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Midsegment Theorem

Midsegment of a triangle joins the midpoints of two sides and is half the length of the side it is parallel to.

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Midsegment Theorem

What if you created a repeated design using the same shape (or shapes) of different sizes? This would be called a fractal. Below, is an example of the first few steps of one. What does the next figure look like? How many triangles are in each figure (green and white triangles)? Is there a pattern? After completing this Concept, you'll be able to better understand how these fractals are created.

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CK-12 Foundation: Chapter5MidsegmentTheoremA

James Sousa: Introduction to the Midsegments of a Triangle

James Sousa: Determining Unknown Values Using Properties of the Midsegments of a Triangle


A midsegment is a line segment that connects two midpoints of adjacent sides of a triangle. For every triangle there are three midsegments. The Midsegment Theorem states that the midsegment of a triangle is half the length of the side it is parallel to.

Example A

Draw the midsegment \begin{align*}\overline{DF}\end{align*} between \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{BC}\end{align*}. Use appropriate tic marks.

Find the midpoints of \begin{align*}\overline{AB}\end{align*} and \begin{align*}\overline{BC}\end{align*} using your ruler. Label these points \begin{align*}D\end{align*} and \begin{align*}F\end{align*}. Connect them to create the midsegment.

Don’t forget to put the tic marks, indicating that \begin{align*}D\end{align*} and \begin{align*}F\end{align*} are midpoints, \begin{align*}\overline{AD} \cong \overline{DB}\end{align*} and \begin{align*}\overline{BF} \cong \overline{FC}\end{align*}.

Example B

Find the midpoint of \begin{align*}\overline{AC}\end{align*} from \begin{align*}\triangle ABC\end{align*}. Label it \begin{align*}E\end{align*} and find the other two midsegments of the triangle.

Example C

Mark everything you have learned from the Midsegment Theorem on \begin{align*}\triangle ABC\end{align*}.

Let’s draw two different triangles, one for the congruent sides, and one for the parallel lines.

Because the midsegments are half the length of the sides they are parallel to, they are congruent to half of each of those sides (as marked). Also, this means that all four of the triangles in \begin{align*}\triangle ABC\end{align*}, created by the midsegments are congruent by SSS.

As for the parallel midsegments and sides, several congruent angles are formed. In the picture to the right, the pink and teal angles are congruent because they are corresponding or alternate interior angles. Then, the purple angles are congruent by the \begin{align*}3^{rd}\end{align*} Angle Theorem.

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

Example D

\begin{align*}M, N,\end{align*} and \begin{align*}O\end{align*} are the midpoints of the sides of the triangle.


a) \begin{align*}MN\end{align*}

b) \begin{align*}XY\end{align*}

c) The perimeter of \begin{align*}\triangle XYZ\end{align*}

Use the Midsegment Theorem.

a) \begin{align*}MN = OZ = 5\end{align*}

b) \begin{align*}XY = 2(ON) = 2 \cdot 4 = 8\end{align*}

c) The perimeter is the sum of the three sides of \begin{align*}\triangle XYZ\end{align*}.

\begin{align*}XY + YZ + XZ = 2 \cdot 4 + 2 \cdot 3 + 2 \cdot 5 = 8 + 6 + 10 = 24\end{align*}

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter5MidsegmentTheoremB

Concept Problem Revisited

To the left is a picture of the \begin{align*}4^{th}\end{align*} figure in the fractal pattern. The number of triangles in each figure is 1, 4, 13, and 40. The pattern is that each term increase by the next power of 3.


A line segment that connects two midpoints of the sides of a triangle is called a midsegment. A midpoint is a point that divides a segment into two equal pieces. Two lines are parallel if they never intersect. Parallel lines have slopes that are equal. In a triangle, midsegments are always parallel to one side of the triangle.

Guided Practice

The vertices of \begin{align*}\triangle LMN\end{align*} are \begin{align*}L(4, 5), M(-2, -7)\end{align*} and \begin{align*}N(-8, 3)\end{align*}.

1. Find the midpoints of all three sides, label them \begin{align*}O, P\end{align*} and \begin{align*}Q\end{align*}. Then, graph the triangle, it’s midpoints and draw in the midsegments.

2. Find the slopes of \begin{align*}\overline{NM}\end{align*} and \begin{align*}\overline{QO}\end{align*}.

3. Find \begin{align*}NM\end{align*} and \begin{align*}QO\end{align*}.

4. If the midpoints of the sides of a triangle are \begin{align*}A(1, 5), B(4, -2)\end{align*}, and \begin{align*}C(-5, 1)\end{align*}, find the vertices of the triangle.


1. Use the midpoint formula 3 times to find all the midpoints.

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

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

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

The graph would look like the graph below.

2. The slope of \begin{align*}\overline{NM}\end{align*} is \begin{align*}\frac{-7-3}{-2-(-8)}=\frac{-10}{6}=-\frac{5}{3}\end{align*}.

The slope of \begin{align*}\overline{QO}\end{align*} is \begin{align*}\frac{-1-4}{1-(-2)}=-\frac{5}{3}\end{align*}.

From this we can conclude that \begin{align*}\overline{NM} \ || \ \overline{QO}\end{align*}. If we were to find the slopes of the other sides and midsegments, we would find \begin{align*}\overline{LM} \ || \ \overline{QP}\end{align*} and \begin{align*}\overline{NL} \ || \ \overline{PO}\end{align*}. This is a property of all midsegments.

3. Now, we need to find the lengths of \begin{align*}\overline{NM}\end{align*} and \begin{align*}\overline{QO}\end{align*}. Use the distance formula.

\begin{align*}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\end{align*}

Note that \begin{align*}QO\end{align*} is half of \begin{align*}NM\end{align*}.

4. The easiest way to solve this problem is to graph the midpoints and then apply what we know from the Midpoint Theorem.

Now that the points are plotted, find the slopes between all three.

slope \begin{align*}AB = \frac{5+2}{1-4}=-\frac{7}{3}\end{align*}

slope \begin{align*}BC = \frac{-2-1}{4+5}=\frac{-3}{9}=-\frac{1}{3}\end{align*}

slope \begin{align*}AC = \frac{5-1}{1+5}=\frac{4}{6}=\frac{2}{3}\end{align*}

Using the slope between two of the points and the third, plot the slope triangle on either side of the third point and extend the line. Repeat this process for all three midpoints. For example, use the slope of \begin{align*}AB\end{align*} with point \begin{align*}C\end{align*}.

The green lines in the graph to the left represent the slope triangles of each midsegment. The three dotted lines represent the sides of the triangle. Where they intersect are the vertices of the triangle (the blue points), which are (-8, 8), (10, 2) and (-2, 6).

Interactive Practice


\begin{align*}R, S, T,\end{align*} and \begin{align*}U\end{align*} are midpoints of the sides of \begin{align*}\triangle XPO\end{align*} and \begin{align*}\triangle YPO\end{align*}.

  1. If \begin{align*}OP = 12\end{align*}, find \begin{align*}RS\end{align*} and \begin{align*}TU\end{align*}.
  2. If \begin{align*}RS = 8\end{align*}, find \begin{align*}TU\end{align*}.
  3. If \begin{align*}RS = 2x\end{align*}, and \begin{align*}OP = 20\end{align*}, find \begin{align*}x\end{align*} and \begin{align*}TU\end{align*}.
  4. If \begin{align*}OP = 4x\end{align*} and \begin{align*}RS = 6x - 8\end{align*}, find \begin{align*}x\end{align*}.
  5. Is \begin{align*}\triangle XOP \cong \triangle YOP\end{align*}? Why or why not?

For questions 6-13, find the indicated variable(s). You may assume that all line segments within a triangle are midsegments.

  1. The sides of \begin{align*}\triangle XYZ\end{align*} are 26, 38, and 42. \begin{align*}\triangle ABC\end{align*} is formed by joining the midpoints of \begin{align*}\triangle XYZ\end{align*}.
    1. Find the perimeter of \begin{align*}\triangle ABC\end{align*}.
    2. Find the perimeter of \begin{align*}\triangle XYZ\end{align*}.
    3. 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 \begin{align*}\triangle ABC\end{align*} below, find the midpoints of each side.

  1. \begin{align*}A(5, -2), B(9, 4)\end{align*} and \begin{align*}C(-3, 8)\end{align*}
  2. \begin{align*}A(-10, 1), B(4, 11)\end{align*} and \begin{align*}C(0, -7)\end{align*}
  3. \begin{align*}A(0, 5), B(4, -1)\end{align*} and \begin{align*}C(-2, -3)\end{align*}
  4. \begin{align*}A(2, 4), B(8, -4)\end{align*} and \begin{align*}C(2, -4)\end{align*}

For questions 19-22, \begin{align*}\triangle CAT\end{align*} has vertices \begin{align*}C(x_1,y_1), A(x_2,y_2)\end{align*} and \begin{align*}T(x_3,y_3)\end{align*}.

  1. Find the midpoints of sides \begin{align*}\overline{CA}\end{align*} and \begin{align*}\overline{AT}\end{align*}. Label them \begin{align*}L\end{align*} and \begin{align*}M\end{align*} respectively.
  2. Find the slopes of \begin{align*}\overline{LM}\end{align*} and \begin{align*}\overline{CT}\end{align*}.
  3. Find the lengths of \begin{align*}\overline{LM}\end{align*} and \begin{align*}\overline{CT}\end{align*}.
  4. What have you just proven algebraically?




Congruent figures are identical in size, shape and measure.
Midpoint Formula

Midpoint Formula

The midpoint formula says that for endpoints (x_1, y_1) and (x_2, y_2), the midpoint is @$\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)@$.


A midsegment connects the midpoints of two sides of a triangle or the non-parallel sides of a trapezoid.

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