MathsCBSE Class 9 · Lines and Angles

Parallel Lines, Transversals and Triangle Angle Sum: Class 9

Updated 17 October 2026 · Fact-checked

When a transversal cuts two parallel lines, corresponding angles are equal, alternate interior angles are equal and co-interior angles add up to $180^\circ$; the converses give tests for parallel lines. Using a line through a vertex parallel to the opposite side, the three angles of a triangle add up to $180^\circ$, and an exterior angle equals the sum of the two interior opposite angles.

Notice only two angle sizes appear: orange angles are all α, blue ones are 180° − α, so corresponding and alternate angles match and co-interior add to 180°.

Illustration drawn for QuestionBuddy.

Understand Parallel Lines, Transversals and the Angle Sum of a Triangle

A transversal is a line that cuts two (or more) lines at distinct points. It makes eight angles. With the transversal $t$ cutting lines $l$ and $m$ :

Corresponding angles lie in matching positions at the two crossings (for example, both above the line and to the right of the transversal).
Alternate interior angles lie between the lines, on opposite sides of the transversal.
Co-interior (same-side interior) angles lie between the lines, on the same side of the transversal.

**If $l ∥ m$ **, then corresponding angles are equal, alternate interior angles are equal and co-interior angles are supplementary ( $18 0^{\circ}$ ). These are the only facts needed for angle chasing between parallel lines. Using vertically opposite angles and linear pairs, one angle between parallel lines determines all eight: four equal to $α$ and four equal to $18 0^{\circ} - α$ .

Converses (tests for parallel lines). If a transversal makes a pair of corresponding angles equal, or a pair of alternate interior angles equal, or a pair of co-interior angles supplementary, then the two lines are parallel. The converse of the corresponding angles axiom is accepted as an axiom. The converses for alternate and co-interior angles then follow from it, using vertically opposite angles or a linear pair. They can also be shown by contradiction, but not through the triangle angle sum. Lines parallel to the same line are parallel to each other.

Angle sum property of a triangle. In $△ A B C$ draw through $A$ a line $P Q ∥ B C$ . Then $∠ P A B = ∠ A B C$ and $∠ Q A C = ∠ A C B$ (alternate angles), and $∠ P A B + ∠ B A C + ∠ C A Q = 18 0^{\circ}$ (a straight line). Hence $∠ A + ∠ B + ∠ C = 18 0^{\circ}$ .

Exterior angle property. If the side $B C$ of $△ A B C$ is produced to $D$ , the exterior angle $∠ A C D$ equals the sum of the two interior opposite angles: $∠ A C D = ∠ A + ∠ B$ , since both $∠ A C D$ and $∠ A + ∠ B$ equal $18 0^{\circ} - ∠ C$ .

Useful results. If the bisectors of $∠ B$ and $∠ C$ of a triangle meet at $I$ , then $∠ B I C = 9 0^{\circ} + \frac{1}{2} ∠ A$ . The angle between the altitude and the bisector from $A$ is $\frac{1}{2} ∣∠ B - ∠ C ∣$ . The bisectors of two co-interior angles between parallel lines meet at $9 0^{\circ}$ .

Writing a solution. Name the angles with three letters, quote the property for every step (for example "alternate angles, $l ∥ m$ "), and keep the unknowns in one equation at a time. For algebra with parallel lines, set up an equation from equal or supplementary angles first and check the resulting angles against the figure.

Parallel Lines, Transversals and the Angle Sum of a Triangle: formulas to remember

Corresponding angles: $l ∥ m \Rightarrow ∠1 = ∠5$; Same position at the two crossings.
Alternate interior angles: $l ∥ m \Rightarrow ∠3 = ∠5$; Between the lines, opposite sides.
Co-interior angles: $l ∥ m \Rightarrow ∠3 + ∠6 = 18 0^{\circ}$; Between the lines, same side.
Angle sum of a triangle: $∠ A + ∠ B + ∠ C = 18 0^{\circ}$; Proved with a parallel through a vertex.
Exterior angle: $exterior angle = sum of interior opposite angles$; Produce one side.
Angle at the incentre: $∠ B I C = 9 0^{\circ} + \frac{1}{2} ∠ A$; $I$ is where the bisectors of $B$ and $C$ meet.

How to solve Parallel Lines, Transversals and the Angle Sum of a Triangle questions

Identify the pair of angles, quote the property, and make one equation.

1Mark the transversal and decide whether the given angles are corresponding, alternate or co-interior.
2If the lines are known to be parallel, use equal or supplementary angles; if the question asks to prove parallel lines, use the converse.
3For a triangle, write the angle sum, and for a produced side, write the exterior angle equation.
4Use algebra when expressions are given: form one equation, solve, and substitute to get every angle.
5Check the result: each linear pair adds to $18 0^{\circ}$ and each triangle adds to $18 0^{\circ}$ .
6Write the reason for every step in brackets.

Parallel Lines, Transversals and the Angle Sum of a Triangle in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Corresponding and co-interior angles

Suppose a straight road crosses two parallel canal banks, making an angle of 58 degrees with the first bank on its upper-left side.

Find the angle it makes with the second bank at the matching position, and the angle on the same side between the banks at the second crossing.

Try it first, then show the solutionSolution

Matching position gives a corresponding angle: $5 8^{\circ}$ (parallel banks).
The angle on the same side between the banks (co-interior with the first) is $18 0^{\circ} - 5 8^{\circ} = 12 2^{\circ}$ .

Answer: The corresponding angle is $5 8^{\circ}$ and the co-interior angle is $12 2^{\circ}$ .

Memory hook: F-shape equal, C-shape add to 180, Z-shape equal.

Tricks to remember Parallel Lines, Transversals and the Angle Sum of a Triangle

Short memory aids. Say each one aloud once, then use it on a question the same day.

F, Z, C shapes
Corresponding angles sit in an F shape (equal), alternate angles in a Z shape (equal), co-interior angles in a C shape (add to 180).
Use it when: Spotting the pair quickly.
Exterior is the sum of far angles
An exterior angle equals the two interior angles that are not next to it.
Use it when: Triangle questions with a produced side.
Parallel through the vertex
To prove an angle sum or chase angles, draw a line through a point parallel to a given line.
Use it when: Proofs and zigzag figures.

Parallel Lines, Transversals and the Angle Sum of a Triangle: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Algebra with parallel lines · 3 marks · Short answerBoard level

A transversal cuts two lines $l$ and $m$ . (i) If $l ∥ m$ and a pair of corresponding angles are $(3 x + 15)^{\circ}$ and $(5 x - 25)^{\circ}$ , find $x$ . (ii) If a pair of co-interior angles are $(2 y + 10)^{\circ}$ and $(3 y - 20)^{\circ}$ , find $y$ and the angles.

Try it first, then show the solutionSolution

(i) Corresponding angles are equal: $3 x + 15 = 5 x - 25$ . (½)
$2 x = 40$ , so $x = 20$ and each angle is $7 5^{\circ}$ . (1)
(ii) Co-interior angles add to $18 0^{\circ}$ : $2 y + 10 + 3 y - 20 = 180$ . (½)
$5 y = 190$ , so $y = 38$ ; the angles are $8 6^{\circ}$ and $9 4^{\circ}$ . (1)

Answer: $x = 20$ (angles $7 5^{\circ}$ ); $y = 38$ (angles $8 6^{\circ}$ and $9 4^{\circ}$ ).

Example 2: Triangle in a ratio · 3 marks · Short answerBoard level

The angles of a triangle are in the ratio $2 : 3 : 4$ . Find them, and find the exterior angle at the vertex of the largest angle.

Try it first, then show the solutionSolution

Angles $2 k, 3 k, 4 k$ with $9 k = 18 0^{\circ}$ , so $k = 2 0^{\circ}$ . (1)
The angles are $4 0^{\circ}, 6 0^{\circ}, 8 0^{\circ}$ . (1)
Exterior angle at the $8 0^{\circ}$ vertex $= 18 0^{\circ} - 8 0^{\circ} = 10 0^{\circ} = 4 0^{\circ} + 6 0^{\circ}$ . (1)

Answer: The angles are $4 0^{\circ}, 6 0^{\circ}, 8 0^{\circ}$ ; the exterior angle at the largest is $10 0^{\circ}$ .

Example 3: Angle at the intersection of bisectors · 5 marks · Long answerHarder board level

In $△ A B C$ the bisectors of $∠ B$ and $∠ C$ meet at $I$ . Prove that $∠ B I C = 9 0^{\circ} + \frac{1}{2} ∠ A$ , and find $∠ B I C$ when $∠ A = 5 0^{\circ}$ .

Try it first, then show the solutionSolution

In $△ A B C$ , $∠ B + ∠ C = 18 0^{\circ} - ∠ A$ . (1)
$∠ I B C = \frac{1}{2} ∠ B$ and $∠ I C B = \frac{1}{2} ∠ C$ , so $∠ I B C + ∠ I C B = \frac{1}{2} (18 0^{\circ} - ∠ A) = 9 0^{\circ} - \frac{1}{2} ∠ A$ . (1)
In $△ B I C$ : $∠ B I C = 18 0^{\circ} - (9 0^{\circ} - \frac{1}{2} ∠ A) = 9 0^{\circ} + \frac{1}{2} ∠ A$ . (1½)
For $∠ A = 5 0^{\circ}$ : $∠ B I C = 9 0^{\circ} + 2 5^{\circ} = 11 5^{\circ}$ . (½)
Check: $∠ B + ∠ C = 13 0^{\circ}$ , halves sum to $6 5^{\circ}$ , and $18 0^{\circ} - 6 5^{\circ} = 11 5^{\circ}$ . (1)

Answer: $∠ B I C = 9 0^{\circ} + \frac{1}{2} ∠ A$ ; for $∠ A = 5 0^{\circ}$ it is $11 5^{\circ}$ .

Example 4: Exterior angle in a ratio · 3 marks · Short answerHarder board level

Side $B C$ of $△ A B C$ is produced to $D$ and $∠ A C D = 12 0^{\circ}$ . If $∠ A : ∠ B = 3 : 5$ , find the three angles of the triangle.

Try it first, then show the solutionSolution

Exterior angle $=$ sum of interior opposite angles: $∠ A + ∠ B = 12 0^{\circ}$ . (1)
$∠ A = \frac{3}{8} \times 12 0^{\circ} = 4 5^{\circ}$ and $∠ B = \frac{5}{8} \times 12 0^{\circ} = 7 5^{\circ}$ . (1)
$∠ A C B = 18 0^{\circ} - 12 0^{\circ} = 6 0^{\circ}$ (linear pair); check $45 + 75 + 60 = 180$ . (1)

Answer: $∠ A = 4 5^{\circ}$ , $∠ B = 7 5^{\circ}$ , $∠ C = 6 0^{\circ}$ .

Example 5: Bisectors with a transversal · 4 marks · Case-basedHOTS / competency

Case: a transversal $E F$ cuts the parallel lines $A B$ and $C D$ at $G$ and $H$ ( $E$ above $A B$ , $F$ below $C D$ ). $GP$ bisects $∠ E GB$ and $H Q$ bisects $∠ G H D$ , where $P$ and $Q$ lie to the right of the transversal. (i) If $∠ E GB = 7 0^{\circ}$ , find $∠ G H D$ . (ii) Find the angles into which each bisector divides the angle. (iii) Prove that $GP ∥ H Q$ . (iv) Find $∠ B G H$ .

Try it first, then show the solutionSolution

(i) $∠ E GB$ and $∠ G H D$ are corresponding angles with $A B ∥ C D$ , so $∠ G H D = 7 0^{\circ}$ . (1)
(ii) Each bisector divides its angle into $3 5^{\circ}$ and $3 5^{\circ}$ . (½)
(iii) $∠ E GP = 3 5^{\circ}$ and $∠ G H Q = 3 5^{\circ}$ are corresponding angles for lines $GP$ and $H Q$ cut by the transversal $E F$ ; they are equal, so $GP ∥ H Q$ (converse of the corresponding angles property). (1½)
(iv) $∠ B G H = 18 0^{\circ} - 7 0^{\circ} = 11 0^{\circ}$ (linear pair); check that $∠ B G H + ∠ G H D = 11 0^{\circ} + 7 0^{\circ} = 18 0^{\circ}$ as co-interior angles. (1)

Answer: (i) $7 0^{\circ}$ ; (ii) $3 5^{\circ}$ each; (iii) proved; (iv) $11 0^{\circ}$ .

Where students lose marks in Parallel Lines, Transversals and the Angle Sum of a Triangle

The slips examiners see most often, and what to write instead.

Mistake: Equating co-interior angles
Confusing co-interior with alternate angles.
Wrong working
$(2 y + 10) = (3 y - 20)$ for co-interior angles.
Right working
$(2 y + 10) + (3 y - 20) = 180$ .
Do this: Check the shape: C-shape adds to 180.
Usually costs: The whole answer.
Mistake: Using the wrong exterior angle sum
Adding the adjacent interior angle.
Wrong working
Exterior angle at $C$ $= ∠ A + ∠ C$ .
Right working
Exterior angle at $C$ $= ∠ A + ∠ B$ .
Do this: Use only the two interior opposite angles.
Usually costs: 2 marks.
Mistake: No parallel condition stated
Using equal corresponding angles without saying the lines are parallel.
Wrong working
$∠1 = ∠5$ (corresponding angles).
Right working
$∠1 = ∠5$ (corresponding angles, $l ∥ m$ ).
Do this: Include the condition in the reason.
Usually costs: ½ mark per step.

Exam tips

Name angles with three letters in proofs to avoid ambiguity.
Write " $l ∥ m$ " in the reason whenever a parallel-lines property is used.
In a triangle question with algebra, form the angle-sum equation first.
If asked to prove lines parallel, use a converse (equal corresponding or alternate angles, or supplementary co-interior angles).

Practice questions on Parallel Lines, Transversals and the Angle Sum of a Triangle

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
In $△ A B C$ , $∠ A = 7 0^{\circ}$ and $∠ B = 5 0^{\circ}$ . The exterior angle at $C$ (formed by producing $B C$ ) is
(a) $6 0^{\circ}$
(b) $12 0^{\circ}$
(c) $13 0^{\circ}$
(d) $11 0^{\circ}$
Hint
The exterior angle equals the sum of the interior opposite angles.
Answer
(b) $12 0^{\circ}$ , since $7 0^{\circ} + 5 0^{\circ} = 12 0^{\circ}$ .
Q2 · 3 marks · Short answerBoard level
Two lines are cut by a transversal and a pair of alternate interior angles are $(4 x + 10)^{\circ}$ and $(6 x - 30)^{\circ}$ . If the lines are parallel, find $x$ and the angles. What can you say about the transversal?
Hint
Alternate interior angles are equal.
Answer
$4 x + 10 = 6 x - 30$ gives $x = 20$ , and each angle is $9 0^{\circ}$ . Each of the angles is a right angle, so the transversal is perpendicular to both parallel lines.
Q3 · 3 marks · Short answerHarder board level
The bisectors of two co-interior angles formed by a transversal across two parallel lines meet at $P$ . Find $∠ P$ and give the reasons.
Hint
Co-interior angles add to $18 0^{\circ}$ ; look at the triangle formed.
Answer
Let the co-interior angles be $2 a$ and $2 b$ with $2 a + 2 b = 18 0^{\circ}$ , so $a + b = 9 0^{\circ}$ . In the triangle formed by the transversal and the two bisectors, the base angles are $a$ and $b$ , so $∠ P = 18 0^{\circ} - (a + b) = 9 0^{\circ}$ .
Q4 · 5 marks · Long answerHarder board level
In $△ A B C$ , $∠ B = 7 0^{\circ}$ and $∠ C = 3 0^{\circ}$ . $A D$ is the altitude to $B C$ and $A E$ bisects $∠ A$ ( $D$ and $E$ on $B C$ ). Find $∠ B A E$ , $∠ B A D$ and $∠ D A E$ , and show that $∠ D A E = \frac{1}{2} (∠ B - ∠ C)$ .
Hint
Find $∠ A$ first; in $△ A B D$ , $∠ A D B = 9 0^{\circ}$ .
Answer
$∠ A = 18 0^{\circ} - 10 0^{\circ} = 8 0^{\circ}$ , so $∠ B A E = 4 0^{\circ}$ . In $△ A B D$ : $∠ B A D = 9 0^{\circ} - 7 0^{\circ} = 2 0^{\circ}$ . $∠ D A E = 4 0^{\circ} - 2 0^{\circ} = 2 0^{\circ}$ . In general $∠ D A E = \frac{1}{2} (18 0^{\circ} - B - C) - (9 0^{\circ} - B) = \frac{1}{2} (B - C)$ ; here $\frac{1}{2} (70 - 30) = 2 0^{\circ}$ .
Q5 · 4 marks · Case-basedHOTS / competency
Case: a roof truss is an isosceles triangle $A B C$ with $A B = A C$ and apex angle $∠ A = 4 0^{\circ}$ . A horizontal beam $D E$ with $D$ on $A B$ and $E$ on $A C$ is parallel to the base $B C$ . (i) Find $∠ B$ and $∠ C$ . (ii) Find $∠ A D E$ and $∠ A E D$ . (iii) Find $∠ D E C$ . (iv) Find the exterior angle at $B$ formed by producing $C B$ , and verify it with the exterior angle property.
Hint
Base angles of an isosceles triangle are equal; $D E ∥ B C$ gives corresponding angles.
Answer
(i) $∠ B = ∠ C = \frac{18 0 ^{\circ} - 4 0 ^{\circ}}{2} = 7 0^{\circ}$ . (ii) $∠ A D E = ∠ B = 7 0^{\circ}$ and $∠ A E D = ∠ C = 7 0^{\circ}$ (corresponding angles). (iii) $∠ D E C = 18 0^{\circ} - 7 0^{\circ} = 11 0^{\circ}$ (linear pair). (iv) The exterior angle at $B$ is $18 0^{\circ} - 7 0^{\circ} = 11 0^{\circ}$ , and $∠ A + ∠ C = 4 0^{\circ} + 7 0^{\circ} = 11 0^{\circ}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Corresponding angles for parallel lines?Show answer
Equal.
2.Co-interior angles for parallel lines?Show answer
Add to $18 0^{\circ}$ .
3.Triangle angle sum?Show answer
$18 0^{\circ}$ .
4.Exterior angle equals?Show answer
The sum of the two interior opposite angles.
5. $∠ B I C$ in terms of $∠ A$ ?Show answer
$9 0^{\circ} + \frac{1}{2} ∠ A$ .

Parallel Lines, Transversals and the Angle Sum of a Triangle: frequently asked questions

Do the angle properties work if the lines are not parallel?

No. The equality or sum properties hold only when the lines are parallel. The converses tell you the lines are parallel when the angle condition holds.

How do I choose between alternate and corresponding angles?

Either gives the same answer; use whichever pair you can spot as an F or Z shape. Both rely on the lines being parallel.

Is the angle sum of a triangle always 180 degrees?

In the plane geometry of this course, yes. It depends on the parallel postulate, and fails on curved surfaces like a sphere.