Triangle Theorems: Insights into the Triangle Midsegment Theorem

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

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

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

Reflection Across

A P

Preimage

Image

C

B

A

A

P

P

$M (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
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 $A (b, c)$	$E (\frac{a + b}{2}, \frac{0 + c}{2})$	$E (\frac{a + b}{2}, \frac{c}{2})$

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})

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 $E (\frac{a + b}{2}, \frac{c}{2})$	$D E = \frac{a + b}{2} - \frac{b}{2}$	$D E = \frac{1}{2} a$

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

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

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

Figure $1$ is an equilateral triangle. A relationship exists from one figure to the next. Each shaded triangle is formed by connecting the midsegments of opposite sides in every white triangle from one figure to the next.

triangles where the midpoints of a triangle are connected

The perimeter of Figure $1$ is $120$ inches. What is the total perimeter of every shaded triangle we would find in Figure $10$ (not pictured)? Round the total perimeter of Figure $10$ to the nearest inch.

We could find a pattern for the perimeter of the additional shaded triangles from figure to figure. That will help us calculate the total perimeter of each shaded triangle in Figure 10. To do so, let's begin by examining the first triangle where shaded triangles have yet to be added.

Examine Figure 1

The perimeter of Figure 1 is 120 inches. Since there are three sides to a triangle, we can divide 120 by 3 to determine that each side is 40 inches.

From Figure 1 to Figure 2

Now let's get to the fun part. In Figure 2, each vertex of the shaded triangle intercepts a midpoint of a side in Figure 1. Using the Triangle Midsegment Theorem, we can find the length of the shaded triangle's sides.

Since the perimeter P of a triangle is the sum of its side lengths, we can calculate the perimeter of the shaded triangle. P=20+20+20 ⇓ P= 60 inches

From Figure 2 to Figure 3

In Figure 3, three more triangles have been added — with each side measuring at half the length of Figure 2's shaded triangle side length.

Notice that the side length of each of the smaller shaded triangles can be written as a fraction of the shaded triangle from Figure 2. That is, their side lengths can be written as the fraction 202. We have purposefully rewritten the fraction to include the common number of 20. Now, let's find the perimeter of one of these triangles. P=20/2+20/2+20/2 ⇓ P= 20+20+20/2 There are three of these triangles. Therefore, multiply the perimeter of one triangle by 3 to obtain their combined perimeter. P=3( 20+20+20/2) ⇓ P= 3/2(60) inches

From Figure 3 to Figure 4

It gets more fun. Let's analyze what happens when we go from Figure 3 to Figure 4.

In Figure 4, nine more triangles are included. Again, writing in a fraction that uses the values of the first shaded triangle from Figure 2, each of the additional triangles in Figure 4 are measured to be 204. That is, they are a quarter of the side of the first shaded triangle in Figure 2. Let's calculate the combined perimeter of the additional shaded triangles. P=9(20+20+20/4) ⇓ P= (3/2)^2(60) inches

Recognizing the Pattern

If we look at the right-hand side of the equations describing the perimeters of each figure's additional shaded regions, we can identify a geometric sequence. That is, a geometric sequence is a sequence where the ratio of any term to its preceding term is constant, and each term of a geometric sequence is multiplied by a common ratio to get the next term. 60, 3/2(60), (3/2)^2(60)

Finding the Sum

The geometric sequence has a first term of 60 and a common ratio of 32. To calculate the sum of a finite geometric sequence, we use the following formula. S_n=a_1(r^n-1)/r-1 In this formula, a_1 is the first term, r is the common ratio, and n is the number of terms. Notice that between Figure 2 and Figure 10 we have 9 terms which means n= 9. By substituting a_1= 60, r= 32, and n= 9, we can find the sum of the triangle's perimeters in the tenth figure.

Emilia's little sister drew a star on her bed frame. Emilia thinks it is a good chance to share some math with her. Emilia wants to find the sum of the measures of $∠ a,$ $∠ b,$ $∠ c,$ $∠ d,$ and $∠ e ?$

To determine the sum of the measures of the five angles, we will focus on one of the triangles. Let's arbitrarily choose the triangle with angle b. We can then introduce two more angles in the diagram to find the value of b.

According to the Interior Angles Theorem, the sum of the measures of ∠ b, ∠ f, and ∠ g equals 180^(∘). m∠ b+m∠ f+m∠ g = 180^(∘) Examining the diagram, we also see that ∠ f and ∠ g are the exterior angles of two separate triangles.

According to the Triangle Exterior Angle Theorem, the measure of the exterior angle of a triangle equals the sum of the measures of the triangle's non-adjacent angles. m∠ g= m∠ d+m∠ a m∠ f= m∠ e+m∠ c Now that we have expressions for ∠ g and ∠ f, we can substitute these into the first equation.

Emilia can now teach her little sister that the sum of the angle measures is 180^(∘).

	16 Theory slides
	13 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Theorems About Triangles

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

Examine Figure 1

From Figure 1 to Figure 2

From Figure 2 to Figure 3

From Figure 3 to Figure 4

Recognizing the Pattern

Finding the Sum

Theorems About Triangles

Recommended exercises

$\overline{B C} ∥ \overline{D E}$

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

$\overline{B C} ∥ \overline{D E}$

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