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# Medians

## Line segment that joins a vertex and the midpoint of the opposite side of a triangle.

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Medians

### Medians

In a triangle, the line segment that joins a vertex and the midpoint of the opposite side is called a median.

LO¯¯¯¯¯¯¯\begin{align*}\overline{LO}\end{align*} is the median from L\begin{align*}L\end{align*} to the midpoint of NM¯¯¯¯¯¯¯¯¯¯\begin{align*}\overline{NM}\end{align*}.

If you draw all three medians they will intersect at one point called the centroid.

The centroid is the “balancing point” of a triangle. This means that if you were to cut out the triangle, the centroid is its center of gravity so you could balance it there.

The Median Theorem states that the medians of a triangle intersect at a point called the centroid that is two-thirds of the distance from the vertices to the midpoint of the opposite sides.

So if G\begin{align*}G\end{align*} is the centroid, then:

AGDGAnd by substitution:DG=23AD, CG=23CF, EG=23BE=13AD, FG=13CF, BG=13BE=12AG, FG=12CG, BG=12EG\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\\ \text{And by substitution}: \quad DG &= \frac{1}{2} AG, \ FG = \frac{1}{2} CG, \ BG = \frac{1}{2} EG\end{align*}

What if you were told that J\begin{align*}J\end{align*}, K\begin{align*}K\end{align*}, and L\begin{align*}L\end{align*} were the midpoints of the sides of FGH\begin{align*}\triangle FGH\end{align*} and that M\begin{align*}M\end{align*} was the centroid of FGH\begin{align*}\triangle FGH\end{align*}? Given the length of JK\begin{align*}JK\end{align*}, how could you find the lengths of JM\begin{align*}JM\end{align*} and KM\begin{align*}KM\end{align*}?

### Examples

#### Example 1

B, D\begin{align*}B, \ D\end{align*}, and F\begin{align*}F\end{align*} are the midpoints of each side and G\begin{align*}G\end{align*} is the centroid. If CG=16\begin{align*}CG = 16\end{align*}, find GF\begin{align*}GF\end{align*} and CF\begin{align*}CF\end{align*}.

Use the Median Theorem.

CG16CF=23CF=23CF=24\begin{align*}CG&=\frac{2}{3}CF \\ 16 &=\frac{2}{3}CF \\ CF &=24\\\end{align*}.

Therefore, GF=8\begin{align*}GF=8\end{align*}.

#### Example 2

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

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

#### Example 3

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

If JM=18\begin{align*}JM = 18\end{align*}, find JN\begin{align*}JN\end{align*} and NM\begin{align*}NM\end{align*}. If HN=14\begin{align*}HN = 14\end{align*}, find NK\begin{align*}NK\end{align*} and HK\begin{align*}HK\end{align*}.

Use the Median Theorem.

JN=2318=12\begin{align*}JN = \frac{2}{3} \cdot 18 = 12\end{align*}. NM=JMJN=1812\begin{align*}NM = JM - JN = 18 - 12\end{align*}. NM=6.\begin{align*}NM = 6.\end{align*}

14=23HK\begin{align*}14 = \frac{2}{3} \cdot HK\end{align*}

1432=HK=21\begin{align*}14 \cdot \frac{3}{2} = HK = 21\end{align*}. NK\begin{align*}NK\end{align*} is a third of 21, NK=7\begin{align*}NK = 7\end{align*}.

#### Example 4

H\begin{align*}H\end{align*} is the centroid of ABC\begin{align*}\triangle ABC\end{align*} and DC=5y16\begin{align*}DC = 5y - 16\end{align*}. Find x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*}.

To solve, use the Median Theorem. Set up and solve equations.

12BH=HF3x+63x+68BH=2HF=2(2x1)=4x2=xHC=23DC32HC=DC 32(2y+8)=5y16 3y+12=5y16  28=2y14=y\begin{align*}\frac{1}{2} BH= HF & \longrightarrow BH = 2HF && HC = \frac{2}{3} DC \longrightarrow \frac{3}{2} HC = DC\\ 3x + 6 &= 2(2x - 1) && \quad \ \frac{3}{2} (2y + 8) = 5y - 16\\ 3x + 6 &= 4x - 2 && \qquad \ 3y + 12 = 5y - 16\\ 8 &= x && \qquad \ \ \qquad 28 = 2y \longrightarrow 14 = y\end{align*}

#### Example 5

B, D\begin{align*}B, \ D\end{align*}, and F\begin{align*}F\end{align*} are the midpoints of each side and G\begin{align*}G\end{align*} is the centroid. If BG=5\begin{align*}BG = 5\end{align*}, find GE\begin{align*}GE\end{align*} and BE\begin{align*}BE\end{align*}

Use the Median Theorem.

\begin{align*}BG&=\frac{1}{3}BE \\ 5 &=\frac{1}{3}BE \\ BE &=15\\\end{align*}.

Therefore, \begin{align*}GE=10\end{align*}.

### Review

For questions 1-4, \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.

1. If \begin{align*}CG = 16\end{align*}, find \begin{align*}GF\end{align*} and \begin{align*}CF\end{align*}
2. If \begin{align*}AD = 30\end{align*}, find \begin{align*}AG\end{align*} and \begin{align*}GD\end{align*}
3. If \begin{align*}GF = x\end{align*}, find \begin{align*}GC\end{align*} and \begin{align*}CF\end{align*}
4. 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*}.

Multi-step Problems Find the equation of a median in the \begin{align*}x-y\end{align*} plane.

1. Plot \begin{align*}\triangle ABC: \ A(-6, 4), \ B(-2, 4)\end{align*} and \begin{align*}C(6, -4)\end{align*}
2. Find the midpoint of \begin{align*}\overline{AC}\end{align*}. Label it \begin{align*}D\end{align*}.
3. Find the slope of \begin{align*}\overline{BD}\end{align*}.
4. Find the equation of \begin{align*}\overline{BD}\end{align*}.
5. Plot \begin{align*}\triangle DEF: \ D(-1, 5), \ E(1, 0), \ F(6, 3)\end{align*}
6. Find the midpoint of \begin{align*}\overline{EF}\end{align*}. Label it \begin{align*}G\end{align*}.
7. Find the slope of \begin{align*}\overline{DG}\end{align*}.
8. Find the equation of \begin{align*}\overline{DG}\end{align*}.

Determine whether the following statement is true or false.

1. The centroid is the balancing point of a triangle.

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Color Highlighted Text Notes

### Vocabulary Language: English

centroid

The centroid is the point of intersection of the medians in a triangle.

Median

The median of a triangle is the line segment that connects a vertex to the opposite side's midpoint.