4.19: Midsegment Theorem
- Page ID
- 4816
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Midsegment of a triangle joins the midpoints of two sides and is half the length of the side it is parallel to.
A line segment that connects two midpoints of the sides of a triangle is called a midsegment. \(\overline{DF}\) is the midsegment between \(\overline{AB}\) and \(\overline{BC}\).
![f-d_7a6fdd253dc7cd459a6328b4683b6685e37937d39abc1a821e21c3cf+IMAGE_TINY+IMAGE_TINY.png](https://k12.libretexts.org/@api/deki/files/1504/f-d_7a6fdd253dc7cd459a6328b4683b6685e37937d39abc1a821e21c3cf%252BIMAGE_TINY%252BIMAGE_TINY.png?revision=1&size=bestfit&width=450)
The tic marks show that \(D\) and \(F\) are midpoints. \(\overline{AD}\cong \overline{DB}\) and \(\overline{BF}\cong \overline{FC}\). For every triangle there are three midsegments.
![f-d_e4bf19775f25fca673edadd5dc98ade9dad694f6d67e572b2b111afd+IMAGE_TINY+IMAGE_TINY.png](https://k12.libretexts.org/@api/deki/files/1505/f-d_e4bf19775f25fca673edadd5dc98ade9dad694f6d67e572b2b111afd%252BIMAGE_TINY%252BIMAGE_TINY.png?revision=1&size=bestfit&width=450)
There are two important properties of midsegments that combine to make the Midsegment Theorem. The Midsegment Theorem states that the midsegment connecting the midpoints of two sides of a triangle is parallel to the third side of the triangle, and the length of this midsegment is half the length of the third side. So, if \(\overline{DF}\) is a midsegment of \(\Delta ABC\), then \(DF=\dfrac{1}{2}AC=AE=EC\) and \(\overline{DF} \parallel \overline{AC}\).
![f-d_223aa51dc6b4e68bbc20a3665d6597ec145252ae4546c6d52ae237cc+IMAGE_TINY+IMAGE_TINY.png](https://k12.libretexts.org/@api/deki/files/1506/f-d_223aa51dc6b4e68bbc20a3665d6597ec145252ae4546c6d52ae237cc%252BIMAGE_TINY%252BIMAGE_TINY.png?revision=1&size=bestfit&width=450)
Note that there are two important ideas here. One is that the midsegment is parallel to a side of the triangle. The other is that the midsegment is always half the length of this side.
What if you were given \(\Delta FGH\) and told that \(\overline{JK}\) was its midsegment? How could you find the length of \(JK\) given the length of the triangle's third side, \(FH\)?
Example \(\PageIndex{1}\)
Find the value of \(x\) and AB. \(A\) and \(B\) are midpoints.
![f-d_b5882a9b2e17a9efb1a0bc04537a87463176c41d73f87a1f36d8e72e+IMAGE_TINY+IMAGE_TINY.png](https://k12.libretexts.org/@api/deki/files/1507/f-d_b5882a9b2e17a9efb1a0bc04537a87463176c41d73f87a1f36d8e72e%252BIMAGE_TINY%252BIMAGE_TINY.png?revision=1&size=bestfit&width=450)
Solution
\(AB=34\div 2=17\). To find \(x\), set \(3x−1\) equal to 17.
\(\begin{align*} 3x−1&=17 \\ 3x&=18 \\ x&=6\end{align*}\)
Example \(\PageIndex{2}\)
True or false: If a line passes through two sides of a triangle and is parallel to the third side, then it is a midsegment.
Solution
This statement is false. A line that passes through two sides of a triangle is only a midsegment if it passes through the midpoints of the two sides of the triangle.
Example \(\PageIndex{3}\)
The vertices of \(\Delta LMN\) 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
To solve this problem, use the midpoint formula 3 times to find all the midpoints. Recall that the midpoint formula is \(\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\).
\(L\) and \(M=\left(\dfrac{4+(−2)}{2}, \dfrac{5+(−7)}{2}\right)=(1,−1),\: point\: O\)
\(M\) and \(N=\left(\dfrac{−2+(−8)}{2},\dfrac{−7+3}{2}\right)=(−5,−2),\: point\: P\)
\(L\) and \(N=\left(\dfrac{4+(−8)}{2}, \dfrac{5+3}{2}\right)=(−2,4),\: point\: Q\)
![f-d_8a9d8e4bdbbb1b6825c4d7d618b47343ba54f174d435c4e2c29d7cee+IMAGE_TINY+IMAGE_TINY.png](https://k12.libretexts.org/@api/deki/files/1508/f-d_8a9d8e4bdbbb1b6825c4d7d618b47343ba54f174d435c4e2c29d7cee%252BIMAGE_TINY%252BIMAGE_TINY.png?revision=1&size=bestfit&width=450)
Example \(\PageIndex{4}\)
![f-d_223aa51dc6b4e68bbc20a3665d6597ec145252ae4546c6d52ae237cc+IMAGE_TINY+IMAGE_TINY.png](https://k12.libretexts.org/@api/deki/files/1506/f-d_223aa51dc6b4e68bbc20a3665d6597ec145252ae4546c6d52ae237cc%252BIMAGE_TINY%252BIMAGE_TINY.png?revision=1&size=bestfit&width=450)
Mark all the congruent segments on \(\Delta ABC\) with midpoints \(D\), \(E\), and \(F\).
Solution
Drawing in all three midsegments, we have:
![f-d_31c4ae81be92cfb678bb438a2b50cc826368bb6f064866fe182ff74b+IMAGE_TINY+IMAGE_TINY.png](https://k12.libretexts.org/@api/deki/files/1509/f-d_31c4ae81be92cfb678bb438a2b50cc826368bb6f064866fe182ff74b%252BIMAGE_TINY%252BIMAGE_TINY.png?revision=1&size=bestfit&width=450)
Also, this means the four smaller triangles are congruent by SSS.
Now, mark all the parallel lines on \(\Delta ABC\), with midpoints \(D\), \(E\), and \(F\).
![f-d_5ff85da6406c46e1b200d0e870306af335ce7db019c0b63c74dfec2b+IMAGE_TINY+IMAGE_TINY.png](https://k12.libretexts.org/@api/deki/files/1510/f-d_5ff85da6406c46e1b200d0e870306af335ce7db019c0b63c74dfec2b%252BIMAGE_TINY%252BIMAGE_TINY.png?revision=1&size=bestfit&width=450)
Example \(\PageIndex{5}\)
\(M\), \(N\), and \(O\) are the midpoints of the sides of \(\Delta \(x\)YZ\).
![f-d_e3142e219ecaad6b5b0ed87dd22e99ecaff22908272b2137944f9131+IMAGE_TINY+IMAGE_TINY.png](https://k12.libretexts.org/@api/deki/files/1511/f-d_e3142e219ecaad6b5b0ed87dd22e99ecaff22908272b2137944f9131%252BIMAGE_TINY%252BIMAGE_TINY.png?revision=1&size=bestfit&width=450)
Solution
Find \(MN\), \(XY\), and the perimeter of \(\Delta \(x\)YZ\).
Use the Midsegment Theorem:
\(MN=OZ=5\)
\(XY=2(ON)=2\cdot 4=8\)
Add up the three sides of \(\Delta XYZ\) 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.
Review
Determine whether each statement is true or false.
- The endpoints of a midsegment are midpoints.
- A midsegment is parallel to the side of the triangle that it does not intersect.
- There are three congruent triangles formed by the midsegments and sides of a triangle.
- There are three midsegments in every triangle.
R, S, T, and U are midpoints of the sides of \(\Delta XPO\) and \(\Delta YPO\)
![f-d_2441e8d9c75f7d5d4c227b492ef6167257c0f3240c09b33e419e9e90+IMAGE_TINY+IMAGE_TINY.png](https://k12.libretexts.org/@api/deki/files/1512/f-d_2441e8d9c75f7d5d4c227b492ef6167257c0f3240c09b33e419e9e90%252BIMAGE_TINY%252BIMAGE_TINY.png?revision=1&size=bestfit&width=450)
- If \(OP=12\), find \(RS\) and \(TU\).
- If \(RS=8\), find \(TU\).
- If \(RS=2x\), and \(OP=20\), find \(x\) and \(TU\).
- If \(OP=4x\) and \(RS=6x−8\), find \(x\).
For questions 9-15, find the indicated variable(s). You may assume that all line segments within a triangle are midsegments.
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Figure \(\PageIndex{11}\) -
Figure \(\PageIndex{12}\) -
Figure \(\PageIndex{13}\) -
Figure \(\PageIndex{14}\) -
Figure \(\PageIndex{15}\) -
Figure \(\PageIndex{16}\) -
Figure \(\PageIndex{17}\) - The sides of \(\Delta XYZ\) are 26, 38, and 42. \(\Delta ABC\) is formed by joining the midpoints of \(\Delta XYZ\).
- What are the lengths of the sides of \(\Delta ABC\)?
- Find the perimeter of \(\Delta ABC\).
- Find the perimeter of \(\Delta XYZ\).
- 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 \(\Delta ABC\) below find the midpoints of each side.
- \(A(5,−2),\: B(9,4)\: and\: C(−3,8)\)
- \(A(−10,1),\: B(4,11)\: and \:C(0,−7)\)
- \(A(−1,3),\: B(5,7)\: and\: C(9,−5)\)
- \(A(−4,−15),\: B(2,−1)\: and\: C(−20,11)\)
Review (Answers)
To see the Review answers, open this PDF file and look for section 5.1.
Resources
Vocabulary
Term | Definition |
---|---|
midsegment | A midsegment connects the midpoints of two sides of a triangle or the non-parallel sides of a trapezoid. |
Congruent | Congruent figures are identical in size, shape and measure. |
Midpoint Formula | The midpoint formula says that for endpoints \((x_1,y_1)\) and \((x_2,y_2)\), the midpoint is (\dfrac{x_1+x_2}{2}, \frac{y_1+y_2}{2})\). |
Additional Resources
Video: Determining Unknown Values Using Properties of the Midsegments of a Triangle
Activities: Midsegment Theorem Discussion Questions
Study Aids: Bisectors, Medians, Altitudes Study Guide
Practice: Midsegment Theorem
Real World: Midsegment Theorem