Visual Tools
Calculators
Tables
Mathematical Keyboard
Converters
Other Tools

Parallel Lines and a Transversal

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 lines. Learn more about the controls
  2. Drag EE or FF to tilt the transversal, AA or BB to turn the lines (they stay parallel), CC to move the second line; the readout under the canvas keeps α=α′\alpha = \alpha' and β+α′=180°\beta + \alpha' = 180°. Learn more about why it holds
  3. Labels toggles the point names, Measures writes each angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene
Proof?A transversal crosses two parallel lines at P and Q. The parallel postulate says the corresponding angles at P and Q are equal. Vertical angles then make the alternate angles equal, and a linear pair makes the co-interior angles add up to 180°. Drag the ends of the transversal or the lines: the three relations hold every time.Parallel lines and a transversal
Display
Drag E or F to tilt the transversal, A, B or C to move the lines
ABCEFDPQ
α 112.1° = α′ 112.1° · β 67.9° + α′ = 180.0°
Step 1 / 5

Step explanations

1Two parallel lines and a transversal
Given
The lines AB and CD are parallel (arrow marks); the transversal EF crosses them at P and at Q. To prove: the pairs of angles at P and Q are related: corresponding equal, alternate equal, co-interior supplementary.
The arrow marks say the two lines are parallel. Learn more about the statement

Getting Started with the Visualizer
The Statement
Corresponding Angles
Alternate Angles
Co-interior Angles
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 tilt the transversal or move the lines to see that the three relations between the angles hold whatever the crossing looks like. 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 angle's size beside its arc
3Draggable points — E and F tilt the transversal, A and B turn the lines, C slides the second line; the hollow points D, P, Q are constructed and follow
4Readout — the line under the figure: α, α′ and β + α′, live while you drag
5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original lines
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 parallel lines — AB and CD, with one arrow mark in the middle of each: the lines are parallel, the one assumption of the theorem
2The angles at P — α above right, its vertical angle below left, and β below right
3The angle at Q — α′: corresponding to α, alternate to its vertical angle, co-interior with β
4Justification tags — Given, Parallel postulate, Vertical angles, Straight angle, Summary
5Replay — at the last step Play becomes Replay and Next is off

The Statement

Two lines in the plane are parallel when they never meet, however far they are extended. A third line that crosses both is a transversal. At the two crossings, PP on the first line and QQ on the second, the transversal makes eight angles, and the theorem says how they are related when the lines are parallel:

• Corresponding angles, in the same position at the two crossings (both upper right, say), are equal: α=α′\alpha = \alpha'.
• Alternate angles, between the parallels on opposite sides of the transversal, are equal: ∠APQ=α′\angle APQ = \alpha'. These are the alternate interior angles that the proof of the triangle angle sum relies on.
• Co-interior angles, between the parallels on the same side of the transversal, are supplementary: β+α′=180°\beta + \alpha' = 180°.

This is Proposition 29 of Book I of Euclid's Elements, the first proposition in which he uses his fifth postulate, the parallel postulate. The three relations are really one fact seen three ways, and the proof shows how the other two follow from the first.

The readout under the canvas shows α\alpha, α′\alpha' and β\beta in degrees, live while you drag.
ABCEFDPQ
Opening scene, frozen

Two parallel lines (arrow marks) and the transversal EF, crossing them at P and Q.


Corresponding Angles

The angles α\alpha at PP and α′\alpha' at QQ sit in the same position at their crossings: above the line, to the right of the transversal. Such a pair is called corresponding angles. The first step of the proof is that for parallel lines they are equal:

α=α′\alpha = \alpha'


Where does this come from? It cannot be proved from the other postulates of plane geometry; it is the parallel postulate itself, or rather one of its equivalent forms. Euclid's own wording speaks of co-interior angles: if they add up to less than two right angles, the lines meet on that side. School courses usually take the corresponding-angles form as the axiom, and that is what the player does here, tagging the step "Parallel postulate". On a sphere, where no lines are parallel in Euclid's sense, the statement fails; in the flat plane it is the one assumption everything about parallels rests on.
αα′ABCEFDPQ
Corresponding angles, frozen

α at P and α′ at Q, in the same position at the two crossings: equal by the parallel postulate.


Alternate Angles

Now look at the angle at PP below the line and to the left of the transversal, ∠APQ\angle APQ. It lies opposite α\alpha across the crossing point PP: the two are vertical angles, so they are equal, and therefore

∠APQ=α=α′\angle APQ = \alpha = \alpha'


The angles ∠APQ\angle APQ and α′\alpha' both lie between the two parallels, one on each side of the transversal: that is what makes them alternate angles. The letter Z drawn through AA, PP, QQ, DD shows the pair at its two corners, which is why they are sometimes called Z-angles. The player marks ∠APQ\angle APQ with the same colour and label as α\alpha, since it is equal to it.
αα′αABCEFDPQ
Alternate angles, frozen

The vertical angle of α at P, below the line on the left, equals α′: alternate angles.


Co-interior Angles

The last relation uses the straight line at PP. The angles β=∠QPB\beta = \angle QPB and ∠APQ\angle APQ share the side PQPQ and together fill the straight angle APBAPB, so they are a linear pair and add up to 180°180°. Since ∠APQ=α′\angle APQ = \alpha',

β+α′=180°\beta + \alpha' = 180°


The angles β\beta and α′\alpha' both lie between the parallels, on the same side of the transversal: co-interior angles (also called consecutive or same-side interior angles). They are supplementary. The letter C through BB, PP, QQ, DD marks the pair, which is why they are sometimes called C-angles.
αα′αβABCEFDPQ
Co-interior angles, frozen

β at P and α′ at Q, between the parallels on the same side: β + α′ = 180°.


Why It Holds

Put the three steps together. The parallel postulate gives one equality, α=α′\alpha = \alpha'; vertical angles carry it to the alternate pair; a linear pair turns it into the supplementary co-interior pair. The end-of-proof mark ∎ in the log closes the argument. The tags name the three facts used: the parallel postulate, vertical angles, and the straight angle. The first is an axiom; the other two have their own proofs.

The same three relations hold for the other angles at the crossings, by the same reasoning with the roles of left and right, or above and below, exchanged: every one of the eight angles at PP and QQ is either α′\alpha' or 180°−α′180° - \alpha'. Knowing one angle at a transversal gives all eight.

The converses are true as well (Euclid I.27 and I.28): if a transversal makes equal alternate angles, or equal corresponding angles, or supplementary co-interior angles, with two lines, then the lines are parallel. That is how parallel lines are recognised and constructed in practice.
αα′αβABCEFDPQ
The three relations, frozen

All four marked angles: corresponding equal, alternate equal, co-interior supplementary.


Dragging the Figure

Drag EE or FF to tilt the transversal: α\alpha changes, and α′\alpha' changes with it, staying equal, while β\beta moves the opposite way so that β+α′\beta + \alpha' stays 180°180°. Tilt the transversal until it is perpendicular to the lines and all eight angles become right angles. Drag AA or BB to turn the first line; the second turns with it, since it is kept parallel. Drag CC to slide the second line up or down.

A drag that would take the transversal off either line, or make it almost parallel to them, is refused: the crossings PP and QQ must exist and lie on the drawn segments.
αα′αβABCEFDPQ
After dragging E and F, frozen

The transversal tilted the other way: α and α′ have changed together, β the other way; the relations hold.


Consequences

• The triangle angle sum. The proof of the triangle angle sum draws a parallel through one vertex and uses the equal alternate angles twice; this page is the justification behind its "Alternate angles" tag.
• Parallelograms. Each pair of opposite sides is a pair of parallels cut by the other sides, so opposite angles of a parallelogram are equal and adjacent angles are supplementary.
• Recognising parallels. By the converses, a pair of equal corresponding angles, or equal alternate angles, is enough to know two lines are parallel; this is how a set square draws parallel lines.
• Distances. Alternate angles and the congruence of triangles show that two parallels are everywhere the same distance apart.