Parallel Lines and Transversals
When a transversal cuts a pair of parallel lines, eight angles are formed — and they line up in three special families: alternate (equal, Z-shape), co-interior (supplementary, C-shape) and corresponding (equal, F-shape).
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Draw two horizontal parallel lines, about 5 cm apart. Now draw a slanted line cutting through BOTH of them — this is called a "transversal". Count how many angles are formed at the crossings. Which ones look equal? Which ones look like they'd add to $180^{\circ}$?
A transversal is a line that crosses two (or more) other lines. When the lines being crossed are parallel, the transversal creates EIGHT angles — four at each crossing — and they fall into three special pairs: alternate, co-interior, corresponding.
Two parallel lines (mark with arrows or use $\parallel$ symbol) are cut by a transversal. At each crossing four angles form. Compared across the two crossings: alternate angles are equal (form a Z), co-interior angles sum to $180^{\circ}$ (form a C), and corresponding angles are equal (form an F).
Know
- What a transversal is
- The three pairs: alternate, co-interior, corresponding
- The Z, C and F shape memory aids
Understand
- Why parallel lines force these angle relationships
- How the three families connect (e.g. alternate + co-int. $= 180^{\circ}$ on same line)
- When a diagram does NOT imply parallel lines
Can Do
- Identify pairs of alt., co-int. and corr. angles
- Find unknown angles using parallel-line rules
- Cite the reason "(alt. angles, $AB \parallel CD$)" in working
Alternate angles sit on OPPOSITE sides of the transversal and BETWEEN the two parallel lines. They trace out a Z (or backward Z). They are EQUAL.
To spot them: look for the "inside corners" on different sides of the transversal — they make a Z. $\angle x = \angle y$ (alt. angles, $AB \parallel CD$). If one is $70^{\circ}$, the other is also $70^{\circ}$.
Book notes · Card 4
- Alternate angles sit on opposite sides of the transversal, between the parallel lines (Z-shape).
- Alternate angles are EQUAL when the lines are parallel.
- Reason name: "(alt. angles, $AB \parallel CD$)".
True or False: Alternate angles formed by a transversal cutting two parallel lines are equal in size.
Co-interior angles sit on the SAME side of the transversal, BETWEEN the two parallel lines. They trace out a C. They are supplementary — they sum to $180^{\circ}$.
Co-interior angles share the SAME side of the transversal. Both are between the parallel lines. $\angle x + \angle y = 180^{\circ}$ (co-int. angles, $AB \parallel CD$). If one is $110^{\circ}$, the other is $70^{\circ}$. (You met these in the parallelogram lesson — same idea.)
Book notes · Card 5
- Co-interior angles: same side of transversal, between the parallel lines (C-shape).
- They are SUPPLEMENTARY: sum to $180^{\circ}$ (NOT equal).
- Reason name: "(co-int. angles, $AB \parallel CD$)".
Two parallel lines are cut by a transversal. One co-interior angle is $115^{\circ}$. The other is:
Corresponding angles are in the SAME POSITION at each crossing — e.g. both above the parallel line and both to the right of the transversal. They trace out an F. They are EQUAL.
To check if two angles correspond: are they on the SAME side of the transversal AND in the same position (both above, or both below) at their respective crossings? If yes, $\angle x = \angle y$ (corr. angles, $AB \parallel CD$). The classic visual: an "F" formed by following the transversal and the parallel line.
Book notes · Card 6
- Corresponding angles: same position at each crossing (F-shape).
- Corresponding angles are EQUAL when lines are parallel.
- Reason name: "(corr. angles, $AB \parallel CD$)".
Two parallel lines are cut by a transversal. Corresponding angles are in size (one word).
Watch Me Solve It · 3 examples
- 1Spot the relationshipBoth angles are BETWEEN the lines, on OPPOSITE sides ⇒ they form a Z ⇒ alternate.
- 2Apply the rule$x = 72^{\circ}$ (alt. angles, $AB \parallel CD$).
- 3State$x = 72^{\circ}$.Equal because alternate angles are equal whenever the lines are parallel.
- 1Spot the relationshipSame side, between the lines ⇒ C-shape ⇒ co-interior.
- 2Apply the rule$x + 108 = 180$ (co-int. angles, $\ell \parallel m$).
- 3Solve$x = 180 - 108 = 72^{\circ}$.Co-interior angles sum to $180^{\circ}$, not equal — check the C.
- 1Spot the relationshipBoth above and on the right ⇒ same position ⇒ F-shape ⇒ corresponding.
- 2Apply the rule$x = 85^{\circ}$ (corr. angles, $AB \parallel CD$).
- 3State$x = 85^{\circ}$.Corresponding angles are equal when the lines are parallel.
Common Pitfalls
Alternate (Z)
- Opposite sides of transversal
- Between the parallel lines
- EQUAL
- (alt. angles, $AB \parallel CD$)
Co-interior (C)
- Same side of transversal
- Between the parallel lines
- SUM $180^{\circ}$
- (co-int. angles, $AB \parallel CD$)
Corresponding (F)
- Same position at each crossing
- EQUAL
- (corr. angles, $AB \parallel CD$)
Reminders
- Rules need parallel lines!
- Mark parallels with arrows or $\parallel$
- Cite parallel-line names in reason
How are you completing this lesson?
Brain Trainer · 4 problems
Four quick drills. Solve, then reveal.
-
1 Alt. angles — one is $54^{\circ}$. The other is?
Alt. angles equal.$54^{\circ}$ -
2 Co-int. angles — one is $123^{\circ}$. The other is?
$180 - 123$.$57^{\circ}$ -
3 Corr. angles — one is $98^{\circ}$. The other is?
Corr. angles equal.$98^{\circ}$ -
4 An angle of $63^{\circ}$ is above the top parallel line. Find the angle below the BOTTOM parallel line directly under it (i.e. vertically opp. to the corresponding angle).
Corr. = $63^{\circ}$, vert. opp. = $63^{\circ}$.$63^{\circ}$
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. Lines $AB \parallel CD$ are cut by transversal $EF$. At the top crossing the angle above-right is $78^{\circ}$. Find, with a reason in each case:
(a) The angle below the top line, same side — on the right.
(b) The corresponding angle on the bottom line.
(c) The co-interior angle to the original $78^{\circ}$.
Q7. Two parallel lines are cut by a transversal. Two co-interior angles are $(2x + 10)^{\circ}$ and $(3x - 20)^{\circ}$.
(a) Set up an equation.
(b) Solve for $x$.
(c) State the actual angle values.
Q8. For each statement, decide TRUE or FALSE and explain why in one sentence.
(a) Vertically opposite angles are always equal, even without parallel lines.
(b) If alternate angles are equal, then the two lines must be parallel.
(c) Two co-interior angles must each be obtuse.
Quick Check
1. B — A line that crosses two other lines.
2. A — Equal.
3. C — $115^{\circ}$.
4. A — F (equal).
5. D — The two lines being cut must be parallel.
Show Your Working Model Answers
Q6 (3 marks): (a) $180 - 78 = 102^{\circ}$ (angles on a straight line at the crossing) [1]. (b) Corresponding angle below-right on bottom line $= 78^{\circ}$ (corr. angles, $AB \parallel CD$) [1]. (c) Co-interior angle $= 180 - 78 = 102^{\circ}$ (co-int. angles, $AB \parallel CD$) [1].
Q7 (3 marks): (a) $(2x + 10) + (3x - 20) = 180$ (co-int. angles) [1]. (b) $5x - 10 = 180 \Rightarrow 5x = 190 \Rightarrow x = 38$ [1]. (c) Angles are $86^{\circ}$ and $94^{\circ}$ (check $86 + 94 = 180$) [1].
Q8 (3 marks): (a) TRUE — vertically opposite angles are always equal at any crossing, regardless of parallelism [1]. (b) TRUE — equal alternate angles is the converse: if they're equal, the lines must be parallel [1]. (c) FALSE — co-interior pairs sum to $180^{\circ}$, so they could be $90^{\circ}$ each, or one acute and one obtuse — only the SUM is fixed [1].
Three-Lines Puzzle
(a) Three lines $\ell_1, \ell_2, \ell_3$ all run in the same direction (all parallel to each other). A transversal cuts all three. At $\ell_1$ it makes an angle of $40^{\circ}$ above-right. What angle does it make above-right at $\ell_2$ and $\ell_3$? Justify. (b) Two parallel lines $AB \parallel CD$ are cut by transversal $XY$ at points $P$ (on $AB$) and $Q$ (on $CD$). Triangle $PQR$ is formed where $R$ lies on $CD$ to the right of $Q$. If $\angle APQ = 65^{\circ}$ (above $AB$, left of transversal) and $\angle PQR = 50^{\circ}$, find $\angle QPR$. (c) Show how the angle-sum of a triangle ($180^{\circ}$) can be PROVED using alternate angles between parallel lines.
Reveal solution
(a) All three above-right angles equal $40^{\circ}$ (corresponding angles transfer through every pair of parallel lines). (b) $\angle APQ$ and $\angle PQR'$ (where $R'$ is to the LEFT of $Q$) are alternate, so $\angle PQR' = 65^{\circ}$. Then $\angle PQR = 180 - 65 = 115^{\circ}$... but actually if $\angle PQR = 50^{\circ}$ is GIVEN as something between $Q$ and another arrangement, $\angle QPR = 180 - 65 - 50 = 65^{\circ}$ via ∠ sum of $\triangle$. (c) Through vertex $A$ of $\triangle ABC$, draw a line $\ell$ parallel to $BC$. The angles $\angle B$ and one angle at $A$ are alternate (equal), the angles $\angle C$ and another angle at $A$ are alternate (equal), and the three angles at $A$ make up a straight line $= 180^{\circ}$. So $\angle A + \angle B + \angle C = 180^{\circ}$.
Transversal
A line crossing two (or more) lines.
Alt. angles (Z)
EQUAL; opposite sides, between lines.
Co-int. (C)
SUM $180^{\circ}$; same side, between lines.
Corr. (F)
EQUAL; same position at each crossing.
Need parallel
All three rules need $AB \parallel CD$.
Reason
"(alt./co-int./corr. angles, $AB \parallel CD$)"
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