Visual Tools
Calculators
Tables
Mathematical Keyboard
Converters
Other Tools

The Angle Bisectors Meet at the Incentre

How to use
  1. ▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
  2. Drag AA, BB or CC; the bisectors, II and the feet follow, and the readout under the canvas keeps the three distances from II equal. Learn more about why it holds
  3. Labels toggles the point names, Measures writes each marked half-angle beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene
Proof?The bisectors of the angles at A and B meet at I. Drop perpendiculars from I to the three sides: ID, IE, IF. Being on the bisector at A makes IE = IF; being on the bisector at B makes IF = ID. So ID = IE, and the right triangles CID and CIE agree in the hypotenuse and a leg, so CI bisects the angle at C. The circle about I with radius ID touches all three sides. Drag any vertex.Incentre
Display
Drag A, B or C; the bisectors, I and the feet follow
ααββABCI
ID = 1.53, IE = 1.53, IF = 1.53
Step 1 / 6

Step explanations

1Two angle bisectors meet at I
Given
AI bisects the angle at A (two angles α) and BI bisects the angle at B (two angles β); they meet at I. To prove: I is equidistant from the three sides, and the bisector of the angle at C passes through I as well.
I and the feet are computed; drag a vertex and they follow. Learn more about the statement

Getting Started with the Visualizer
The Statement
The Feet of the Perpendiculars
The First Bisector
The Second Bisector
The Third Bisector
Why It Holds
Dragging the Figure
Consequences



Getting Started with the Visualizer

DemoA run through the proof player
Step 0 of 7
Step through the proof, then drag the vertices and watch the three angle bisectors keep meeting at one point, with the circle about it still touching all three sides. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.
Opening scene: the controls
1234567
1Proof — the theorem loaded in the player; the ? gives its idea in one line
2Display — Labels shows the point names, Measures writes each marked half-angle beside its arc
3Draggable points — A, B and C move freely; the hollow crossing point I and, later, the feet D, E, F follow
4Readout — the line under the figure: the distances ID, IE, IF in the frame's units, live while you drag
5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure
6Speed and step counter — how fast Play advances, and which of the 6 steps is on screen
7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it
Proof complete: step 6 of 6
12345
1The incentre I — where the three angle bisectors meet, equidistant from the three sides
2The incircle — centre I, radius ID; it touches each side at the foot of a perpendicular
3The third bisector CI — through I by the converse: equal distances to CB and CA put I on the bisector at C
4Justification tags — Given, Construction, AAS congruence (twice), Pythagoras + SSS, Tangent ⟂ radius
5Replay — at the last step Play becomes Replay and Next is off

The Statement

Each angle of a triangle has a bisector, the line that cuts it into two equal halves. The theorem says that the three bisectors pass through one point II, which is at the same distance from the three sides:

ID=IE=IFID = IE = IF


where DD, EE, FF are the feet of the perpendiculars from II to the sides. The circle with centre II and radius IDID therefore touches all three sides; it is the inscribed circle, or incircle, and II is the incentre. It is the twin of the circumcentre theorem, with "equidistant from the sides" in place of "equidistant from the vertices".

The key fact, that a point on a bisector is equidistant from the two sides of the angle, was already proved on the angle bisector theorem page; here it is used twice and then turned round. In the player II and the three feet are hollow, computed from the vertices. The readout under the canvas shows the three distances, in the frame's own units, live while you drag.
ααββABCI
Opening scene, frozen

The bisectors at A and B, with their pairs of equal angles, meeting at the hollow point I.


The Feet of the Perpendiculars

Take the bisectors of the angles at AA and at BB. They meet at a point inside the triangle; call it II. The construction drops the three perpendiculars from II to the sides: to BCBC with foot DD, to CACA with foot EE, and to ABAB with foot FF. The lengths IDID, IEIE, IFIF are the distances from II to the three sides, since the perpendicular is the shortest segment from a point to a line.

The whole proof is about these three lengths: two of them are equal because II is on one bisector, two because it is on the other, and the three equalities together put II on the third bisector.
ααββABCIDEF
The feet, frozen

The perpendiculars ID, IE, IF from I to the three sides.


The First Bisector

II lies on the bisector of the angle at AA. The right triangles AIEAIE and AIFAIF share the hypotenuse AIAI, have equal angles α\alpha at AA, and have right angles at EE and FF; by the angle sum their third angles agree too, and with the common side that makes them congruent (angle-angle-side). So

IE=IFIE = IF


the distances from II to the two sides of the angle at AA are equal. The player fills the two triangles and marks the equal legs with one tick each.
ααββABCIDEF
The first bisector, frozen

AIE and AIF on the hypotenuse AI with equal angles α: IE = IF.


The Second Bisector

II also lies on the bisector of the angle at BB. The same argument with the right triangles BIFBIF and BIDBID, which share the hypotenuse BIBI and have equal angles β\beta at BB, gives

IF=IDIF = ID


Together with the first step, ID=IE=IFID = IE = IF: the crossing point of two bisectors is equidistant from all three sides. That is the first half of the theorem.
ααββABCIDEF
The second bisector, frozen

BIF and BID on the hypotenuse BI with equal angles β: IF = ID.


The Third Bisector

Now join CC to II and look at the right triangles CIDCID and CIECIE. They share the hypotenuse CICI and have equal legs ID=IEID = IE. By the Pythagorean theorem their remaining legs are equal too, CD2=CI2−ID2=CI2−IE2=CE2CD^2 = CI^2 - ID^2 = CI^2 - IE^2 = CE^2, so the triangles agree in all three sides and are congruent by side-side-side. Their angles at CC are therefore equal:

∠ICD=∠ICE=γ\angle ICD = \angle ICE = \gamma


which says that CICI bisects the angle at CC. So the bisector of the third angle passes through II: all three bisectors are concurrent. This is the converse of the first two steps, a point equidistant from the two sides of an angle lies on its bisector, and it is proved by the same pair of right triangles read the other way.
ααββγγABCIDEF
The third bisector, frozen

CID and CIE with equal legs on the hypotenuse CI: the angles γ at C are equal, so CI bisects C.


Why It Holds

The circle with centre II and radius ID=IE=IFID = IE = IF passes through the three feet DD, EE, FF. At each foot the side is perpendicular to the radius, and a line perpendicular to a radius at its end is a tangent; so each side touches the circle at one point, and the circle lies inside the triangle. It is the inscribed circle, drawn at the last step. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given bisectors, a construction, angle-angle-side twice, Pythagoras with side-side-side, and the tangent-radius theorem.

As with the circumcircle, the proof shows uniqueness too: the centre of any circle touching all three sides is equidistant from them, so it lies on all three bisectors, and two of those already meet in a single point.
ααββγγABCIDEF
The conclusion, frozen

The incircle about I, touching the three sides at D, E, F.


Dragging the Figure

Drag any vertex: the bisectors swing, the feet slide along the sides, the incircle grows and shrinks, and the readout keeps the three distances equal. Unlike the circumcentre, the incentre never leaves the inside of the triangle, because each bisector runs from its vertex into the interior. Make the triangle long and thin and the incircle becomes tiny; make it equilateral and the incentre coincides with the circumcentre and the centroid.

A drag that would flatten the triangle, or carry a vertex out of the frame, is refused.
ααββγγABCIDEF
After dragging A, frozen

A pulled right: a leaning triangle with a smaller incircle, still touching all three sides.


Consequences

• The inradius. The incircle's radius is r=Areasr = \dfrac{\text{Area}}{s}, with ss the half perimeter: the segments from II cut the triangle into three triangles of heights rr on the bases aa, bb, cc. Heron's formula for the area is on the sines and cosines law page.
• Tangent lengths. The sides are tangents from the vertices, so by the two tangents page the pieces from each vertex to the points of contact are equal: s−as - a, s−bs - b, s−cs - c.
• Excircles. The bisector of one angle together with the bisectors of the two exterior angles at the other vertices meet at an excentre, the centre of a circle touching one side and the extensions of the other two; there are three.
• Always inside. The incentre lies inside every triangle; the circumcentre does not.
• The angle at the incentre. ∠BIC=90°+12∠A\angle BIC = 90° + \tfrac{1}{2}\angle A, from the angle sum in triangle BICBIC with half-angles β\beta and γ\gamma.