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# 5.1: Midsegments

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Define midsegment.
• Use the Midsegment Theorem.

## Review Queue

Find the midpoint between the given points.

1. (-4, 1) and (6, 7)
2. (5, -3) and (11, 5)
3. Find the equation of the line between (-2, -3) and (-1, 1).
4. Find the equation of the line that is parallel to the line from #3 through (2, -7).

Know What? A fractal is a repeated design using the same shape (or shapes) of different sizes. Below, is an example of the first few steps of a fractal. Draw the next figure in the pattern.

## Defining Midsegment

Midsegment: A line segment that connects two midpoints of the sides of a triangle.

$\overline{DF}$ is the midsegment between $\overline{AB}$ and $\overline{BC}$.

The tic marks show that $D$ and $F$ are midpoints.

$\overline{AD} \cong \overline{DB}$ and $\overline{BF} \cong \overline{FC}$

Example 1: Draw the midsegment $\overline{DE}$ between $\overline{AB}$ and $\overline{AC}$ for $\triangle ABC$ above.

Solution: Find the midpoints of $\overline{AB}$ and $\overline{AC}$ using your ruler. Label these points $D$ and $E$. Connect them to create the midsegment.

Example 2: You now have all three midpoints of $\triangle ABC$. Draw in midsegment $\overline{DF}$ and $\overline{FE}$.

Solution:

For every triangle there are three midsegments.

## Midsegments in the $x-y$ Plane

Let’s transfer what we know about midpoints in the $x-y$ plane to midsegments in the $x-y$ plane. We will need to use the midpoint formula, $\left ( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right )$.

Example 3: The vertices of $\triangle LMN$ are $L(4, 5), \ M(-2, -7)$ and $N(-8, 3)$. Find the midpoints of all three sides, label them $O, \ P$ and $Q$. Then, graph the triangle, plot the midpoints and draw the midsegments.

Solution: Use the midpoint formula 3 times to find all the midpoints.

$L$ and $M = \left ( \frac{4 + (-2)}{2}, \frac{5 + (-7)}{2} \right ) = (1, -1)$ point $O$

$M$ and $N = \left ( \frac{-2 + (-8)}{2}, \frac{-7 + 3}{2} \right ) = (-5, -2)$, point $P$

$L$ and $N = \left ( \frac{4 + (-8)}{2}, \frac{5 + 3}{2} \right ) = (-2, 4)$, point $Q$

The graph is to the right.

Example 4: Find the slopes of $\overline{NM}$ and $\overline{QO}$.

Solution: The slope of $\overline{NM}$ is $\frac{-7-3}{-2-(-8)} = \frac{-10}{6} = - \frac{5}{3}$. The slope of $\overline{QO}$ is $\frac{-1-4}{1-(-2)} = - \frac{5}{3}$.

From this we can conclude that $\overline{NM} \| \overline{QO}$. If we were to find the slopes of the other sides and midsegments, we would find $\overline{LM} \| \overline{QP}$ and $\overline{NL} \| \overline{PO}$.

Example 5: Find $NM$ and $QO$.

Solution: Now, we need to find the lengths of $\overline{NM}$ and $\overline{QO}$. Use the distance formula.

$NM &= \sqrt{(-7 -3)^2 + (-2 - (-8))^2} = \sqrt{(-10)^2 + 6^2} = \sqrt{100 + 36} = \sqrt{136} \approx 11.66\\QO &= \sqrt{(1 - (-2))^2 + (-1 - 4)^2} = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34} \approx 5.83$

From this we can conclude that $QO$ is half of $NM$. If we were to find the lengths of the other sides and midsegments, we would find that $OP$ is half of $NL$ and $QP$ is half of $LM$.

## The Midsegment Theorem

The conclusions drawn in Examples 4 and 5 can be combined into the Midsegment Theorem.

Midsegment Theorem: The midsegment of a triangle is half the length of the side it is parallel to.

If $\overline{DF}$ is a midsegment of $\triangle ABC$, then $DF = \frac{1}{2} AC = AE = EC$ and $\overline{DF} \| \overline{AC}$.

Example 6a: Mark all the congruent segments on $\triangle ABC$ with midpoints $D, \ E$, and $F$.

Solution: Drawing in all three midsegments, we have:

Also, this means the four triangles are congruent by SSS.

Example 6b: Mark all the parallel lines on $\triangle ABC$, with midpoints $D, \ E$, and $F$.

Solution:

To play with the properties of midsegments, go to http://www.mathopenref.com/trianglemidsegment.html.

Example 7: $M, \ N$, and $O$ are the midpoints of the sides of the triangle.

Find

a) $MN$

b) $XY$

c) The perimeter of $\triangle XYZ$

Solution: Use the Midsegment Theorem.

a) $MN = OZ = 5$

b) $XY = 2(ON) = 2 \cdot 4 = 8$

c) Add up the three sides of $\triangle XYZ$ to find the perimeter.

$XY + YZ + XZ = 2 \cdot 4 + 2 \cdot 3 + 2 \cdot 5 = 8 + 6 + 10 = 24$

Remember: No line segment over $MN$ means length or distance.

Example 8: Algebra Connection Find the value of $x$ and $AB$. $A$ and $B$ are midpoints.

Solution: $AB = 34 \div 2 = 17$. To find $x$, set $3x - 1$ equal to 17.

$3x - 1 & = 17\\3x & = 18\\x & =6$

Know What? Revisited To the left is a picture of the $4^{th}$ figure in the fractal pattern.

## Review Questions

• Questions 1-5 use the definition of a midsegment and the Midsegment Theorem.
• Questions 6-9 and 18 are similar to Example 7.
• Questions 10-17 are similar to Example 8.
• Questions 19-22 are similar to Example 3.
• Questions 23-30 are similar to Examples 3, 4, and 5.

Determine if each statement is true or false.

1. The endpoints of a midsegment are midpoints.
2. A midsegment is parallel to the side of the triangle that it does not intersect.
3. There are three congruent triangles formed by the midsegments and sides of a triangle.
4. If a line passes through two sides of a triangle and is parallel to the third side, then it is a midsegment.
5. There are three midsegments in every triangle.

$R, \ S, \ T$, and $U$ are midpoints of the sides of $\triangle XPO$ and $\triangle YPO$.

1. If $OP = 12$, find $RS$ and $TU$.
2. If $RS = 8$, find $TU$.
3. If $RS = 2x$, and $OP = 20$, find $x$ and $TU$.
4. If $OP = 4x$ and $RS = 6x - 8$, find $x$.

For questions 10-17, find the indicated variable(s). You may assume that all line segments within a triangle are midsegments.

1. The sides of $\triangle XYZ$ are 26, 38, and 42. $\triangle ABC$ is formed by joining the midpoints of $\triangle XYZ$.
1. What are the lengths of the sides of $\triangle ABC$?
2. Find the perimeter of $\triangle ABC$.
3. Find the perimeter of $\triangle XYZ$.
4. What is the relationship between the perimeter of a triangle and the perimeter of the triangle formed by connecting its midpoints?

Coordinate Geometry Given the vertices of $\triangle ABC$ below find the midpoints of each side.

1. $A(5, -2), \ B(9, 4)$ and $C(-3, 8)$
2. $A(-10, 1), \ B(4, 11)$ and $C(0, -7)$
3. $A(-1, 3), \ B(5, 7)$ and $C(9, -5)$
4. $A(-4, -15), \ B(2, -1)$ and $C(-20, 11)$

Multi-Step Problem The midpoints of the sides of a triangle are $A(1, 5), \ B(4, -2)$, and $C(-5, 1)$. Answer the following questions. The graph is below.

1. Find the slope of $AB, \ BC$, and $AC$.
2. The side that passes through $A$ should be parallel to which midsegment? ($\triangle ABC$ are all midsegments of a triangle).
3. Using your answer from #24, take the slope of $\overline{BC}$ and use the “rise over run” in either direction to create a parallel line to $\overline{BC}$ that passes through $A$. Extend it with a ruler.
4. Repeat #24 and #25 with $B$ and $C$. What are coordinates of the larger triangle?

Multi-Step Problem The midpoints of the sides of $\triangle RST$ are $G(0, -2), \ H(9, 1)$, and $I(6, -5)$. Answer the following questions.

1. Find the slope of $GH, \ HI$, and $GI$.
2. Plot the three midpoints and connect them to form midsegment triangle, $\triangle GHI$.
3. Using the slopes, find the coordinates of the vertices of $\triangle RST$. (#22 above)
4. Find $GH$ using the distance formula. Then, find the length of the sides it is parallel to. What should happen?

## Review Queue Answers

1. $\left ( \frac{-4+6}{2}, \frac{1+7}{2} \right ) = (1, 4)$
2. $\left ( \frac{5+11}{2}, \frac{-3+5}{2} \right ) = (8, 1)$
3. $m=\frac{-3-1}{-2-(-1)} = \frac{-4}{-1}=4\!\\y=mx+b\!\\-3=4(-2)+b\!\\b=5, \ y=4x+5$
4. $-7=4(2)+b\!\\b=-15, \ y=4x-15$

8 , 9 , 10

## Date Created:

Feb 22, 2012

Dec 11, 2014
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