MathsICSE Class 10 · Tangent and Secant Properties

The Alternate Segment Theorem: Tangent–Chord Angles (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

The angle between a tangent and a chord drawn from the point of contact equals the angle that chord subtends in the alternate segment, the segment on the other side of the chord. If tangent $TA$ makes $50^\circ$ with chord $AB$, any point $C$ in the alternate segment gives $\angle ACB = 50^\circ$.

The angle between tangent and chord equals the angle the chord makes at any point C on the circle in the alternate segment.

Illustration drawn for QuestionBuddy.

Understand Alternate Segment Theorem

This is the last of the tangent properties in the CISCE syllabus, and its proof is to be taught.

The theorem. Let $X A Y$ be the tangent at $A$ and $A B$ a chord. Then

$∠ B A Y =$ the angle in the segment on the other side of $A B$ from $Y$ (for example $∠ A C B$ ), and
$∠ B A X =$ the angle in the segment on the other side of $A B$ from $X$ (for example $∠ A D B$ ).

Since $∠ B A X + ∠ B A Y = 18 0^{\circ}$ , this matches $∠ A C B + ∠ A D B = 18 0^{\circ}$ for cyclic quadrilateral $A C B D$ .

Proof (for $∠ B A Y$ acute). Draw the diameter $A P$ and join $P B$ .

1. $∠ P A Y = 9 0^{\circ}$ (tangent ⟂ radius), so $∠ B A Y = 9 0^{\circ} - ∠ P A B$ .
2. $∠ A B P = 9 0^{\circ}$ (angle in a semicircle), so in $△ A B P$ , $∠ A P B = 9 0^{\circ} - ∠ P A B$ .
3. Hence $∠ B A Y = ∠ A P B$ , and $∠ A P B = ∠ A C B$ for any $C$ in the same segment as $P$ .

For the obtuse angle $∠ B A X$ , use $∠ B A X = 18 0^{\circ} - ∠ B A Y$ and the cyclic quadrilateral $A C B D$ .

Picking the right segment. Look at the angle between the tangent and the chord. The equal angle is on the far side of the chord, at a point on the circle, with its arms going to the two ends of the chord. A quick check: the two angles must 'open' over the same chord.

Links with other theorems.

With the angle at the centre: the tangent–chord angle is half the angle the chord subtends at the centre, so $∠ B A Y = \frac{1}{2} ∠ A O B$ and $∠ O A B = 9 0^{\circ} - ∠ B A Y$ .
With equal tangents: tangents $P A$ and $P B$ give $∠ P A B = ∠ P B A =$ the angle in the alternate segment of chord $A B$ .
With isosceles triangles: if $A B = A C$ in an inscribed triangle, the tangent at $A$ is parallel to $B C$ .
It provides the equal angle in the proof of $P T^{2} = P A \times P B$ .

In the papers. 2024 Q7(i) (3 marks) ended with $∠ B C Q = ∠ B A C = 5 0^{\circ}$ ; 2024 Q3(ii) (4 marks) needed $∠ QN M = 4 2^{\circ}$ and $∠ S N M = 2 5^{\circ}$ from the tangents; 2025 Q9(iii) (4 marks) used $∠ Z Y X = ∠ A X Z = 6 6^{\circ}$ . For both 2024 questions the examiners report candidates who did not apply the theorem.

Alternate Segment Theorem: formulas to remember

Alternate segment theorem: $∠ B A Y = ∠ A C B$; $X A Y$ tangent at $A$ ; $C$ in the segment on the other side of $A B$ from $Y$ .
Other side of the chord: $∠ B A X = ∠ A D B = 18 0^{\circ} - ∠ A C B$; $D$ in the other segment.
Tangent–chord angle and centre: $∠ B A Y = \frac{1}{2} ∠ A O B$; $∠ A O B$ on the same arc.

How to solve Alternate Segment Theorem questions

Spot every tangent–chord angle, then jump across the chord to the alternate segment.

1Find the point of contact and the chords that start there.
2For each tangent–chord angle, identify which side of the chord it is on.
3Copy the angle to a point on the circle on the other side of that chord, with arms to the chord's ends.
4Continue with the triangle angle sum, cyclic quadrilaterals, isosceles triangles or the angle at the centre.
5State '(alternate segment theorem)' or '(angles in alternate segment)' as the reason.

Alternate Segment Theorem 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.

Tangent–chord angle on a circular track

Suppose a straight road touches a circular running track at gate $A$ . A straight path runs from $A$ across the track to a water point $B$ on the far edge, making $5 5^{\circ}$ with the road.

A coach stands at $C$ on the edge of the track, on the other side of the path from that $5 5^{\circ}$ angle. At what angle does the coach see $A$ and $B$ ?

Try it first, then show the solutionSolution

The road is a tangent at $A$ and $A B$ is a chord.
$∠ A C B = 5 5^{\circ}$ (alternate segment theorem).

Answer: $5 5^{\circ}$

Memory hook: Road–path angle jumps across the path.

Tangent parallel to the base

Imagine a triangular flower bed $A B C$ with its corners on a circular fence and $A B = A C$ . A straight hedge touches the fence at $A$ .

Will the hedge run parallel to the side $B C$ ?

Try it first, then show the solutionSolution

The angle between the hedge and $A B$ equals $∠ A C B$ (alternate segment theorem).
$∠ A C B = ∠ A B C$ (since $A B = A C$ ), so the hedge makes with $A B$ an angle equal to $∠ A B C$ : these are alternate angles.

Answer: Yes, the hedge is parallel to $B C$ .

Memory hook: Isosceles apex: tangent ∥ base.

Tricks to remember Alternate Segment Theorem

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

Jump the chord
The tangent–chord angle reappears on the other side of the chord, standing on that same chord.
Use it when: Finding the alternate segment quickly.
Half the centre
Tangent–chord angle = half the centre angle on that chord, just like an angle at the circumference.
Use it when: Questions that mix $O$ with a tangent.
Proof in three moves
Diameter, semicircle, complement: draw $A P$ , use $∠ A B P = 9 0^{\circ}$ , then both angles equal $9 0^{\circ} - ∠ P A B$ .
Use it when: Writing the proof under exam pressure.

Alternate Segment Theorem: solved examples

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

Example 1: Direct use · 1 mark · MCQBoard level

$T A$ is a tangent at $A$ to a circle and $A B$ is a chord with $∠ T A B = 5 8^{\circ}$ . $C$ is a point on the circle on the other side of $A B$ from $T$ . Then $∠ A C B$ is:
(a) $3 2^{\circ}$
(b) $5 8^{\circ}$
(c) $11 6^{\circ}$
(d) $12 2^{\circ}$

Try it first, then show the solutionSolution

$∠ A C B = ∠ T A B = 5 8^{\circ}$ (alternate segment theorem).

Answer: (b)

Example 2: Both tangent–chord angles at one point · 3 marks · Short answerBoard level

$△ A B C$ is inscribed in a circle, and $P C Q$ is the tangent at $C$ , with $P$ on the same side as $A$ . $∠ B C Q = 4 8^{\circ}$ and $∠ A C P = 6 4^{\circ}$ . Find $∠ B A C$ , $∠ A B C$ and $∠ A C B$ .

Try it first, then show the solutionSolution

$∠ B A C = ∠ B C Q = 4 8^{\circ}$ (alternate segment theorem, chord $B C$ ).
$∠ A B C = ∠ A C P = 6 4^{\circ}$ (alternate segment theorem, chord $A C$ ).
$∠ A C B = 18 0^{\circ} - 4 8^{\circ} - 6 4^{\circ} = 6 8^{\circ}$ (angles on the straight line $P C Q$ , or the angle sum of the triangle).

Answer: $4 8^{\circ}$ , $6 4^{\circ}$ , $6 8^{\circ}$

Example 3: Tangent, chord and centre · 3 marks · Short answerHarder board level

$O$ is the centre of a circle, $T A$ is the tangent at $A$ and $A B$ is a chord with $∠ A O B = 12 8^{\circ}$ . $T$ is on the same side of $A B$ as the minor arc, and $D$ is a point on the minor arc $A B$ . Find $∠ T A B$ , $∠ A D B$ and $∠ O A B$ .

Try it first, then show the solutionSolution

The angle at the circumference on the major arc is $\frac{1}{2} \times 12 8^{\circ} = 6 4^{\circ}$ , so $∠ T A B = 6 4^{\circ}$ (alternate segment theorem).
$∠ A D B = 18 0^{\circ} - 6 4^{\circ} = 11 6^{\circ}$ (cyclic quadrilateral with a point on the major arc).
$∠ O A B = 9 0^{\circ} - 6 4^{\circ} = 2 6^{\circ}$ (tangent ⟂ radius). Check: $\frac{18 0 ^{\circ} - 12 8 ^{\circ}}{2} = 2 6^{\circ}$ ✓

Answer: $∠ T A B = 6 4^{\circ}$ , $∠ A D B = 11 6^{\circ}$ , $∠ O A B = 2 6^{\circ}$

Example 4: Proving the theorem · 4 marks · Long answerHarder board level

(a) $T A$ is the tangent at $A$ to a circle with centre $O$ , $A B$ is a chord, and $C$ is a point in the segment on the other side of $A B$ from $T$ . Prove that $∠ T A B = ∠ A C B$ .
(b) If $∠ T A B = 3 7^{\circ}$ and $D$ is a point on the other arc $A B$ , find $∠ A C B$ and $∠ A D B$ .

Try it first, then show the solutionSolution

(a) Draw the diameter $A P$ and join $P B$ . $∠ T A P = 9 0^{\circ}$ (tangent ⟂ radius), so $∠ T A B = 9 0^{\circ} - ∠ P A B$ .
$∠ A B P = 9 0^{\circ}$ (angle in a semicircle), so $∠ A P B = 9 0^{\circ} - ∠ P A B$ (angle sum of $△ A B P$ ).
So $∠ T A B = ∠ A P B$ . $P$ and $C$ are in the same segment, so $∠ A P B = ∠ A C B$ (angles in the same segment). Hence $∠ T A B = ∠ A C B$ .
(b) $∠ A C B = 3 7^{\circ}$ ; $∠ A D B = 18 0^{\circ} - 3 7^{\circ} = 14 3^{\circ}$ (opposite angles of cyclic quadrilateral $A C B D$ ).

Answer: (b) $∠ A C B = 3 7^{\circ}$ , $∠ A D B = 14 3^{\circ}$

Example 5: Two tangents and a diameter · 4 marks · Long answerHOTS / competency

$P Q$ and $P S$ are tangents from $P$ to a circle with centre $O$ , touching it at $Q$ and $S$ . $M N$ is a diameter, with $M$ on the minor arc $QS$ . $∠ P QM = 3 5^{\circ}$ and $∠ P S M = 2 7^{\circ}$ . Find (a) $∠ O QM$ , (b) $∠ QN S$ , (c) $∠ QO S$ , (d) $∠ QM S$ , (e) $∠ QP S$ .

Try it first, then show the solutionSolution

(a) $∠ O QP = 9 0^{\circ}$ (tangent ⟂ radius), so $∠ O QM = 9 0^{\circ} - 3 5^{\circ} = 5 5^{\circ}$ .
(b) $∠ QN M = ∠ P QM = 3 5^{\circ}$ and $∠ S N M = ∠ P S M = 2 7^{\circ}$ (alternate segment theorem), so $∠ QN S = 6 2^{\circ}$ .
(c) $∠ QO S = 2 \times 6 2^{\circ} = 12 4^{\circ}$ (angle at the centre).
(d) $∠ QM S = 18 0^{\circ} - 6 2^{\circ} = 11 8^{\circ}$ (opposite angles of cyclic quadrilateral $QM S N$ ).
(e) $∠ QP S = 18 0^{\circ} - 12 4^{\circ} = 5 6^{\circ}$ (quadrilateral $O QP S$ has two right angles).

Answer: (a) $5 5^{\circ}$ (b) $6 2^{\circ}$ (c) $12 4^{\circ}$ (d) $11 8^{\circ}$ (e) $5 6^{\circ}$

Example 6: Case study: lamp posts beside a road · 4 marks · Case-basedHOTS / competency

A straight road $A X B$ touches a circular park at gate $X$ . Lamp posts $Y$ and $Z$ stand on the park's boundary with $Z Y = X Y$ , and $Y$ lies on the far side of the chord $X Z$ from $A$ . $∠ A X Z = 7 0^{\circ}$ .
(a) Find $∠ X Y Z$ .
(b) Find $∠ Y X Z$ and $∠ B X Y$ .
(c) $O$ is the centre of the park. Find $∠ X O Z$ .

Try it first, then show the solutionSolution

(a) $∠ X Y Z = ∠ A X Z = 7 0^{\circ}$ (alternate segment theorem, chord $X Z$ ).
(b) $X Y = Z Y$ , so $∠ Y X Z = ∠ Y Z X = \frac{18 0 ^{\circ} - 7 0 ^{\circ}}{2} = 5 5^{\circ}$ .
$∠ B X Y = ∠ X Z Y = 5 5^{\circ}$ (alternate segment theorem, chord $X Y$ ). Check: $7 0^{\circ} + 5 5^{\circ} + 5 5^{\circ} = 18 0^{\circ}$ along $A X B$ ✓
(c) $∠ X O Z = 2∠ X Y Z = 14 0^{\circ}$ (angle at the centre).

Answer: (a) $7 0^{\circ}$ (b) $5 5^{\circ}$ , $5 5^{\circ}$ (c) $14 0^{\circ}$

Where students lose marks in Alternate Segment Theorem

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

Mistake: Not applying the theorem at all
A tangent–chord angle is the only way to reach some angles in the figure.
Wrong working
Trying to find $∠ QN M$ with the angle sum alone.
Right working
$∠ QN M = ∠ P QM = 4 2^{\circ}$ (alternate segment theorem).
Do this: Whenever a tangent meets a chord, write down the equal angle across the chord at once.
Usually costs: Reported in 2024 Q3(ii); errors in (b) were carried into (c) and (d).
Mistake: Choosing the wrong segment
The equal angle lies on the far side of the chord from the tangent–chord angle.
Wrong working
$∠ T A B = 5 8^{\circ}$ , so the angle at $D$ on the near arc is $5 8^{\circ}$ .
Right working
On the near arc the angle is $18 0^{\circ} - 5 8^{\circ} = 12 2^{\circ}$ .
Do this: Shade the alternate segment before copying the angle.
Usually costs: The angle and every answer that depends on it.
Mistake: Reasons missing or angles named loosely
Circle-theorem marks need the theorem and three-letter angle names.
Wrong working
' $∠ Y = 6 6^{\circ}$ '
Right working
' $∠ Z Y X = ∠ A X Z = 6 6^{\circ}$ (alternate segment theorem)'
Do this: Name every angle with three letters and give its reason.
Usually costs: Reported in 2025 Q9(iii): candidates did not state reasons.

Exam tips

2024 Q7(i) (3 marks): $∠ B C Q = ∠ B A C = 5 0^{\circ}$ by the alternate segment theorem; the report lists missing reasons as a common error.
2024 Q3(ii) (4 marks): tangent angles $4 2^{\circ}$ and $2 5^{\circ}$ gave $∠ QN S = 6 7^{\circ}$ , then $∠ QO S = 13 4^{\circ}$ and $∠ QM S = 11 3^{\circ}$ . The examiners suggest proving and demonstrating the theorem in class.
2025 Q9(iii) (4 marks): $∠ Z Y X = ∠ A X Z = 6 6^{\circ}$ , combined with an isosceles triangle and the tangent–radius right angle.
The proof can be asked: learn the 'diameter through the point of contact' method.

Practice questions on Alternate Segment Theorem

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
$T A$ is a tangent at $A$ , and $C$ is a point on the circle on the other side of chord $A B$ from $T$ . If $∠ A C B = 5 3^{\circ}$ , then $∠ T A B$ is:
(a) $3 7^{\circ}$
(b) $5 3^{\circ}$
(c) $10 6^{\circ}$
(d) $12 7^{\circ}$
Hint
Alternate segment theorem.
Answer
(b)
Q2 · 3 marks · Short answerBoard level
$△ A B C$ is inscribed in a circle and $T_{1} A T_{2}$ is the tangent at $A$ , with $T_{1}$ on the same side as $B$ . $∠ T_{1} A B = 7 2^{\circ}$ and $∠ A B C = 5 0^{\circ}$ . Find $∠ A C B$ , $∠ B A C$ and $∠ T_{2} A C$ .
Hint
Each tangent–chord angle equals the triangle angle at the far vertex.
Answer
$∠ A C B = 7 2^{\circ}$ (alternate segment); $∠ B A C = 18 0^{\circ} - 7 2^{\circ} - 5 0^{\circ} = 5 8^{\circ}$ ; $∠ T_{2} A C = ∠ A B C = 5 0^{\circ}$ (alternate segment).
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The angle between a tangent and a chord through the point of contact equals the angle that the chord subtends at the centre.
Reason (R): The angle between a tangent and a chord through the point of contact equals the angle in the alternate segment.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Compare with the angle at the circumference.
Answer
(b): the tangent–chord angle is half the angle at the centre.
Q4 · 3 marks · Short answerHarder board level
$△ A B C$ is inscribed in a circle with $A B = A C$ and $∠ B A C = 4 4^{\circ}$ . $T A$ is the tangent at $A$ , with $T$ on the same side of $A$ as $B$ . Find $∠ T A B$ , and prove that the tangent at $A$ is parallel to $B C$ .
Hint
Base angles of the isosceles triangle are $6 8^{\circ}$ .
Answer
$∠ T A B = ∠ A C B = 6 8^{\circ}$ (alternate segment). It equals $∠ A B C = 6 8^{\circ}$ , an alternate angle with transversal $A B$ , so the tangent is parallel to $B C$ .
Q5 · 3 marks · Short answerHarder board level
$P T$ is a tangent at $T$ and $P A B$ is a secant, with $A$ between $P$ and $B$ . $∠ P T A = 3 4^{\circ}$ and $∠ T P A = 4 0^{\circ}$ . Find $∠ T B A$ , $∠ P A T$ , $∠ T A B$ and $∠ B T A$ .
Hint
$∠ T B A$ is in the alternate segment of chord $T A$ .
Answer
$∠ T B A = 3 4^{\circ}$ (alternate segment); $∠ P A T = 18 0^{\circ} - 3 4^{\circ} - 4 0^{\circ} = 10 6^{\circ}$ ; $∠ T A B = 7 4^{\circ}$ (straight line); $∠ B T A = 18 0^{\circ} - 3 4^{\circ} - 7 4^{\circ} = 7 2^{\circ}$ .
Q6 · 4 marks · Long answerHarder board level
$A B C D$ is a cyclic quadrilateral and $T D S$ is the tangent at $D$ , with $T$ on the side of $A$ . $∠ T D A = 3 8^{\circ}$ and $∠ S D C = 4 6^{\circ}$ . Find $∠ A C D$ , $∠ D A C$ , $∠ A D C$ and $∠ A B C$ .
Hint
Use chords $D A$ and $D C$ in turn.
Answer
$∠ A C D = 3 8^{\circ}$ and $∠ D A C = 4 6^{\circ}$ (alternate segment); $∠ A D C = 18 0^{\circ} - 3 8^{\circ} - 4 6^{\circ} = 9 6^{\circ}$ ; $∠ A B C = 18 0^{\circ} - 9 6^{\circ} = 8 4^{\circ}$ (cyclic quadrilateral).
Q7 · 4 marks · Long answerHOTS / competency
Two circles touch externally at $P$ . A straight line through $P$ meets the first circle again at $A$ and the second at $B$ . Prove that the tangents at $A$ and $B$ are parallel.
Hint
Draw the common tangent $X P Y$ at $P$ and compare tangent–chord angles on chords $P A$ and $P B$ .
Answer
Let the tangents at $A$ and $B$ be $A L$ and $B M$ , on opposite sides of $A B$ . For the first circle, the tangents at $A$ and $P$ make equal angles with chord $A P$ (both equal the angle in the alternate segment), so $∠ L A P = ∠ X P A$ . For the second, $∠ M B P = ∠ Y P B$ . $∠ X P A = ∠ Y P B$ (vertically opposite), so $∠ L A P = ∠ M B P$ : alternate angles, hence $A L ∥ B M$ .

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.State the alternate segment theorem.Show answer
The angle between a tangent and a chord through the point of contact equals the angle in the alternate segment.
2.Tangent–chord angle $4 0^{\circ}$ . Angle in the other (near) segment?Show answer
$14 0^{\circ}$
3.Tangent–chord angle compared with the angle at the centre on that chord?Show answer
Half of it.
4.Which construction is used in the proof?Show answer
The diameter through the point of contact.
5.Tangent at the apex of an isosceles inscribed triangle is…?Show answer
Parallel to the base.

Alternate Segment Theorem: frequently asked questions

What is the alternate segment theorem?

If a tangent and a chord meet at the point of contact, the angle between them equals the angle the chord makes at any point of the circle in the alternate segment, the part of the circle on the other side of the chord.

How do I find the alternate segment?

Look at which side of the chord the tangent–chord angle lies. The alternate segment is the part of the circle on the opposite side of that chord.

Is the alternate segment theorem proof in the ICSE syllabus?

Yes. The CISCE syllabus says proofs of the tangent and secant theorems are to be taught, and the 2024 examiners' report asks teachers to prove and demonstrate this one.