Math Core

Cut a triangle out of cardboard and try to balance it on a pencil tip. There is exactly one point where it balances, and you can find it with nothing more than midpoints. This lesson covers two more kinds of special segments, medians and altitudes, and the points where each set of three meets.

Medians and the centroid

Definition

Median of a triangle

A median of a triangle is a segment from a vertex to the midpoint of the opposite side. Every triangle has three medians.

Like the perpendicular bisectors and angle bisectors from the last lesson, the three medians are always concurrent. The point where they meet is the centroid. It's the triangle's center of mass, the balance point of a flat triangle with even thickness.

The three medians AD, BE and CF (dashed) meet at the centroid G.

The centroid doesn't sit in the middle of each median. It divides every median in a fixed ratio.

Centroid Theorem

The centroid of a triangle is two-thirds of the way from each vertex to the midpoint of the opposite side. For median $\overline{AD}$ with centroid $G$ :

AG = \frac{2}{3}AD, \qquad GD = \frac{1}{3}AD, \qquad AG = 2 \cdot GD.

The longer piece (vertex to centroid) is always twice the shorter piece (centroid to midpoint).

You can see this in the figure. $A(0, 0)$ and the midpoint $D(6, 3)$ are the ends of median $\overline{AD}$ , and $G(4, 2)$ is exactly $\dfrac{2}{3}$ of the way: $\dfrac{2}{3}$ of $6$ is $4$ and $\dfrac{2}{3}$ of $3$ is $2$ .

Worked example: Using the 2 : 1 ratio

$G$ is the centroid of $\triangle ABC$ , and $\overline{AD}$ is a median.

If $AD = 18$ , find $AG$ and $GD$ .
If $GD = 7$ , find $AG$ and $AD$ .

Solution.

$AG = \dfrac{2}{3}(18) = 12$ and $GD = \dfrac{1}{3}(18) = 6$ . Check: $12 = 2 \cdot 6$ and $12 + 6 = 18$ .
$AG = 2 \cdot GD = 14$ , so $AD = 14 + 7 = 21$ .

Worked example: Centroid problems with algebra

$G$ is the centroid of $\triangle ABC$ and $\overline{AD}$ is a median. If $AG = 3x - 1$ and $GD = x + 2$ , find $AD$ .

Solution. The vertex-to-centroid piece is twice the centroid-to-midpoint piece:

\begin{aligned} 3x - 1 &= 2(x + 2) \\ 3x - 1 &= 2x + 4 \\ x &= 5 \end{aligned}

So $AG = 14$ and $GD = 7$ , and $AD = 14 + 7 = 21$ .

Common mistake

The ratio is $2 : 1$ from the vertex, not from the midpoint. The piece touching the vertex is the long one. Also watch what the question asks for: the whole median, the long piece and the short piece are three different numbers.

The centroid on a coordinate grid

Because the centroid is two-thirds of the way along each median, its coordinates turn out to be the averages of the vertices' coordinates.

Centroid formula

The centroid of the triangle with vertices $(x_1, y_1)$ , $(x_2, y_2)$ and $(x_3, y_3)$ is

\left( \frac{x_1 + x_2 + x_3}{3}, \; \frac{y_1 + y_2 + y_3}{3} \right).

For the triangle in the figure, $\left(\dfrac{0 + 9 + 3}{3}, \dfrac{0 + 0 + 6}{3}\right) = (4, 2)$ , matching the point $G$ .

Worked example: Finding a centroid

Find the centroid of the triangle with vertices $(1, 2)$ , $(7, 4)$ and $(4, 9)$ .

Solution. Average the coordinates:

\left( \frac{1 + 7 + 4}{3}, \; \frac{2 + 4 + 9}{3} \right) = \left( \frac{12}{3}, \frac{15}{3} \right) = (4, 5).

Altitudes and the orthocenter

Definition

Altitude of a triangle

An altitude of a triangle is the perpendicular segment from a vertex to the line containing the opposite side. Its length is the triangle's height for that base.

In an acute triangle, every altitude lands inside the triangle.

In this acute triangle, the altitude CH (dashed) from C meets side AB at a right angle.

In an obtuse triangle, the altitudes from the two acute angles fall outside the triangle. You have to extend the opposite side to meet them.

In obtuse triangle ABC, the altitude from C meets the line AB extended past B, outside the triangle.

In a right triangle, the two legs are themselves altitudes, since each leg is perpendicular to the other.

The lines containing the three altitudes are concurrent. Their meeting point is the orthocenter. It is inside an acute triangle, at the right-angle vertex of a right triangle, and outside an obtuse triangle.

Worked example: Finding an orthocenter with coordinates

Find the orthocenter of the triangle with vertices $A(0, 0)$ , $B(6, 0)$ and $C(2, 4)$ .

The three altitudes of triangle ABC (dashed) meet at the orthocenter H(2, 2).Open in grapher →

Solution. Find two altitudes and intersect them.

$\overline{AB}$ is horizontal, so the altitude from $C$ is vertical: $x = 2$ .
$\overline{BC}$ has slope $\dfrac{4 - 0}{2 - 6} = -1$ . The altitude from $A$ is perpendicular, so its slope is $1$ , and it passes through $(0, 0)$ : $y = x$ .

At $x = 2$ , $y = 2$ . The orthocenter is $H(2, 2)$ .

Check with the third altitude: $\overline{AC}$ has slope $2$ , so the altitude from $B$ has slope $-\dfrac{1}{2}$ : $y = -\dfrac{1}{2}(x - 6)$ . At $x = 2$ , $y = -\dfrac{1}{2}(-4) = 2$ . It passes through $(2, 2)$ too.

Tip

A median goes to the midpoint of the opposite side; an altitude goes to the opposite side at a right angle. They are the same segment only in special cases, such as the segment from the vertex angle of an isosceles triangle.

The four centers at a glance

Center	Where three of these meet	Special property	Always inside?
Circumcenter	perpendicular bisectors	equidistant from the vertices	No
Incenter	angle bisectors	equidistant from the sides	Yes
Centroid	medians	balance point; $2 : 1$ along each median	Yes
Orthocenter	altitudes	none beyond concurrency	No

Practice

Practice 1

In $\triangle ABC$ , $\overline{BM}$ is a median and $AC = 16$ . What is $AM$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

A median of a triangle is $27$ units long. How far is the centroid from the vertex at the end of that median?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

$G$ is the centroid of $\triangle PQR$ , and $\overline{PS}$ is a median. If $GS = 5$ , what is $PS$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

$G$ is the centroid of $\triangle ABC$ and $\overline{AD}$ is a median. If $AG = 5x - 2$ and $GD = 2x + 1$ , what is $AD$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Find the centroid of the triangle with vertices $(-3, 2)$ , $(5, -4)$ and $(1, 8)$ .

Enter a point like (2, -3)

Practice 6

Where is the orthocenter of a right triangle?

Practice 7

Segment $\overline{KM}$ goes from vertex $K$ of $\triangle JKL$ to side $\overline{JL}$ , and $\overline{KM} \perp \overline{JL}$ . What is $\overline{KM}$ definitely?

Practice 8

Find the orthocenter of the triangle with vertices $A(0, 0)$ , $B(10, 0)$ and $C(4, 6)$ .

Enter a point like (2, -3)