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# Incenter of a Triangle — Properties, Formula, Examples

[Geometry](/content/tag/geometry/index.html)

**TL;DR**

The incenter of a triangle is the single point where its three angle bisectors meet, and it sits equidistant from all three sides — making it the center of the largest circle that fits inside the triangle. This article covers the definition, the key properties, the coordinate and angle formulas, the inradius, and worked examples, plus how the incenter differs from the circumcenter.

**Last updated on June 24, 2026 10 min read**

## What is The Incenter of A Triangle?

The incenter of a triangle is the **point of intersection of the three interior angle bisectors** of the triangle. An angle bisector is the ray that cuts an angle into two equal halves; every triangle has three, and they always cross at a single point — that point is the incenter, usually written III.

What makes III special is its distance to the sides. The incenter is **equidistant from all three sides** of the triangle. That equal distance is the radius of the **incircle** — the largest circle that fits inside the triangle, touching each side at exactly one point. The radius is called the **inradius**, written rrr.

Two facts follow immediately and are worth holding onto. First, the incenter **always lies inside** the triangle — unlike some other triangle centers, it never escapes the figure, no matter how stretched the triangle is. Second, the sides are **tangent** to the incircle, so each perpendicular from III to a side has the same length rrr.

## How The Incenter is Equidistant From The Sides

The "equidistant from the sides" property isn't a coincidence — it falls straight out of the [angle bisector theorem](/content/math/geometry/angle-bisector-theorem/index.html), which says any point on an angle's bisector is equally far from the two arms of that angle.

Walk the logic in one line each:

1. III lies on the bisector of ∠A, so III is equidistant from sides AB and AC.
2. III lies on the bisector of ∠B, so III is equidistant from sides AB and BC.
3. Chaining these, III is equidistant from all three sides — ID=IE=IF=r.

This is also why the third bisector _must_ pass through the same point: once III is equidistant from all three sides, it's automatically equidistant from AC and BC, which is exactly the condition for lying on the bisector of ∠C.

## Properties of The Incenter

The incenter packs several reliable properties, each useful in problems.

- **Always interior.** III lies inside the triangle for every triangle — acute, right, or obtuse.
- **Equidistant from the sides.** The perpendicular distances ID=IE=IF=r, the inradius.
- **Center of the incircle.** The incircle is tangent to all three sides; III is its center.
- **Equal tangent segments.** From each vertex, the two tangent lengths to the incircle are equal — a property that solves many "find the unknown side" problems.
- **Area relationship.** The triangle's area equals r times its semiperimeter: Area=r⋅s, where s=(a+b+c)/2. This is the standard way to _find_ the inradius.

## The Incenter Formulas

There are two formulas worth knowing — one for the incenter's coordinates, one for the angle it subtends.

**The coordinate formula.** If the vertices are A(x1,y1), B(x2,y2), C(x3,y3), and the side lengths opposite them are a, b, c, then:

I=(a x1 + b x2 + c x3)/(a + b + c,; a y1 + b y2 + c y3)/(a + b + c)

The variable glossary: a is the side opposite vertex A (so a=BC), b=C A, c=A B. Each vertex coordinate is _weighted by the length of the opposite side_ — that weighting is what pulls the incenter toward where the triangle is "tightest." Where does it come from? The incenter is the weighted average of the vertices using the opposite side lengths as weights, which is exactly what places it equidistant from the three sides.

**The angle formula.** The angle subtended at the incenter by two vertices relates to the triangle's angles:

∠BIC=90∘+∠A/2.

This says the angle at III looking across to B and C is always more than a right angle — another consequence of III sitting inside.

## Examples of the Incenter of a Triangle

The examples run from a direct angle calculation to coordinates and the inradius.

### Example 1

**In △ABC, ∠A=50∘. Find ∠BIC, the angle at the incenter.**

Use the angle formula directly.

∠BIC=90∘+∠A/2

Final answer: ∠BIC=115∘.

### Example 2

**A triangle has angles ∠A=80∘ and ∠B=60∘. The incenter is III. Find ∠AIB.**

1. ∠C=180∘−80∘−60∘=40∘.
2. ∠AIB=90∘+∠C/2.

Final answer: ∠AIB=110∘.

### Example 3

**Find the incenter of the triangle with vertices A(0,0), B(6,0), C(0,8).**

1. a=BC=(6−0)²+(0−8)²=10.
2. b=CA=8.
3. c=AB=6.

Now using the coordinate formula:

I_x = (a x1 + b x2 + c x3)/(a + b + c) and I_y = (a y1 + b y2 + c y3)/(a + b + c).

Final answer: I=(2,2).

### Example 4

**For the same triangle — A(0,0), B(6,0), C(0,8) — find the inradius.**

Use the area relationship Area=r⋅s:

1. Area=12×6×8=24.
2. s=(a+b+c)/2=12.
3. r=Area/s=2.

Final answer: r=2 units.

### Example 5

**A triangle has sides a=13, b=14, c=15. Find its inradius.**

1. s=(13+14+15)/2=21.
2. Area=s(s−a)(s−b)(s−c)=84.
3. r=Area/s=4.

Final answer: r=4 units.

### Example 6

**A circular fountain must sit inside a triangular plaza with sides 9 m, 12 m, and 15 m. What is the largest radius the fountain can have, and where is its center?**

1. The sides satisfy the Pythagorean theorem; the plaza is right-angled.
2. Area=12×9×12=54.
3. s=(9+12+15)/2=18.
4. r=Area/s=3 m.

Final answer: the fountain can have radius 3 m, centered at the incenter.

## Why The Incenter Matters

> "One point that every side can reach equally — that is what the bisectors are searching for."

- **It solves the inscribed-circle problem.** Whenever something round has to fit snugly inside a triangular boundary — the incenter is the center and the inradius is the size.
- **It encodes "fairness" by distance.** A facility that should be equally far from three straight boundaries sits at the incenter.
- **It links area, perimeter, and radius.** The relation Area=r⋅s ties three different measurements of a triangle into one equation.

## Where The Incenter Trips Students Up

- **Mistake 1: Confusing the incenter with the circumcenter**. The incenter is equidistant from the three **sides**; the circumcenter is equidistant from the three **vertices**.
- **Mistake 2: Using equal weights in the coordinate formula**. The correct way is: I=(a x1 + b x2 + c x3)/(a + b + c).
- **Mistake 3: Mismatching side labels**. By convention a is the side opposite A.

## Conclusion

- The **incenter of a triangle** is where the three angle bisectors meet.
- It is **equidistant from all three sides** and is the center of the inscribed circle (incircle).
- The incenter **always lies inside** the triangle.
- Its coordinates weight each vertex by the **opposite side length**: (a x1 + b x2 + c x3)/(a + b + c); the inradius comes from Area=r⋅s.
- Don't confuse it with the **circumcenter**, which is equidistant from the vertices.