as

E^{'}

— lies on

\overline{B C} .

This proof will be done using indirect reasoning. In this method, it is temporarily assumed that the negation of the statement is true.

Assume That

E^{'}

Does Not Lie on

\overline{B C}

Assume that $E^{'},$ the image of $E$ after the translation along $\overline{D B},$ does not lie on $\overline{B C} .$ This means that $E^{'}$ lies either above or below $\overline{B C} .$ 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 $B E^{'}$ and $\overline{E C} .$

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

Next, it will be proven that $△ A D E$ is congruent to $△ E E^{'} F .$

Show That

△ A D E ≅ △ E E^{'} F

It is given that $A D = D B$ because $D$ is the midpoint of $\overline{A B} .$ Additionally, since $△ A D E$ is translated along $\overline{D B},$ it can be concluded that $D B = E E^{'} .$ Then, by the Transitive Property of Equality, $A D = E E^{'} .$

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

Recall that $A E = D E^{'} .$ Additionally, $\overline{D E}$ is the common side of $△ A D E$ and $△ E^{'} E D .$ All pieces of information can now be summarized.

$A D = E E^{'}$
$A E = D E^{'}$
$\overline{D E}$ is the common side of $△ A D E$ and $△ E^{'} E D .$

Using all the information,

△ A D E

is congruent to

△ E^{'} E D

by the Side-Side-Side Congruence Theorem.

△ A D E ≅ △ E^{'} E D

Because corresponding angles of congruent triangles are congruent,

∠ A D E

is congruent to

∠ D E E^{'} .

Since translations preserve angles,

\overline{D E}

is parallel to

\overline{B F} .

By the Alternate Interior Angles Theorem,

∠ D E E^{'}

is congruent to

∠ E E^{'} F .

∠ A D E ≅ ∠ D E E^{'} and ∠ D E E^{'} ≅ ∠ E E^{'} F

Therefore, by the Transitive Property of Congruence,

∠ A D E ≅ ∠ E E^{'} F .

Since

\overline{D E}

is parallel to

\overline{B F}

and the image of

E

is translated in the same direction as the image of

D,

∠ B D E^{'}

is congruent to

∠ E^{'} E F .

Additionally,

∠ B D E^{'}

is congruent to

∠ D A E .

∠ E^{'} E F ≅ ∠ B D E^{'} and ∠ B D E^{'} ≅ ∠ D A E

One more time, by the Transitive Property of Congruence,

∠ E^{'} E F ≅ ∠ D A E .

Summarize the obtained information about the triangles $△ A D E$ and $△ E E^{'} F .$

$A D = E E^{'}$
$∠ A D E ≅ ∠ E E^{'} F$
$∠ D A E ≅ ∠ E^{'} E F$

Therefore, by the Angle-Side-Angle Congruence Theorem, $△ A D E$ is congruent to $△ E E^{'} F .$

Contradiction

It has been proven that

△ A D E ≅ △ E E^{'} F .

Because corresponding sides of congruent triangles are congruent, it follows that

E F

is equal to

A E .

E F = A E

This means that

E F

is equal to

E C,

because

E

is the midpoint of

\overline{A C} .

However, since

F

lies between

E

and

C,

it cannot be true that

E F = E C .

Therefore, the temporary assumption leads to a contradiction.

Assumption of the Theorem	Indirect Assumption
$A E = E F = E C$	$E F < E C$
$E F = E C$ and $E F < E C$ $\times$

Therefore, this contradiction verifies that the image of $E$ must lie on $\overline{B C} .$

It has been proven that

E^{'}

lies on

\overline{B C} .

Because translations preserve angles,

∠ A D E

is congruent to

∠ D B E^{'} .

By the Converse Corresponding Angles Theorem, $\overline{B C}$ is parallel to $\overline{D E} .$

$\overline{B C} ∥ \overline{D E}$

$D E = \frac{1}{2} B C$

It has been previously obtained that

\overline{B C}

and

\overline{D E}

are parallel. Now, another rigid motion to

△ B D E^{'}

will be applied.

Rotate

△ B D E^{'}

Around the Midpoint of

\overline{D E^{'}}

Rotate

△ B D E^{'}

counterclockwise about the midpoint of

\overline{D E^{'}}

so that

E^{'}

is mapped onto

D .

It can be noted that the triangle is rotated

18 0^{\circ} .

Since rotations preserve angles and lengths, this rotation maps

B

onto

E .

Therefore,

\overline{B D}

is mapped onto

\overline{E E^{'}} .

The Midsegment Theorem proof first rotation

As a result of the rotation, it can be concluded that

△ B D E^{'}

and

△ B^{''} D^{''} E^{''},

△ E E^{'} D,

are congruent triangles.

Rotate

△ D E D^{''}

Around the Midpoint of

\overline{D^{''} E}

Rotate

△ D E D^{''}

counterclockwise about the midpoint of

\overline{D^{''} E}

so that

E

is mapped onto

D^{''} .

It can be noted that the triangle is again rotated

18 0^{\circ} .

Since rotations preserve angles and lengths, this rotation maps

D

onto

C .

Therefore,

\overline{D D^{''}}

and

\overline{D E}

are mapped onto

\overline{C E}

and

\overline{C E^{'}},

respectively.

The Midsegment Theorem proof second rotation

As a result of the rotation, it can be concluded that

△ D E D^{''}

and

△ E E^{'} C^{''}

are congruent triangles.

By the Segment Addition Postulate, the length of

\overline{B C}

can be calculated by adding the lengths of smaller segments.

B C = B E^{'} + C E^{'}

Because corresponding sides of congruent triangles are congruent,

\overline{D E}

is congruent to

\overline{B E^{'}}

and

\overline{D E}

is congruent to

\overline{C E^{'}} .

Therefore,

D E = B E^{'}

and

D E = C E^{'} .

By the Substitution Property of Equality,

B C

can be expressed in terms of

D E .

B C = D E + D E \Leftrightarrow B C = 2 D E

Finally, by the Division Property of Equality, the second statement of the theorem is obtained.

$D E = \frac{1}{2} B C$

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.

\frac{3}{2} = 1.5 ft

The width of the seat is

1.5

feet.

Closure

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.

∙ A C = C F ∙ A B = B D ∙ D E = E F ∙ \overline{B C} ∥ \overline{D F} ∙ \overline{C E} ∥ \overline{A D}

From here, what are the lengths of

\overline{A D}

and

\overline{B C} ?

Hint

Use the Triangle Midsegment Theorem.

Solution

By the definition of a midsegment, both $\overline{B C}$ and $\overline{C E}$ are midsegments of $△ A D F .$ By the Triangle Midsegment Theorem, $C E$ is half of $A D,$ and $B C$ is half of $D F .$

C E B C = \frac{1}{2} A D = \frac{1}{2} D F

Knowing that $C E$ is $43$ inches and $D F$ is $68$ inches, these values can be substitute into these equations to find $A D$ and $B C .$

43 B C = \frac{1}{2} A D \Rightarrow A D = 86 = \frac{1}{2} (68) \Rightarrow B C = 34

Therefore, the length of $\overline{A D}$ is $86$ inches and the length of $\overline{B C}$ is $34$ inches.

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Catch-Up and Review

Proof

Hint

Solution

Proof

Proof

Hint

Solution

Proof

Proof

Hint

Solution

Proof

BC∥DE

DE=21​BC

Proof

BC∥DE

DE=21​BC

Hint

Solution

Hint

Solution

$\overline{B C} ∥ \overline{D E}$

$D E = \frac{1}{2} B C$

$\overline{B C} ∥ \overline{D E}$

$D E = \frac{1}{2} B C$