# 4.23: Medians

- Page ID
- 4820

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

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

\overline{LO}\) is the median from L\) to the midpoint of \overline{NM}\).

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 \(G\) is the centroid, then:

\(AG=\dfrac{2}{3} AD, CG=\dfrac{2}{3} CF,\:EG=\dfrac{2}{3} BE\)

\(DG=\dfrac{1}{3} AD, FG=\dfrac{1}{3} CF,\:BG=\dfrac{1}{3} BE\)

\(And\: by \:substitution:DG =\dfrac{1}{2} AG,\:FG=\dfrac{1}{2} CG,\:BG=\dfrac{1}{2} EG\)

Example \(\PageIndex{1}\)

\(B\), \(D\), and \(F\) are the midpoints of each side and \(G\) is the centroid. If \(CG=16\), find \(GF\) and \(CF\).

**Solution**

Use the Median Theorem.

\(\begin{align*} CG&=\dfrac{2}{3} CF \\ 16&=\dfrac{2}{3} CF \\ CF&=24.\end{align*}\)

Therefore, \(GF=8\).

Example \(\PageIndex{2}\)

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

**Solution**

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 \(\PageIndex{3}\)

\(I\), \(K\), and \(M\) are midpoints of the sides of \(\Delta HJL\).

**Solution**

If \(JM=18\), find \(JN\) and \(NM\). If \(HN=14\), find \(NK\) and \(HK\).

Use the Median Theorem.

\(JN=\dfrac{2}{3} \cdot 18=12. NM=JM−JN=18−12\). \(NM=6\).

\(14=\dfrac{2}{3} \cdot HK\)

\(14\cdot \dfrac{3}{2} =HK=21\). \(NK\) is a third of 21, \(NK=7\).

Example \(\PageIndex{4}\)

H is the centroid of \(\Delta ABC\) and \(DC=5y−16\). Find \(x\) and \(y\).

**Solution**

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

\(\begin{align*} \dfrac{1}{2} BH=HF &\rightarrow BH=2HF &\qquad HC=\dfrac{2}{3} DC &\rightarrow \dfrac{3}{2} HC=DC \\ 3x+6&=2(2x−1) &\qquad \dfrac{3}{2} (2y+8)&=5y−16\\ 3x+6&=4x−2 &\qquad 3y+12 &=5y−16 \\ 8&=x &\qquad 28&=2y\rightarrow 14=y\end{align*} \)

Example \(\PageIndex{5}\)

\(B\), \(D\), and \(F\) are the midpoints of each side and G is the centroid. If \(BG=5\), find \(GE\) and \(BE\)

**Solution**

Use the Median Theorem.

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

Therefore, \(GE=10\).

## Review

For questions 1-4, \(B\), \(D\), and \(F\) are the midpoints of each side and \(G\) is the centroid. Find the following lengths.

- If \(CG=16\), find \(GF\) and \(CF\)
- If \(AD=30\), find \(AG\) and \(GD\)
- If \(GF=x\), find \(GC\) and \(CF\)
- If \(AG=9x\) and \(GD=5x−1\), find \(x\) and \(AD\).

**Multi-step Problems** Find the equation of a median in the x−y\) plane.

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

Determine whether the following statement is true or false.

- The centroid is the balancing point of a triangle.

## Review (Answers)

To see the Review answers, open this PDF file and look for section 5.4.

## Resources

## Vocabulary

Term | Definition |
---|---|

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. |

## Additional Resources

Interactive Element

Video: The Medians of a Triangle

Activities: Medians Discussion Questions

Study Aids: Bisectors, Medians, Altitudes Study Guide

Real World: Medians