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Perpendicular Bisector Theorem

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 PP along the perpendicular bisector, or AA or BB; the readout under the canvas shows PAPA, PBPB and the angle at MM. Learn more about why it holds
  3. 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?The perpendicular bisector of AB goes through the midpoint M at right angles to AB. For P on it, the triangles PMA and PMB have AM = MB, equal right angles at M and the common side PM, so they are congruent (SAS) and PA = PB. Conversely, if PA = PB, the same two triangles are congruent by SSS, the angles at M are equal and make a straight angle, so each is 90° and P is on the perpendicular bisector. Drag P along the line, or A or B.Perpendicular bisector
Display
Drag P along the perpendicular bisector, or A or B
ABMP
PA = 3.41, PB = 3.41, ∠PMA = 90°
Step 1 / 6

Step explanations

1A segment and its perpendicular bisector
Given
M is the midpoint of AB (AM = MB, one tick each), and the dashed line through M is perpendicular to AB: it is the perpendicular bisector of AB. P is any point on it. To prove: PA = PB, and conversely, every point equally far from A and B lies on this line.
Drag P along the dashed line. Learn more about the statement

Getting Started with the Visualizer
The Statement
Two Triangles
Side, Angle, Side
Equal Distances
The Converse
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 PP along the dashed perpendicular bisector, or drag AA or BB, and watch the two distances in the readout stay equal while the angle at MM stays 90°90°. 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
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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, A and B move the segment; the hollow midpoint M follows
4Readout — the line under the figure: PA, PB and the angle at M, 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
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1The bisector — the dashed line through M at right angles to AB
2The two triangles — PMA and PMB, shaded either side of PM
3The ticks — one tick for AM = MB, two for PA = PB
4Justification tags — Given, Construction, SAS, Corresponding parts, SSS, Linear pair
5Replay — at the last step Play becomes Replay and Next is off

The Statement

The perpendicular bisector of a segment ABAB is the line through its midpoint MM at right angles to it. The theorem describes it by distances:

P is on the perpendicular bisector of AB  ⟺  PA=PBP \text{ is on the perpendicular bisector of } AB \iff PA = PB


The forward direction says that every point of the bisector is equally far from AA and BB; the converse says that every point equally far from AA and BB is on the bisector. Together they make the perpendicular bisector the locus of points equidistant from AA and BB.

This is the fact behind the circumcentre, the chord bisector and the kite diagonals pages. In the player AA and BB are free and PP slides along the perpendicular at MM; the readout shows PAPA, PBPB and the angle at MM.
ABMP
Opening scene, frozen

The segment AB, its midpoint M and the dashed perpendicular bisector; P on it.


Two Triangles

Join PP to AA, to BB and to the midpoint MM. The segment PMPM, which lies along the bisector, cuts the triangle PABPAB into the two triangles PMAPMA and PMBPMB. They share the side PMPM, and the whole proof, in both directions, compares these two triangles.
ABMP
Two triangles, frozen

PA, PB and PM drawn; the triangles PMA and PMB shaded.


Side, Angle, Side

For PP on the perpendicular bisector, the triangles PMAPMA and PMBPMB have

AM=MB,∠PMA=∠PMB=90°,PM=PMAM = MB, \qquad \angle PMA = \angle PMB = 90°, \qquad PM = PM


two sides and the angle between them. By SAS they are congruent.
ABMP
Side, angle, side, frozen

AM = MB, right angles at M, PM common: PMA ≅ PMB.


Equal Distances

Corresponding parts of congruent triangles are equal, so the third sides agree:

PA=PBPA = PB


Every point of the perpendicular bisector is equally far from the two ends of the segment.
ABMP
Equal distances, frozen

PA = PB, two ticks each.


The Converse

Now forget the bisector and start from a point PP with PA=PBPA = PB; let MM be the midpoint of ABAB. The triangles PMAPMA and PMBPMB have PA=PBPA = PB, AM=MBAM = MB and the common side PMPM: three pairs of equal sides, so by SSS they are congruent, and

∠PMA=∠PMB\angle PMA = \angle PMB


This is the same argument as on the isosceles base angles page, where the triangle PABPAB with PA=PBPA = PB is isosceles and PMPM is its median.
ABMP
The converse, frozen

From PA = PB: SSS makes the angles at M equal.


Why It Holds

The angles PMAPMA and PMBPMB lie side by side along the line ABAB, so they make a straight angle, 180°180°. They are equal, so each is half of it:

∠PMA=∠PMB=90°\angle PMA = \angle PMB = 90°


PMPM is perpendicular to ABAB at its midpoint, so PP is on the perpendicular bisector. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, a construction, SAS, corresponding parts, SSS and the linear pair.

The two directions use the same two triangles: SAS when the right angle is known, SSS when the equal distances are known.
ABMP
The conclusion, frozen

Equal angles in a straight line: 90° each, P on the bisector.


Dragging the Figure

Drag PP along the dashed line: the two segments PAPA and PBPB grow or shrink together, and the readout shows them equal at every position while the angle at MM stays 90°90°. Drag AA or BB to turn or stretch the segment; the bisector turns with it and PP keeps its place along it.

The player keeps PP on the bisector above ABAB and the whole dashed line inside the frame; a drag that would push it out is refused.
ABMP
After dragging P, frozen

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


Consequences

• The circumcentre. The perpendicular bisectors of two sides of a triangle meet at a point equally far from all three vertices, so the third bisector passes through it too: the circumcentre.
• Chords. The centre of a circle is equally far from the ends of any chord, so it lies on the chord's perpendicular bisector: the chord bisector theorem from the other side.
• Kites and rhombuses. In a kite two vertices are each equally far from the other two, so the line through them is the perpendicular bisector of the other diagonal: the kite diagonals are perpendicular.
• Construction. Two arcs of equal radius about AA and about BB meet in two points equally far from both; the line through them is the perpendicular bisector, which is how it is drawn with ruler and compass.
• Reflection. Reflecting in the perpendicular bisector swaps AA and BB; every point of the mirror line is its own image and so is equally far from a point and its image.