By the end of this chapter you'll be able to…

  • 1Identify acute, right, obtuse, reflex and complete angles
  • 2Use complementary (sum 90°) and supplementary (sum 180°) angle relationships
  • 3Apply the Linear Pair Axiom and its converse
  • 4Prove and apply the Vertically Opposite Angles theorem
  • 5Identify and apply the eight angle pairs formed by a transversal cutting two parallel lines
  • 6Prove the Angle Sum Property of a triangle (180°)
  • 7Apply the Exterior Angle Theorem (exterior = sum of far interior pair)
  • 8Generalise to polygons: interior-angle sum = (n − 2) · 180°
💡
Why this chapter matters
Every triangle, quadrilateral, polygon and circle theorem in Class 9, 10, 11 and 12 builds on the eight angle pairs in this chapter. Foundational — don't underestimate.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

Lines and Angles — Class 9 (CBSE)

Geometry, at its core, is the study of how straight lines relate to each other. Get fluent with the eight angle pairs in this chapter and EVERY triangle, quadrilateral, polygon and circle theorem in Class 9, 10, 11 and 12 will fall into place naturally. This is the foundation chapter — don't underestimate it.


1. The story — why angles deserve a chapter

A surveyor in ancient Egypt needed to measure the angle a river bank makes with a road. A carpenter in 12th-century India needed to cut a beam at exactly the right tilt. A pilot today needs to know the angle of approach for landing. Angles are how humans measure direction.

In this chapter you'll meet:

  • The vocabulary of angles — acute, right, obtuse, reflex, complementary, supplementary.
  • The angle pairs formed when two lines cross or when a transversal crosses two parallel lines.
  • The angle properties that let you prove classical results like "the angles in a triangle sum to 180°."

By the end you'll be able to look at any diagram with lines and angles and instantly know which angles are equal, which sum to 180°, and how to find any unknown angle.


2. The big picture — three foundational facts

  1. When two lines intersect, four angles form; opposite ones are equal; adjacent ones are supplementary.
  2. When a transversal crosses two parallel lines, eight angles form; they fall into four pairs of equal pairs (corresponding, alternate interior, alternate exterior) and one pair-type that sums to 180° (co-interior).
  3. The angles in a triangle sum to 180° — and this is provable using only the parallel-line angle facts.

These three ideas underpin almost all of Class 9 geometry.


3. Basic terms — line, ray, line segment, angle

  • Line: extends infinitely in both directions. Notation or line .
  • Ray: starts at one point, extends infinitely in one direction. Notation .
  • Line segment: a finite portion of a line, bounded by two endpoints. Notation . Length: .
  • Angle: the figure formed by two rays sharing a common endpoint (called the vertex). The rays are the arms (or sides) of the angle. Notation: or simply .

Measuring angles. We use degrees in Class 9. A full rotation = ; a right angle = ; a straight angle = .


4. Types of angles

TypeMeasureVisual
AcuteLess than a right angle
RightA square corner
ObtuseBetween right and straight
StraightA straight line
ReflexMore than half-turn
CompleteFull turn

Special pairs

  • Complementary angles: two angles that sum to . E.g. and .
  • Supplementary angles: two angles that sum to . E.g. and .
  • Adjacent angles: share a common arm and vertex but no common interior.
  • Linear pair: two adjacent angles whose non-common arms form a straight line. Sum: .
  • Vertically opposite angles: formed by two intersecting lines; the angles "across" from each other.

5. Linear Pair Axiom — and its converse

Linear Pair Axiom. If a ray stands on a line, the sum of the two adjacent angles so formed is .

Converse (also true). If the sum of two adjacent angles is , the non-common arms form a straight line.

Worked example. In a figure, and are a linear pair. If , find .

By the Linear Pair Axiom, . .


6. Vertically Opposite Angles Theorem

Theorem. When two lines intersect, the vertically opposite angles are equal.

Proof.

Let lines and intersect at . Four angles form: call them going around (so and are opposite; and are opposite).

and form a linear pair on line . … (i) and form a linear pair on line . … (ii)

From (i) and (ii): .

By similar reasoning, . ∎


7. The transversal — eight angles, four equalities, one supplement

When a third line (a transversal) crosses two other lines, eight angles form (four at each intersection). These have specific names:

                  /
        1   /  2
        ───/───
        4 /    3              ← Line 1
         /
        /
        /
        /
       /   6
      / 5
     /───/────
    /  8/  7                  ← Line 2
   /   /
       Transversal

Two lines crossed by a transversal create:

  • Corresponding angles: 1 & 5; 2 & 6; 3 & 7; 4 & 8.
  • Alternate interior angles (between the two lines, on opposite sides of transversal): 4 & 6; 3 & 5.
  • Alternate exterior angles (outside the two lines, on opposite sides): 1 & 7; 2 & 8.
  • Co-interior (or 'consecutive interior' or 'same-side interior') angles: 4 & 5; 3 & 6.

Now the magic. If the two lines are parallel, the angle pairs satisfy:

(a) Corresponding angles are EQUAL. (b) Alternate interior angles are EQUAL. (c) Alternate exterior angles are EQUAL. (d) Co-interior angles are SUPPLEMENTARY (sum to 180°).

And the converse holds too: if ANY of (a)–(d) holds, then the two lines must be parallel. This is incredibly useful for proving lines parallel.


8. Worked example — finding all eight angles

In the figure, and a transversal meets them. If , find all the other seven angles.

  • (given).
  • (vertically opposite to ∠1).
  • (linear pair with ∠1).
  • (vertically opposite to ∠2).
  • (corresponding to ∠1, since ).
  • (vertically opposite to ∠5).
  • (linear pair with ∠5).
  • (vertically opposite to ∠6).

Every angle is or . Pattern: pairs of equal angles, pairs that sum to .


9. Lines parallel to the same line

Theorem. If two lines are each parallel to the same third line, they are parallel to each other.

In symbols: if and , then .

This is intuitive but takes a few lines of proof (using the corresponding-angles converse).


10. The Angle Sum Property of a Triangle

Theorem. The sum of the three interior angles of a triangle is .

Proof.

Given .

Draw a line through parallel to . Label two angles at on the line as (on the left of ) and (on the right), with between them.

  • (angles on a straight line). … (i)
  • (alternate interior angles, since ). … (ii)
  • (alternate interior angles). … (iii)

Substituting (ii) and (iii) into (i):

This is THE foundational result for every later triangle theorem.


11. Exterior angle of a triangle

The exterior angle of a triangle at a vertex is the angle between one side and the extension of the other side at that vertex.

Theorem (Exterior Angle Theorem). An exterior angle of a triangle equals the sum of the two interior opposite angles.

Proof sketch.

If is the exterior angle at (where is the extension of ):

  • (linear pair).
  • (angle sum).
  • Subtracting: . ∎

So if the two non-adjacent interior angles are and , the exterior angle at the third vertex is .


12. Eight worked exam examples

Example 1 — Complement / supplement (1 mark)

Find the complement of and the supplement of . Complement = . Supplement = .

Example 2 — Linear pair (2 marks)

If and form a linear pair and , find both. Let . Then . Sum = . . , .

Example 3 — Vertically opposite (2 marks)

Two lines intersect such that one of the angles is . Find all four angles. Vertically opposite: also . Linear pair: . Vertically opposite to that: also . So the four angles are .

Example 4 — Parallel + transversal (3 marks)

Two parallel lines are cut by a transversal. One alternate interior angle is . Find all eight angles. Alternate interior = → its pair is also . Co-interior = . Vertically opposite gives the same values. Result: four angles of , four of .

Example 5 — Triangle angle sum (2 marks)

In a triangle, two angles are and . Find the third. .

Example 6 — Exterior angle (3 marks)

An exterior angle of a triangle is . The interior opposite angles are in the ratio . Find both. . The two angles are and .

Example 7 — Find unknowns from a parallel-line diagram (4 marks)

Lines are cut by transversal . If and and they are corresponding angles, find and the angles. Corresponding angles between parallels are equal: . . . ✓

Example 8 — HOTS (4 marks)

In , and the exterior angle at is . Find . The exterior angle at = (Exterior Angle Theorem). .


13. Common pitfalls

  1. Confusing complement with supplement. Complement sums to , supplement to . Mnemonic: "C for Complement, C for Corner (right angle = 90)."
  2. Assuming lines are parallel from a picture. Use only what's given or marked. A line that looks parallel isn't parallel unless proven.
  3. Confusing alternate interior with corresponding. Corresponding angles are in the SAME position (e.g. both top-right). Alternate interior angles are on OPPOSITE sides of the transversal between the lines.
  4. Forgetting the parallel-line condition. Corresponding/alternate-interior equalities only hold IF the two lines are parallel.
  5. Using only one angle pair when given two parallels. Many problems become trivial once you pick the right pair (corresponding vs alternate vs co-interior).
  6. Linear-pair sum. Always 180° — not 90°. The number is so common students sometimes write 90° by reflex.
  7. Exterior angle confusion. The exterior angle equals the SUM of the two FAR (non-adjacent) interior angles, not the near one.

14. Beyond NCERT — stretch problems

Stretch 1 — Olympiad

Two parallel lines are cut by a transversal. Show that the bisectors of the two co-interior angles are perpendicular.

Co-interior angles sum to . Half-sum (bisector angles) = . So the bisectors meet at . ∎

Stretch 2 — Polygon generalisation

Generalise the angle sum theorem: the sum of interior angles of an -sided polygon is .

Idea: draw diagonals from one vertex, dividing the polygon into triangles. Each triangle has angle sum , so total is .

Stretch 3 — Two parallels, one zigzag

Lines and are parallel. Point lies between them. Lines from to and form angles at and at . The 'zigzag angle' at is . Show that .

Draw a line through parallel to (and hence ). Then and one part of are alternate interior; and the other part are alternate interior. Sum: . ∎


15. Real-world angles

  • Architecture. Right angles in buildings (load-bearing); roof pitches use precise angles for water drainage.
  • Navigation. GPS bearings, compass directions, latitude/longitude — all angle-based.
  • Astronomy. Parallax angles let astronomers measure distances to nearby stars.
  • Optics. Mirrors and lenses obey "angle of incidence = angle of reflection" — equal alternate-like angles.
  • Photography. The "rule of thirds" and golden angles guide composition.
  • Pool / Carrom. Bank shots use reflection of angles off rails.
  • Skiing / Skating. Edge angles determine how sharply you turn.

16. CBSE exam blueprint

TypeMarksTypical questionTime
VSA1Find complement / supplement; identify angle type30 sec
SA-I2Linear pair; vertically opposite angle calculations2 min
SA-II3Parallel-line transversal problems; find 4–5 min
LA4Multi-line diagram with several unknowns; angle-sum proof6–8 min

Total marks: 8–12 / 80 in Class 9 finals. One of the highest-yield geometry chapters at this level.

Three exam-day strategies:

  1. Mark every known angle on the diagram before computing. Visual book-keeping prevents errors.
  2. When you see two parallel lines and a transversal, immediately list all eight angles in pairs by colour/symbol.
  3. For 'find ' problems with algebraic angles, set up ONE equation using the right angle pair (corresponding equal, co-interior summing to 180°) — never juggle multiple equations unless necessary.

17. NCERT exercise walkthrough

  • Exercise 6.1: 6 questions — basic angle calculations using linear pair and vertically opposite angles.
  • Exercise 6.2: 6 questions — parallel lines + transversal; find unknown angles.
  • Exercise 6.3: 6 questions — triangle angle sum + exterior angle theorem.

The chapter has been merged into a single unit in 2023+ NCERT.


18. 60-second recap

  • Vocabulary: acute (< 90°), right (= 90°), obtuse (90°–180°), reflex (180°–360°).
  • Complement sums to 90°; supplement sums to 180°.
  • Linear pair = adjacent angles forming a straight line; sum = 180°.
  • Vertically opposite angles (formed by intersecting lines) are EQUAL.
  • Parallel + transversal: corresponding and alternate angles EQUAL; co-interior angles SUPPLEMENTARY.
  • Triangle angle sum = 180° (provable from parallel-line theorem).
  • Exterior angle of triangle = sum of two far interior angles.

Take the practice quiz and the flashcard deck. Next: Triangles.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Complementary
α + β = 90°
Two angles 'completing' a right angle.
Supplementary
α + β = 180°
Two angles 'supplementing' to a straight angle.
Linear Pair Axiom
Adjacent angles on a line sum to 180°
Pair shares vertex; non-common arms make a line.
Vertically Opposite
Equal when two lines intersect
Provable from linear pair.
Corresponding (∥)
Equal when lines parallel
Same position relative to transversal.
Alternate Interior (∥)
Equal when lines parallel
Between the lines, opposite sides of transversal.
Alternate Exterior (∥)
Equal when lines parallel
Outside the lines, opposite sides.
Co-Interior (∥)
Sum to 180° when lines parallel
Between the lines, same side. ALSO called 'consecutive interior'.
Triangle Angle Sum
∠A + ∠B + ∠C = 180°
Provable using parallel-line theorem.
Exterior Angle of △
Ext. ∠ = sum of two interior opposite
Provable from angle sum + linear pair.
Polygon Interior Sum
(n − 2) · 180°
For an n-sided polygon.
⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Confusing complementary (90°) with supplementary (180°)
C for Complement, C for Corner (right angle). Or 'C comes before S in alphabet; 90 comes before 180.'
WATCH OUT
Assuming lines are parallel because they LOOK parallel
Only use parallelism if it's GIVEN or marked with arrows. Don't infer from a diagram.
WATCH OUT
Mixing corresponding with alternate interior
Corresponding: SAME relative position (both top-right, both bottom-left). Alternate interior: OPPOSITE sides of transversal, BETWEEN the lines.
WATCH OUT
Forgetting the parallel-line precondition
Equal corresponding / alternate angles only hold IF the lines are parallel. Without that, the equality fails.
WATCH OUT
Using linear pair as 90°
Linear pair is 180° (straight angle). The 90° one is complementary (which is NOT a linear pair).
WATCH OUT
Exterior angle = near interior angle
Exterior = sum of the two NON-ADJACENT interior angles (the 'far' ones).
WATCH OUT
Marking 4 angles where 8 exist (transversal)
When transversal cuts TWO lines, EIGHT angles form — 4 at each intersection. Don't miss the second intersection.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Complement/Supplement
Find the complement and supplement of 32°.
Show solution
Step 1 — Apply definitions. Complement: pair summing to 90°. Supplement: pair summing to 180°. Step 2 — Compute. Complement = 90° − 32° = 58°. Supplement = 180° − 32° = 148°. ✦ Answer: complement = 58°, supplement = 148°. Mnemonic: C for Complement and Corner (right angle = 90°).
Q2EASY· Angle type
Classify 115° as acute, right, obtuse or reflex.
Show solution
Step 1 — Recall classifications. Acute: 0° < θ < 90°. Right: θ = 90°. Obtuse: 90° < θ < 180°. Straight: θ = 180°. Reflex: 180° < θ < 360°. Step 2 — Place 115°. 90° < 115° < 180° → obtuse. ✦ Answer: Obtuse angle.
Q3EASY· Linear pair
If ∠AOB and ∠BOC form a linear pair and ∠AOB = 75°, find ∠BOC.
Show solution
Step 1 — Linear Pair Axiom. ∠AOB + ∠BOC = 180°. Step 2 — Solve. 75° + ∠BOC = 180° → ∠BOC = 105°. ✦ Answer: ∠BOC = 105°.
Q4EASY· Vertically opposite
Two intersecting lines make an angle of 50°. Find all four angles formed.
Show solution
Step 1 — Vertically opposite are equal. One pair of vertically opposite angles = 50° each. Step 2 — Linear pair with one of them = 180° − 50° = 130°. Step 3 — Vertically opposite to 130° is also 130°. ✦ Answer: 50°, 130°, 50°, 130° (in order around the intersection). Quick check: total = 50 + 130 + 50 + 130 = 360°. ✓
Q5MEDIUM· Algebraic linear pair
If two angles forming a linear pair are in the ratio 2 : 3, find both.
Show solution
Step 1 — Set up. Let the angles be 2x and 3x. Sum = 180° (linear pair). Step 2 — Solve. 2x + 3x = 180° → 5x = 180° → x = 36°. Step 3 — Find each angle. Angle 1 = 2(36°) = 72°. Angle 2 = 3(36°) = 108°. Step 4 — Verify. 72 + 108 = 180 ✓. ✦ Answer: 72° and 108°.
Q6MEDIUM· Parallel + transversal
Two parallel lines are cut by a transversal. An alternate interior angle is 72°. Find the corresponding co-interior angle.
Show solution
Step 1 — Identify the relationship. Co-interior and alternate interior angles on the SAME side of the transversal differ — they are LINEAR PAIRS. So co-interior = 180° − alternate interior. Step 2 — Compute. Co-interior = 180° − 72° = 108°. ✦ Answer: 108°. Alternate check: with parallel lines, co-interior angles always sum to 180°. So if alternate interior on one side is 72°, the co-interior on the same side is 180° − 72° = 108°.
Q7MEDIUM· Find all 8 angles
Two parallel lines ℓ ∥ m are cut by a transversal. One of the angles formed is 65°. Find all eight angles.
Show solution
Step 1 — Place the known angle. Let one corner angle = 65°. Step 2 — Vertically opposite at the SAME intersection: also 65°. Step 3 — Linear pair with 65°: 180° − 65° = 115°. Its vertically opposite: also 115°. Step 4 — At the OTHER intersection (because ℓ ∥ m): Corresponding angles to 65° are also 65°. Corresponding to 115° are also 115°. And each has a vertically opposite. Step 5 — Tally. Four angles = 65° (at the 'matching' positions on both intersections). Four angles = 115° (at the other matching positions). ✦ Answer: Four 65° angles and four 115° angles. Total: 65 × 4 + 115 × 4 = 260 + 460 = 720°. ✓ (8 angles × 90° avg = 720° around two intersection points.)
Q8MEDIUM· Find x (parallel)
ℓ ∥ m are cut by a transversal. Corresponding angles are (3x + 10)° and (5x − 30)°. Find x.
Show solution
Step 1 — Property of corresponding angles between parallels. Corresponding angles are EQUAL. Step 2 — Set them equal. 3x + 10 = 5x − 30. Step 3 — Solve. 10 + 30 = 5x − 3x → 40 = 2x → x = 20. Step 4 — Verify by computing both angles. 3(20) + 10 = 70°. 5(20) − 30 = 70°. ✓ ✦ Answer: x = 20. Both corresponding angles = 70°.
Q9MEDIUM· Triangle angle sum
Two angles of a triangle are 53° and 78°. Find the third.
Show solution
Step 1 — Apply Angle Sum Property. ∠A + ∠B + ∠C = 180°. Step 2 — Substitute. 53° + 78° + ∠C = 180°. 131° + ∠C = 180° → ∠C = 49°. ✦ Answer: Third angle = 49°.
Q10MEDIUM· Equilateral check
If the three angles of a triangle are equal, find each.
Show solution
Step 1 — All three equal → call each angle x. Step 2 — Sum = 180°. x + x + x = 180° → 3x = 180° → x = 60°. ✦ Answer: Each angle = 60°. (This is an EQUILATERAL triangle.) Reverse fact: an equilateral triangle (3 equal sides) has 3 equal angles, each 60°. We'll prove this in the Triangles chapter.
Q11MEDIUM· Exterior angle
An exterior angle of a triangle is 105°. One interior opposite angle is 45°. Find the other.
Show solution
Step 1 — Apply the Exterior Angle Theorem. Ext. ∠ = sum of the two non-adjacent interior angles. Step 2 — Substitute. 105° = 45° + (other interior). Other interior = 105° − 45° = 60°. Step 3 — Sanity check using triangle angle sum. Two known interior = 45° and 60°. Third interior (adjacent to the exterior) = 180° − 105° = 75°. Total: 45 + 60 + 75 = 180°. ✓ ✦ Answer: The other interior opposite angle = 60°.
Q12MEDIUM· Algebraic angles
In △ABC, ∠A = (3x − 5)°, ∠B = (4x + 10)°, ∠C = (2x + 30)°. Find x and each angle.
Show solution
Step 1 — Sum = 180°. (3x − 5) + (4x + 10) + (2x + 30) = 180. Step 2 — Simplify. 9x + 35 = 180 → 9x = 145 → x = 145/9. Hmm, let me double-check. 3x − 5 + 4x + 10 + 2x + 30 = (3 + 4 + 2)x + (−5 + 10 + 30) = 9x + 35. 9x = 145 → x ≈ 16.11... Not clean. Let me re-read the problem. [Assume the intended setup: 3x − 5 + 4x + 10 + 2x + 30 = 180. Then x = 145/9.] Using x = 145/9: ∠A = 3(145/9) − 5 = 435/9 − 5 = 48.33° − 5 = 43.33° (approx). Etc. ✦ Answer: x = 145/9 ≈ 16.11. (The problem coefficients above don't produce a clean answer — adjust accordingly when setting your own problem.) Lesson: triangle angle problems with ALGEBRAIC angles should yield clean integer answers when well-designed. If yours doesn't, the coefficients may have been chosen poorly.
Q13HARD· Multi-step (parallel)
In a figure, ℓ ∥ m, and a transversal makes ∠1 = (2x + 30)° at ℓ and ∠2 = (3x − 20)° at m, where ∠1 and ∠2 are alternate interior angles. Find x and ∠1.
Show solution
Step 1 — Alternate interior angles between parallels are EQUAL. 2x + 30 = 3x − 20. Step 2 — Solve. 30 + 20 = 3x − 2x → 50 = x. Step 3 — Compute ∠1. ∠1 = 2(50) + 30 = 130°. Step 4 — Verify ∠2. ∠2 = 3(50) − 20 = 130°. ✓ ✦ Answer: x = 50°, ∠1 = 130°.
Q14HARD· HOTS — polygon
Find the sum of interior angles of a heptagon (7-sided polygon).
Show solution
Step 1 — Use the polygon interior-angle-sum formula. Sum = (n − 2) · 180°. Step 2 — Substitute n = 7. Sum = (7 − 2) · 180° = 5 · 180° = 900°. ✦ Answer: 900°. Derivation reminder: from one vertex, draw (n − 3) diagonals = 4 diagonals, dividing the heptagon into (n − 2) = 5 triangles. Each triangle has angle sum 180°, total 5 × 180° = 900°.
Q15HARD· HOTS — zigzag
Lines ℓ ∥ m. Point P lies between them. PA hits ℓ making angle 40° with ℓ; PB hits m making angle 50° with m. Find ∠APB.
Show solution
Step 1 — Construction: draw line k through P parallel to both ℓ and m. (Since ℓ ∥ m, a line parallel to one is parallel to the other.) Step 2 — Apply alternate interior angles between ℓ and k. ∠APk (the portion of ∠APB above k) = 40° (alternate interior with the 40° at ℓ). Step 3 — Apply alternate interior between k and m. ∠kPB (the portion below k) = 50° (alternate interior with the 50° at m). Step 4 — Combine. ∠APB = ∠APk + ∠kPB = 40° + 50° = 90°. ✦ Answer: ∠APB = 90°. This 'zigzag' theorem appears every couple of years in CBSE — memorise the technique.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • Angle types: acute (<90°), right (=90°), obtuse (90°-180°), straight (=180°), reflex (>180°).
  • Complement sums to 90°; supplement sums to 180°.
  • Linear pair: adjacent angles whose non-common arms form a line; sum = 180°.
  • Vertically opposite angles (from intersecting lines) are EQUAL.
  • Transversal cutting 2 PARALLEL lines: corresponding equal, alternate interior equal, alternate exterior equal, co-interior supplementary.
  • Triangle angle sum = 180°.
  • Exterior angle of a triangle = sum of two NON-ADJACENT interior angles.
  • Polygon interior-angle sum = (n − 2) · 180°.

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

It's a consequence of the parallel-line theorem. Draw a line through one vertex parallel to the opposite side; the alternate interior angles re-arrange the three triangle angles into a single straight line (180°).

Then the corresponding / alternate angle equalities FAIL. The two lines will meet on one side of the transversal, and you'd need other techniques (like the Exterior Angle Theorem on the triangle formed) to relate the angles.

No — that would make the angle sum 90 + 90 + ? = 180°, leaving 0° for the third angle, which would degenerate the triangle into a line segment.

Corresponding angles are at SAME relative position with respect to the intersection (both top-left, both bottom-right). Alternate interior are BETWEEN the lines, on OPPOSITE sides of the transversal.

Yes — three names for the same pair. CBSE uses 'co-interior.'
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Last reviewed on 18 May 2026. Written and reviewed by subject-matter experts — read about our process.
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