Visual Tools
Calculators
Tables
Mathematical Keyboard
Converters
Other Tools

Bisectors of a Linear Pair

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 CC to change the two angles, slide OO along the line, or drag AA or BB; the readout under the canvas shows α\alpha, β\beta and the angle between the bisectors. 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 angles α = ∠AOC and β = ∠COB lie along the line AB, so α + β = 180°. The bisector OP halves α and the bisector OQ halves β, so the angle between the bisectors is α/2 + β/2 = (α + β)/2 = 90°. Drag C to change α and β, or slide O along the line.Bisectors of a linear pair
Display
Drag C to change the two angles, O along the line, or A or B
αβABCOPQ
α = 109°, β = 71°, ∠POQ = 90°
Step 1 / 5

Step explanations

1A linear pair and its two bisectors
Given
O is a point of the line AB and OC is a ray from it. The angles α = ∠AOC and β = ∠COB are a linear pair: they share the arm OC and their other arms OA, OB lie along one line. OP bisects α and OQ bisects β. To prove: ∠POQ = 90°, the two bisectors are perpendicular.
Drag C: the bisectors turn, the angle between them does not. Learn more about the statement

Getting Started with the Visualizer
The Statement
A Straight Angle
The Two Halves
Adding the Halves
Why It Holds
Dragging the Figure
Consequences



Getting Started with the Visualizer

DemoA run through the proof player
Step 0 of 6
Step through the proof, then drag CC to change the two angles, or slide OO along the line, and watch the angle between the two bisectors in the readout stay at 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
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 — C moves freely and sets the two angles, O slides along the line, A and B move its ends; the hollow points P and Q follow on the bisectors
4Readout — the line under the figure: α, β and the angle between the bisectors, 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 5 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 5 of 5
12345
1The two bisectors — OP halves α, OQ halves β; the square mark between them is the right angle
2The halves — α/2 twice and β/2 twice around O, inside the larger arcs of α and β
3The straight angle — the line AB through O: α and β together make 180°
4Justification tags — Given, Straight angle, Bisectors, Adding, Substituting
5Replay — at the last step Play becomes Replay and Next is off

The Statement

Two angles form a linear pair when they share one arm and their other arms lie along one straight line, on either side of the common arm. In the player the ray OCOC stands on the line ABAB at OO, and α=∠AOC\alpha = \angle AOC and β=∠COB\beta = \angle COB are a linear pair. The theorem says that their bisectors are perpendicular:

∠POQ=90°\angle POQ = 90°


where OPOP bisects α\alpha and OQOQ bisects β\beta. It is a two-line proof, and it is the reason behind several facts about triangles and circles: the internal and external bisectors at a vertex of a triangle are perpendicular, which is where the excentres and some right-angle constructions come from.

CC, AA and BB are free and OO slides along the line; the readout shows α\alpha, β\beta and the angle between the bisectors.
αβABCOPQ
Opening scene, frozen

The ray OC on the line AB; α and β a linear pair; the bisectors OP and OQ.


A Straight Angle

The angles α\alpha and β\beta sit side by side along the line ABAB and together fill the straight angle AOBAOB, so they are supplementary:

α+β=180°\alpha + \beta = 180°


This is the same fact that drives the vertical angles page, where two linear pairs share an angle.
αβABCOPQ
A straight angle, frozen

α and β lit: together they fill the straight angle AOB.


The Two Halves

The bisector OPOP cuts α\alpha into two equal parts, ∠AOP=∠POC=α/2\angle AOP = \angle POC = \alpha/2, and the bisector OQOQ cuts β\beta into ∠COQ=∠QOB=β/2\angle COQ = \angle QOB = \beta/2. The angle between the two bisectors is made of the two halves that touch the common arm OCOC:

∠POQ=∠POC+∠COQ=α2+β2\angle POQ = \angle POC + \angle COQ = \frac{\alpha}{2} + \frac{\beta}{2}
αβα/2α/2β/2β/2ABCOPQ
The two halves, frozen

α/2 twice and β/2 twice; the two halves next to OC make ∠POQ.


Adding the Halves

Half of α\alpha plus half of β\beta is half of their sum:

∠POQ=α2+β2=α+β2\angle POQ = \frac{\alpha}{2} + \frac{\beta}{2} = \frac{\alpha + \beta}{2}


The two angles may be anything; only their sum enters, and the sum is fixed by the straight line.
αβα/2α/2β/2β/2ABCOPQ
Adding the halves, frozen

∠POQ = α/2 + β/2 = (α + β)/2.


Why It Holds

Put in α+β=180°\alpha + \beta = 180°:

∠POQ=180°2=90°\angle POQ = \frac{180°}{2} = 90°


The bisectors of a linear pair are perpendicular, whatever the two angles are. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, a straight angle, the bisectors, an addition and a substitution.

In words: halving two angles halves their sum, and half of a straight angle is a right angle.
αβα/2α/2β/2β/2ABCOPQ
The conclusion, frozen

Half of 180° is 90°: the square mark between the bisectors.


Dragging the Figure

Drag CC: the ray OCOC swings, α\alpha grows as β\beta shrinks (or the other way round), both bisectors turn, and the angle between them in the readout stays at 90°90°. Slide OO along the line, or drag AA or BB, to change the picture without changing the argument.

The player keeps CC above the line and both angles at least 20°20°, so that all the marked angles stay readable, and keeps the bisector rays inside the frame. A drag that would break one of these is refused.
αβα/2α/2β/2β/2ABCOPQ
After dragging C, frozen

C leaned to the left: α larger, β smaller, the bisectors still perpendicular.


Consequences

• Internal and external bisectors. At a vertex of a triangle the interior angle and its exterior angle are a linear pair, so the internal and external bisectors are perpendicular; the incentre and the three excentres form a triangle whose altitudes are the internal bisectors.
• Perpendicular lines from any angle. To draw a perpendicular at a point of a line without a set square, draw any ray there and bisect the two angles it makes; either bisector is perpendicular to the other.
• The angle bisectors of a parallelogram. The bisectors of two consecutive angles of a parallelogram meet at a right angle, because consecutive angles are supplementary, like a linear pair.
• Vertical angles. The bisectors of two vertical angles lie on one line, since each makes equal angles with the bisector of the adjacent angle between them, by this theorem.
• Supplementary angles in general. The proof never used that the two angles are adjacent, only that they add up to 180°180°: any two supplementary angles with a common arm have perpendicular bisectors.