# 5.4: Medians

**At Grade**Created by: CK-12

**Practice**Medians

What if your art teacher assigned an art project involving triangles? You decide to make a series of hanging triangles of all different sizes from one long piece of wire. Where should you hang the triangles from so that they balance horizontally?

You decide to plot one triangle on the coordinate plane to find the location of this point. The coordinates of the vertices are (0, 0), (6, 12) and (18, 0). What is the coordinate of this point? After completing this Concept, you'll be able to use medians to help you answer these questions.

### Watch This

CK-12 Foundation: Chapter5MediansA

James Sousa: Medians of a Triangle

James Sousa: Using the Properties of Medians to Solve for Unknown Values

### Guidance

A **median** is the line segment that joins a vertex and the midpoint of the opposite side (of a triangle). The three medians of a triangle intersect at one point, just like the perpendicular bisectors and angle bisectors. This point is called the **centroid**, and is the point of concurrency for the medians of a triangle. Unlike the circumcenter and incenter, the centroid does not have anything to do with circles. It has a different property.

##### Investigation: Properties of the Centroid

Tools Needed: pencil, paper, ruler, compass

1. Construct a scalene triangle with sides of length 6 cm, 10 cm, and 12 cm (Investigation 4-2). Use the ruler to measure each side and mark the midpoint.

2. Draw in the medians and mark the centroid.

Measure the length of each median. Then, measure the length from each vertex to the centroid and from the centroid to the midpoint. Do you notice anything?

3. Cut out the triangle. Place the centroid on either the tip of the pencil or the pointer of the compass. What happens?

From this investigation, we have discovered the properties of the centroid. They are summarized below.

**Concurrency of Medians Theorem:** The medians of a triangle intersect in a point that is two-thirds of the distance from the vertices to the midpoint of the opposite side. The centroid is also the “balancing point” of a triangle.

If \begin{align*}G\end{align*}

\begin{align*}AG &= \frac{2}{3} AD, CG=\frac{2}{3} CF, EG=\frac{2}{3} BE\\
DG &= \frac{1}{3} AD, FG=\frac{1}{3} CF, BG=\frac{1}{3} BE\end{align*}

And, combining these equations, we can also conclude:

\begin{align*}DG=\frac{1}{2} AG, FG=\frac{1}{2} CG, BG=\frac{1}{2} EG\end{align*}

In addition to these ratios, \begin{align*}G\end{align*} is also the balance point of \begin{align*}\triangle ACE\end{align*}. This means that the triangle will balance when placed on a pencil at this point.

#### Example A

Draw the median \begin{align*}\overline{LO}\end{align*} for \begin{align*}\triangle LMN\end{align*} below.

From the definition, we need to locate the midpoint of \begin{align*}\overline{NM}\end{align*}. We were told that the median is \begin{align*}\overline{LO}\end{align*}, which means that it will connect the vertex \begin{align*}L\end{align*} and the midpoint of \begin{align*}\overline{NM}\end{align*}, to be labeled \begin{align*}O\end{align*}. Measure \begin{align*}NM\end{align*} and make a point halfway between \begin{align*}N\end{align*} and \begin{align*}M\end{align*}. Then, connect \begin{align*}O\end{align*} to \begin{align*}L\end{align*}.

#### Example B

Find the other two medians of \begin{align*}\triangle LMN\end{align*}.

Repeat the process from Example A for sides \begin{align*}\overline{LN}\end{align*} and \begin{align*}\overline{LM}\end{align*}. Be sure to always include the appropriate tick marks to indicate midpoints.

#### Example C

\begin{align*}I, K\end{align*}, and \begin{align*}M\end{align*} are midpoints of the sides of \begin{align*}\triangle HJL\end{align*}.

a) If \begin{align*}JM = 18\end{align*}, find \begin{align*}JN\end{align*} and \begin{align*}NM\end{align*}.

b) If \begin{align*}HN = 14\end{align*}, find \begin{align*}NK\end{align*} and \begin{align*}HK\end{align*}.

a) \begin{align*}JN\end{align*} is two-thirds of \begin{align*}JM\end{align*}. So, \begin{align*}JN =\frac{2}{3} \cdot 18=12\end{align*}. \begin{align*}NM\end{align*} is either half of 12, a third of 18 or \begin{align*}18 - 12\end{align*}. \begin{align*}NM = 6\end{align*}.

b) \begin{align*}HN\end{align*} is two-thirds of \begin{align*}HK\end{align*}. So, \begin{align*}14 =\frac{2}{3} \cdot HK\end{align*} and \begin{align*}HK = 14 \cdot \frac{3}{2} = 21\end{align*}. \begin{align*}NK\end{align*} is a third of 21, half of 14, or \begin{align*}21-14\end{align*}. \begin{align*}NK = 7\end{align*}.

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter5MediansB

#### Concept Problem Revisited

The point that you should put the wire through is the centroid. That way, each triangle will balance on the wire.

The triangle that we wanted to plot on the \begin{align*}x-y\end{align*} plane is to the right. Drawing all the medians, it looks like the centroid is (8, 4). To verify this, you could find the equation of two medians and set them equal to each other and solve for \begin{align*}x\end{align*}. Two equations are \begin{align*}y=\frac{1}{2} x\end{align*} and \begin{align*}y=-4x+36\end{align*}. Setting them equal to each other, we find that \begin{align*}x = 8\end{align*} and then \begin{align*}y = 4\end{align*}.

### Vocabulary

A ** median** is the line segment that joins a vertex and the midpoint of the opposite side in a triangle. A

**is a point that divides a segment into two equal pieces. A**

*midpoint***is the point of intersection for the medians of a triangle.**

*centroid*### Guided Practice

1. Find the equation of the median from \begin{align*}B\end{align*} to the midpoint of \begin{align*}\overline{AC}\end{align*} for the triangle in the \begin{align*}x-y\end{align*} plane below.

2. \begin{align*}H\end{align*} is the centroid of \begin{align*}\triangle ABC\end{align*} and \begin{align*}DC = 5y - 16\end{align*}. Find \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.

3. True or false: The median bisects the side it intersects.

**Answers:**

1. To find the equation of the median, first we need to find the midpoint of \begin{align*}\overline{AC}\end{align*}, using the Midpoint Formula.

\begin{align*}\left(\frac{-6+6}{2}, \frac{-4+(-4)}{2}\right)=\left(\frac{0}{2}, \frac{-8}{2}\right)=(0,-4)\end{align*}

Now, we have two points that make a line, \begin{align*}B\end{align*} and the midpoint. Find the slope and \begin{align*}y-\end{align*}intercept.

\begin{align*}m &= \frac{-4-4}{0-(-2)}=\frac{-8}{2}=-4\\ y &= -4x+b\\ -4 &= -4(0)+b\\ -4 &= b\end{align*}

The equation of the median is \begin{align*}y=-4x-4\end{align*}

2. \begin{align*}HF\end{align*} is half of \begin{align*}BH\end{align*}. Use this information to solve for \begin{align*}x\end{align*}. For \begin{align*}y\end{align*}, \begin{align*}HC\end{align*} is two-thirds of \begin{align*}DC\end{align*}. Set up an equation for both.

\begin{align*}\frac{1}{2} BH &= HF \ \text{or} \ BH=2HF && HC=\frac{2}{3} DC \ \text{or} \ \frac{3}{2} HC=DC\\ 3x+6 &= 2(2x-1) && \frac{3}{2} (2y+8)=5y-16\\ 3x+6 &= 4x-2 && \quad 3y+12=5y-16\\ 8 &= x && \qquad \quad \ 28=2y\end{align*}

3. This statement is true. By definition, a median intersects a side of a triangle at its midpoint. Midpoints divide segments into two equal parts.

### Practice

For questions 1-4, find the equation of each median, from vertex \begin{align*}A\end{align*} to the opposite side, \begin{align*}\overline{BC}\end{align*}.

- \begin{align*}A(9, 5), B(2, 5), C(4,1)\end{align*}
- \begin{align*}A(-2, 3), B(-3, -7), C(5, -5)\end{align*}
- \begin{align*}A(-1, 5), B(0, -1), C(6, 3)\end{align*}
- \begin{align*}A(6, -3), B(-5, -4), C(-1, -8)\end{align*}

For questions 5-9, \begin{align*}B, D\end{align*}, and \begin{align*}F\end{align*} are the midpoints of each side and \begin{align*}G\end{align*} is the centroid. Find the following lengths.

- If \begin{align*}BG = 5\end{align*}, find \begin{align*}GE\end{align*} and \begin{align*}BE\end{align*}
- If \begin{align*}CG = 16\end{align*}, find \begin{align*}GF\end{align*} and \begin{align*}CF\end{align*}
- If \begin{align*}AD = 30\end{align*}, find \begin{align*}AG\end{align*} and \begin{align*}GD\end{align*}
- If \begin{align*}GF = x\end{align*}, find \begin{align*}GC\end{align*} and \begin{align*}CF\end{align*}
- If \begin{align*}AG = 9x\end{align*} and \begin{align*}GD = 5x - 1\end{align*}, find \begin{align*}x\end{align*} and \begin{align*}AD\end{align*}.

Use \begin{align*}\triangle ABC\end{align*} with \begin{align*}A(-2, 9), B(6, 1)\end{align*} and \begin{align*}C(-4, -7)\end{align*} for questions 10-15.

- Find the midpoint of \begin{align*}\overline{AB}\end{align*} and label it \begin{align*}M\end{align*}.
- Write the equation of \begin{align*}\overleftrightarrow{CM}\end{align*}.
- Find the midpoint of \begin{align*}\overline{BC}\end{align*} and label it \begin{align*}N\end{align*}.
- Write the equation of \begin{align*}\overleftrightarrow{AN}\end{align*}.
- Find the intersection of \begin{align*}\overleftrightarrow{CM}\end{align*} and \begin{align*}\overleftrightarrow{AN}\end{align*}.
- What is this point called?

Another way to find the centroid of a triangle in the coordinate plane is to find the midpoint of one side and then find the point two thirds of the way from the third vertex to this point. To find the point two thirds of the way from point \begin{align*}A(x_1, y_1)\end{align*} to \begin{align*}B(x_2, y_2)\end{align*} use the formula: \begin{align*}\left(\frac{x_1+2x_2}{3}, \frac{y_1+2y_2}{3}\right)\end{align*}. Use this method to find the centroid in the following problems.

- (-1, 3), (5, -2) and (-1, -4)
- (1, -2), (-5, 4) and (7, 7)
- (2, -7), (-5, 1) and (6, -9)

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### Image Attributions

Here you'll learn the definitions of median and centroid and how to apply them.