The perpendicular bisectors of a triangle meet at one point, the circumcenter of Perpendicular Bisectors and the Circumcenter. Three other families of segments also meet at single points, and each point has its own job.
| Segments | Each one goes… | They meet at the… | Special property |
|---|---|---|---|
| angle bisectors | from a vertex, splitting its angle | incenter | equally far from all three sides; center of the inscribed circle |
| medians | from a vertex to the midpoint of the opposite side | centroid | balance point; $\frac{2}{3}$ of the way along each median from the vertex |
| altitudes | from a vertex, perpendicular to the opposite side (or its extension) | orthocenter | (none needed here) |
| perpendicular bisectors | through the midpoint of a side, at a right angle | circumcenter | equally far from all three vertices |
Why it works
Incenter: every point on an angle bisector is equally far from the two sides of the angle (the two right triangles to the sides are congruent by AAS). So the point where two bisectors meet is equally far from all three sides, and it must lie on the third bisector as well. Centroid: with coordinates, the point $\frac{2}{3}$ of the way from $A$ to the midpoint of $\overline{BC}$ is $\left(\frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3}\right)$. That expression treats all three vertices the same way, so the other two medians pass through the same point.
Key idea
The medians meet at the centroid, which is $\frac{2}{3}$ of the way from each vertex to the opposite midpoint and averages the vertices' coordinates. The angle bisectors meet at the incenter, equally far from the sides.
Worked example: Centroid lengths
$G$ is the centroid of $\triangle ABC$ and $\overline{AM}$ is a median with $GM = 5$. Find $AG$ and $AM$.
- The centroid splits each median $2 : 1$, with the long part at the vertex: $AG = 2 \cdot GM = 10$.
- $AM = AG + GM = 15$.
Answer: $AG = 10$, $AM = 15$
Worked example: Centroid coordinates
Find the centroid of $A(-2, 1)$, $B(4, 5)$ and $C(1, -3)$.
- Average the coordinates: $x = \frac{-2 + 4 + 1}{3} = 1$ and $y = \frac{1 + 5 + (-3)}{3} = 1$.
Answer: $G(1, 1)$
Worked example: The incenter
$I$ is the incenter of $\triangle ABC$, and the distance from $I$ to $\overline{AB}$ is $2x + 3$ while the distance to $\overline{BC}$ is $5x - 6$. Find the radius of the inscribed circle.
- The incenter is equally far from all sides: $2x + 3 = 5x - 6$, so $x = 3$.
- The distance is $2(3) + 3 = 9$, which is the radius of the inscribed circle.
Answer: $9$
Common mistake
Mixing up medians and altitudes, or expecting every center to be inside. A median ends at a midpoint; an altitude ends at a right angle. They coincide only in special cases. In an obtuse triangle the altitudes from the acute angles fall outside, so the orthocenter is outside, and so is the circumcenter. The centroid and the incenter are always inside.