### Midsegment Theorem

A line segment that connects two midpoints of the sides of a triangle is called a **midsegment**. \begin{align*} \overline{DF}\end{align*} is the midsegment between \begin{align*} \overline{AB}\end{align*} and \begin{align*}\overline{BC}\end{align*}.

The tic marks show 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*}. For every triangle there are three midsegments.

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 \begin{align*}\overline{DF}\end{align*} is a midsegment of \begin{align*}\triangle ABC\end{align*}, then \begin{align*}DF = \frac{1}{2} AC = AE = EC\end{align*} and \begin{align*}\overline{DF} \| \overline{AC}\end{align*}.

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 \begin{align*}\triangle FGH \end{align*} and told that \begin{align*} \overline{JK}\end{align*} was its midsegment? How could you find the length of \begin{align*}JK\end{align*} given the length of the triangle's third side, \begin{align*}FH\end{align*}?

### Examples

#### Example 1

Find the value of \begin{align*}x\end{align*} and \begin{align*}AB\end{align*}. \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are midpoints.

\begin{align*}AB = 34 \div 2 = 17\end{align*}. To find \begin{align*}x\end{align*}, set \begin{align*}3x - 1\end{align*} equal to 17.

\begin{align*}3x - 1 & = 17\\ 3x & = 18\\ x & =6\end{align*}

#### Example 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.

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 3

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*}. 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, plot the midpoints and draw the midsegments.

To solve this problem, use the midpoint formula 3 times to find all the midpoints. Recall that the midpoint formula is \begin{align*}\left ( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right ) \end{align*}.

\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*}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*}

\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*}

#### Example 4

Mark all the congruent segments on \begin{align*}\triangle ABC\end{align*} with midpoints \begin{align*}D, \ E\end{align*}, and \begin{align*}F\end{align*}.

Drawing in all three midsegments, we have:

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

Now, mark all the parallel lines on \begin{align*}\triangle ABC\end{align*}, with midpoints \begin{align*}D, \ E\end{align*}, and \begin{align*}F\end{align*}.

#### Example 5

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

Find \begin{align*}MN\end{align*}, \begin{align*}XY\end{align*}, and the perimeter of \begin{align*}\triangle XYZ\end{align*}.

Use the Midsegment Theorem:

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

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

Add up the three sides of \begin{align*}\triangle XYZ\end{align*} to find the perimeter.

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

Remember: No line segment over \begin{align*}MN\end{align*} 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.

\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*}.

- If \begin{align*}OP = 12\end{align*}, find \begin{align*}RS\end{align*} and \begin{align*}TU\end{align*}.
- If \begin{align*}RS = 8\end{align*}, find \begin{align*}TU\end{align*}.
- 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*}.
- If \begin{align*}OP = 4x\end{align*} and \begin{align*}RS = 6x - 8\end{align*}, find \begin{align*}x\end{align*}.

For questions 9-15, find the indicated variable(s). You may assume that all line segments within a triangle are midsegments.

- 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*}.
- What are the lengths of the sides of \begin{align*}\triangle ABC\end{align*}?
- Find the perimeter of \begin{align*}\triangle ABC\end{align*}.
- Find the perimeter of \begin{align*}\triangle XYZ\end{align*}.
- 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.

- \begin{align*}A(5, -2), \ B(9, 4)\end{align*} and \begin{align*}C(-3, 8)\end{align*}
- \begin{align*}A(-10, 1), \ B(4, 11)\end{align*} and \begin{align*}C(0, -7)\end{align*}
- \begin{align*}A(-1, 3), \ B(5, 7)\end{align*} and \begin{align*}C(9, -5)\end{align*}
- \begin{align*}A(-4, -15), \ B(2, -1)\end{align*} and \begin{align*}C(-20, 11)\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 5.1.