Is every point on an angle bisector equidistant from the sides of the angle?

Yes. The perpendiculars from the point to the two sides form two right triangles with equal angles at the vertex and a common hypotenuse, so they are congruent by AAS and the two perpendiculars are equal.

What is the converse of the angle bisector locus theorem?

If a point inside an angle is the same distance from both sides, it lies on the bisector: the two right triangles share their hypotenuse and have equal legs, so they are congruent by HL and the angles at the vertex are equal.

How is this different from the angle bisector theorem?

The angle bisector theorem says the bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides. This page is about distances to the arms of the angle; the angle bisector theorem uses it as one step.

Angle Bisector: Equidistant from the Arms

How to use

▶ 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
Drag $P$ along the bisector, or $O$ , $A$ or $B$ to change the angle; the readout under the canvas shows $PD$ , $PE$ and the two angles at $O$ . Learn more about why it holds
Labels toggles the point names, Measures writes each marked angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene

Proof?Drop the perpendiculars from P to the two arms, with feet D and E. For P on the bisector the right triangles OPD and OPE have equal angles at O, right angles at D and E, and the common side OP, so they are congruent (AAS) and PD = PE. Conversely, if PD = PE, the right triangles share the hypotenuse OP and have equal legs, so they are congruent (HL) and ∠POD = ∠POE. Drag P along the bisector, or O, A or B.Angle bisector locus

DisplayLabelsMeasures

Drag P along the bisector, or O, A or B to change the angle

PD = 1.14,  PE = 1.14,  ∠AOP = 19°,  ∠POB = 19°

Step 1 / 6

Step explanations

1An angle and its bisector

Given
The rays OA and OB make an angle at O, and the dashed ray from O bisects it: the two angles φ are equal. P is a point on the bisector. To prove: P is equally far from the two arms, and conversely every point inside the angle that is equally far from the arms lies on the bisector.

Drag P along the dashed bisector. Learn more about the statement

Getting Started with the Visualizer

Getting Started with the Visualizer

DemoA run through the proof player

Step 0 of 7

Step through the proof, then drag

P

along the dashed bisector, or drag

O

A

B

to change the angle, and watch the two distances in the readout stay equal while the two half angles stay equal. 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

Proof?Angle bisector locusDisplayLabelsMeasuresDrag P along the bisector, or O, A or B to change the angle

PD = 1.14,  PE = 1.14,  ∠AOP = 19°,  ∠POB = 19°

← Back▶ PlayNext →ResetNormal ▾Step 1 / 6

Step explanations

1An angle and its bisector

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 angle's size beside its arc

3Draggable points — P slides along the dashed bisector, O, A and B change the angle; the hollow feet D and E follow

4Readout — the line under the figure: PD, PE and the two angles at O, 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

Proof?Angle bisector locusDisplayLabelsMeasuresDrag P along the bisector, or O, A or B to change the angle

PD = 1.14,  PE = 1.14,  ∠AOP = 19°,  ∠POB = 19°

← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6

Step explanations

1An angle and its bisector

2The perpendiculars PD and PE

Construction
The distance from P to an arm is the length of the perpendicular to it. Drop PD ⟂ OA and PE ⟂ OB, with feet D and E. The claim is PD = PE.

3OPD ≅ OPE

AAS
The triangles OPD and OPE have ∠POD = ∠POE = φ (the bisector), right angles at D and E, and the common side OP, opposite the right angles. Two angles and a side not between them: by AAS they are congruent.

4PD = PE

Corresponding parts
The sides opposite the equal angles at O are equal: PD = PE (one tick each). Every point of the bisector is equally far from the two arms.

5Conversely: PD = PE gives ∠POD = ∠POE

HL congruence
Now start only from a point P inside the angle with PD = PE. The right triangles OPD and OPE share the hypotenuse OP and have equal legs PD = PE. By the hypotenuse-leg criterion they are congruent, so ∠POD = ∠POE.

6P lies on the bisector

Definition
OP makes equal angles with the two arms, so it is the bisector and P lies on it. The bisector of an angle is exactly the set of points inside it that are equally far from its two arms. ∎

12345

1The bisector — the dashed ray from O, with two equal angles φ

2The two right triangles — OPD and OPE, shaded either side of OP

3The perpendiculars — PD and PE, one tick each: equal at the end

4Justification tags — Given, Construction, AAS, Corresponding parts, HL congruence, Definition

5Replay — at the last step Play becomes Replay and Next is off

The Statement

The bisector of an angle is the ray from its vertex that cuts it into two equal angles. The theorem describes it by distances: for a point

P

inside the angle

AOB

, with

D

and

E

the feet of the perpendiculars from

P

to the arms

OA

and

OB

P \text{ is on the bisector} \iff PD = PE

So the bisector is the locus of points inside the angle that are equally far from its arms. It is the angle version of the perpendicular bisector theorem, with "equally far from two points" replaced by "equally far from two lines". It is a different statement from the angle bisector theorem, which is about the ratio in which the bisector of a triangle's angle cuts the opposite side, though that page uses this one.

In the player

O

A

and

B

are free and

P

slides along the bisector; the readout shows

PD

PE

and the two angles at

O

Opening scene, frozen

The angle AOB, its dashed bisector with two equal angles φ, and P on it.

The Perpendiculars

The distance from a point to a line is the length of the perpendicular from it (the perpendicular is shortest). Drop

PD \perp OA

and

PE \perp OB

. The claim is now about two segments:

PD = PE

The perpendiculars, frozen

PD ⟂ OA and PE ⟂ OB, square marks at D and E.

Angle, Angle, Side

Compare the right triangles

OPD

and

OPE

. They have

\angle POD = \angle POE, \qquad \angle ODP = \angle OEP = 90°, \qquad OP = OP

two angles and a side that is not between them, the common hypotenuse

OP

. By AAS they are congruent.

Angle, angle, side, frozen

OPD and OPE shaded: φ, a right angle and the common side OP.

Equal Distances

Corresponding parts of congruent triangles are equal, so the legs opposite the equal angles at

O

agree:

PD = PE

Every point of the bisector is equally far from the two arms.

Equal distances, frozen

PD = PE, one tick each.

The Converse

Now start from a point

P

inside the angle with

PD = PE

, without knowing where it lies. The triangles

OPD

and

OPE

are right-angled at

D

and

E

, share the hypotenuse

OP

, and have equal legs

PD = PE

. By the hypotenuse-leg criterion they are congruent, so

\angle POD = \angle POE

The converse, frozen

From PD = PE: HL on the two right triangles.

Why It Holds

OP

makes equal angles with the two arms, which is what it means to bisect the angle; so

P

lies on the bisector. With the forward direction,

P \text{ on the bisector} \iff PD = PE

The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, a construction, AAS, corresponding parts, HL and the definition of the bisector.

Both directions compare the same two right triangles with the common hypotenuse

OP

: AAS when the equal angles are known, HL when the equal distances are known.

The conclusion, frozen

Equal angles at O: P lies on the bisector.

Dragging the Figure

Drag

P

along the dashed bisector: the perpendiculars and their feet move, and the readout shows

PD

and

PE

equal at every position. Drag

O

A

B

to change the angle; the bisector turns with the arms, and

P

keeps its place along it.

The player keeps the angle between

30°

and

120°

, the feet

D

and

E

on the drawn arms, and everything in the frame; a drag that would break one of these is refused.

After dragging P, frozen

P moved out along the bisector: both distances longer, still equal.

Consequences

• The incentre. The bisectors of two angles of a triangle meet at a point equally far from all three sides, so it lies on the third bisector too: the incentre, the centre of the inscribed circle.
• Excentres. The same argument with the external bisectors gives three more points equally far from the three side lines, the centres of the excircles; internal and external bisectors are perpendicular (bisectors of a linear pair).
• Tangents from a point. A circle touching both arms of an angle has its centre equally far from them, so on the bisector; this is why the centre of a circle lies on the bisector of the angle between two tangents.
• Two crossing lines. The points equally far from two crossing lines fill the two bisectors of the four angles they make, a pair of perpendicular lines.
• The angle bisector theorem. On the angle bisector theorem page the equal distances from the foot of the bisector to the two sides make the two areas proportional to the sides.