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?

### Midsegment Theorem

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.

#### Drawing Midsegments

Draw the midsegment \begin{align*}\overline{DF}\end{align*}

Find the midpoints of \begin{align*}\overline{AB}\end{align*}

Don’t forget to put the tic marks, indicating that \begin{align*}D\end{align*}

#### Finding the Midpoint

Find the midpoint of \begin{align*}\overline{AC}\end{align*}

#### Visualizing the Midsegment Theorem

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*}

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*}

#### Using the Midsegment Theorem

\begin{align*}M, N,\end{align*}

Find

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*}

#### Earlier Problem Revisited

To the left is a picture of the \begin{align*}4^{th}\end{align*}

### Examples

The vertices of \begin{align*}\triangle LMN\end{align*}

#### Example 1

Find the midpoints of all three sides, label them \begin{align*}O, P\end{align*}

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

\begin{align*}L\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.

#### Example 2

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

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.*

#### Example 3

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

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*}.

#### Example 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.

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

### Review

\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*}.
- 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.

- 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*}.
- 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(0, 5), B(4, -1)\end{align*} and \begin{align*}C(-2, -3)\end{align*}
- \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*}.

- 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.
- Find the slopes of \begin{align*}\overline{LM}\end{align*} and \begin{align*}\overline{CT}\end{align*}.
- Find the lengths of \begin{align*}\overline{LM}\end{align*} and \begin{align*}\overline{CT}\end{align*}.
- What have you just proven algebraically?

### Review (Answers)

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