MathsICSE Class 10 · Circles

Angle in a Semicircle and Angles in the Same Segment (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

An angle in a semicircle is a right angle: if $AB$ is a diameter and $C$ is any other point on the circle, then $\angle ACB = 90^\circ$. Angles in the same segment are equal: points on the same arc see a chord at the same angle. The first is proved; the second is used without proof.

Left: any point C on the circle sees diameter AB at 90°. Right: C and D on the same arc see chord AB at equal angles.

Illustration drawn for QuestionBuddy.

Understand Angle in a Semicircle and Angles in the Same Segment

These are the second and third angle properties in the CISCE syllabus for Circles.

1. Angles in the same segment are equal (without proof). A chord $A B$ cuts the circle into two segments. If $C$ and $D$ are on the same side of $A B$ , then $∠ A C B = ∠ A D B$ . Both are half of the same angle at the centre.

If $C$ and $D$ are on opposite sides of $A B$ , the angles are supplementary, not equal ( $A C B D$ is then a cyclic quadrilateral).
The angle in the major segment is acute; the angle in the minor segment is obtuse.

2. The angle in a semicircle is a right angle (proof to be taught). If $A B$ is a diameter with centre $O$ , then $∠ A O B = 18 0^{\circ}$ . For any other point $C$ on the circle, the angle at the centre theorem gives $∠ A C B = \frac{1}{2} \times 18 0^{\circ} = 9 0^{\circ}$ .

A second proof uses radii only: $O A = O C = O B$ , so $∠ O A C = ∠ O C A = x$ and $∠ O B C = ∠ O C B = y$ . In $△ A B C$ , $2 x + 2 y = 18 0^{\circ}$ , so $∠ A C B = x + y = 9 0^{\circ}$ .

What it lets you do.

Pythagoras in a circle. With $A B$ a diameter, $A C^{2} + B C^{2} = A B^{2}$ .
Complementary angles. In $△ A C B$ , $∠ C A B + ∠ C B A = 9 0^{\circ}$ .
Perpendicular to a diameter. If $C D ⊥ A B$ with $D$ on $A B$ , the similar right triangles give $C D^{2} = A D \cdot D B$ .
Spotting a diameter. If a chord subtends $9 0^{\circ}$ at a point on the circle, that chord is a diameter; the circumcentre of a right-angled triangle is the mid-point of its hypotenuse.

Intersecting chords. When chords $A C$ and $B D$ meet at $X$ inside the circle, look for pairs of equal angles in the same segment: $∠ B A C = ∠ B D C$ (chord $B C$ ) and $∠ A B D = ∠ A C D$ (chord $A D$ ). Triangles $A X B$ and $D X C$ then have the same three angles.

Angle in a Semicircle and Angles in the Same Segment: formulas to remember

Angles in the same segment: $∠ A C B = ∠ A D B$; $C$ , $D$ on the same side of chord $A B$ ; without proof.
Angle in a semicircle: $A B a diameter \Rightarrow ∠ A C B = 9 0^{\circ}$; Proof to be taught.
Pythagoras with a diameter: $A C^{2} + B C^{2} = A B^{2}$; $C$ on the circle, $A B$ a diameter.
Perpendicular to the diameter: $C D^{2} = A D \cdot D B$; $C D ⊥ A B$ ; from similar triangles.

How to solve Angle in a Semicircle and Angles in the Same Segment questions

Look for a diameter first, then for two angles standing on the same chord on the same side.

1Mark any diameter: every angle it subtends on the circle is $9 0^{\circ}$ .
2In a right triangle with the diameter as hypotenuse, the other two angles add to $9 0^{\circ}$ ; use Pythagoras for lengths.
3For angles in the same segment, name the chord both angles stand on and check the points are on the same side of it.
4Where chords cross, write the two pairs of equal angles and use vertically opposite angles.
5Give the reason each time: '(angle in a semicircle)' or '(angles in the same segment)'.

Angle in a Semicircle and Angles in the Same Segment 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.

Finding the centre with a set square

Suppose you want to find the centre of a round plate using only a set square and a pencil.

Explain how the right-angle corner helps you find the centre.

Try it first, then show the solutionSolution

Put the right-angle corner on the rim; the two edges cut the rim at $A$ and $B$ . Since $∠ A C B = 9 0^{\circ}$ , $A B$ is a diameter. Draw it.
Repeat at another place on the rim to get a second diameter.
The two diameters cross at the centre.

Answer: Two diameters drawn this way meet at the centre.

Memory hook: A right angle on the rim sits on a diameter.

Pythagoras in a semicircle

Imagine a semicircular window whose straight edge is 1.3 m long. A straight wooden bar runs from one end of that edge to a point $P$ on the curve and is 1.2 m long.

How long is a bar from $P$ to the other end of the straight edge?

Try it first, then show the solutionSolution

The straight edge is a diameter, so the angle at $P$ is $9 0^{\circ}$ (angle in a semicircle).
Length $= 1. 3^{2} - 1. 2^{2} = 1.69 - 1.44 = 0.25$ .

Answer: 0.5 m

Memory hook: Diameter = hypotenuse.

Tricks to remember Angle in a Semicircle and Angles in the Same Segment

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

Diameter means 90
See a diameter? Draw a little square at the third point on the circle.
Use it when: Spotting right angles hidden in a figure.
Same chord, same side, same angle
Three checks before using 'same segment': the same chord, the same side, and both vertices on the circle.
Use it when: Avoiding wrong 'same segment' claims.
Opposite sides? Add to 180
On opposite sides of the chord the angles are supplementary, not equal.
Use it when: Telling this topic apart from cyclic quadrilaterals.

Angle in a Semicircle and Angles in the Same Segment: solved examples

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

Example 1: Same segment · 1 mark · MCQBoard level

$A$ , $B$ , $C$ and $D$ lie on a circle, with $C$ and $D$ on the same side of chord $A B$ . If $∠ A C B = 4 7^{\circ}$ , then $∠ A D B$ is:
(a) $4 7^{\circ}$
(b) $9 4^{\circ}$
(c) $13 3^{\circ}$
(d) $23. 5^{\circ}$

Try it first, then show the solutionSolution

$∠ A D B = ∠ A C B = 4 7^{\circ}$ (angles in the same segment).

Answer: (a)

Example 2: Diameter from two chords · 2 marks · Short answerBoard level

$A B$ is a diameter of a circle and $C$ is a point on the circle with $A C = 6$ cm and $B C = 8$ cm. Find $A B$ and the radius of the circle.

Try it first, then show the solutionSolution

$∠ A C B = 9 0^{\circ}$ (angle in a semicircle).
$A B = 6^{2} + 8^{2} = 10$ cm, so the radius is 5 cm.

Answer: $A B = 10$ cm, radius 5 cm

Example 3: Crossing chords · 3 marks · Short answerHarder board level

Chords $A C$ and $B D$ of a circle meet at $X$ . $∠ B A C = 2 8^{\circ}$ and $∠ A X B = 10 0^{\circ}$ . Find (a) $∠ A B D$ , (b) $∠ B D C$ , (c) $∠ A C D$ .

Try it first, then show the solutionSolution

(a) In $△ A X B$ : $∠ A B D = 18 0^{\circ} - 10 0^{\circ} - 2 8^{\circ} = 5 2^{\circ}$ .
(b) $∠ B D C = ∠ B A C = 2 8^{\circ}$ (angles in the same segment, chord $B C$ ).
(c) $∠ A C D = ∠ A B D = 5 2^{\circ}$ (angles in the same segment, chord $A D$ ).
Check in $△ D X C$ : $∠ D X C = 10 0^{\circ}$ (vertically opposite) and $2 8^{\circ} + 5 2^{\circ} + 10 0^{\circ} = 18 0^{\circ}$ ✓

Answer: (a) $5 2^{\circ}$ (b) $2 8^{\circ}$ (c) $5 2^{\circ}$

Example 4: Perpendicular to a diameter · 4 marks · Long answerHarder board level

$A B$ is a diameter of a circle and $C$ is a point on the circle. $C D ⊥ A B$ , with $D$ on $A B$ , $A D = 4$ cm and $D B = 9$ cm.
(a) Prove that $△ A D C \sim △ C D B$ .
(b) Find $C D$ , $A C$ and $B C$ .

Try it first, then show the solutionSolution

(a) $∠ A C B = 9 0^{\circ}$ (angle in a semicircle). In $△ A D C$ , $∠ A C D = 9 0^{\circ} - ∠ A$ ; in $△ A C B$ , $∠ B = 9 0^{\circ} - ∠ A$ . So $∠ A C D = ∠ C B D$ , and both triangles have a right angle at $D$ : similar (AA).
(b) $\frac{A D}{C D} = \frac{C D}{D B}$ , so $C D^{2} = 4 \times 9 = 36$ and $C D = 6$ cm.
$A C = 4^{2} + 6^{2} = 52 = 213 \approx 7.21$ cm.
$B C = 9^{2} + 6^{2} = 117 = 313 \approx 10.82$ cm. Check: $52 + 117 = 169 = 1 3^{2}$ ✓

Answer: (b) $C D = 6$ cm, $A C = 213$ cm, $B C = 313$ cm

Example 5: A circle on a side as diameter · 4 marks · Long answerHOTS / competency

In $△ A B C$ , $A B = A C = 13$ cm and $B C = 10$ cm. A circle is drawn with $A B$ as diameter and meets $B C$ at $D$ .
(a) Prove that $D$ is the mid-point of $B C$ .
(b) Find $A D$ .

Try it first, then show the solutionSolution

(a) $∠ A D B = 9 0^{\circ}$ (angle in a semicircle), so $A D ⊥ B C$ .
In $△ A D B$ and $△ A D C$ : $A B = A C$ , $A D$ is common and both have a right angle at $D$ , so they are congruent (RHS) and $B D = D C$ .
(b) $B D = 5$ cm, so $A D = 1 3^{2} - 5^{2} = 12$ cm.

Answer: (a) $B D = D C = 5$ cm (b) $A D = 12$ cm

Example 6: Case study: a semicircular arch · 4 marks · Case-basedHOTS / competency

The opening of a garden arch is a semicircle standing on a straight base $A B = 10$ m. A cable runs from $A$ to a point $P$ on the arch and on to $B$ .
(a) What is $∠ A P B$ , wherever $P$ is?
(b) If $P A = 6$ m, find $P B$ .
(c) How high is $P$ above the base?

Try it first, then show the solutionSolution

(a) $A B$ is a diameter, so $∠ A P B = 9 0^{\circ}$ (angle in a semicircle).
(b) $P B = 1 0^{2} - 6^{2} = 8$ m.
(c) Let $h$ be the height. Area of $△ A P B = \frac{1}{2} \times 6 \times 8 = \frac{1}{2} \times 10 \times h$ , so $h = 4.8$ m.

Answer: (a) $9 0^{\circ}$ (b) 8 m (c) 4.8 m

Where students lose marks in Angle in a Semicircle and Angles in the Same Segment

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

Mistake: Wrongly assuming 'same segment'
The two angles must stand on the same chord and lie on the same side of it.
Wrong working
$∠ A C B = ∠ A D B$ with $C$ and $D$ on opposite sides of $A B$ .
Right working
On opposite sides, $∠ A C B + ∠ A D B = 18 0^{\circ}$ .
Do this: Shade the segment and check both vertices are in it.
Usually costs: The angle and any later answer built on it.
Mistake: Missing the right angle
A diameter in the figure gives a $9 0^{\circ}$ angle that is not marked.
Wrong working
Trying to find $∠ A C B$ from other angles when $A B$ is a diameter.
Right working
$∠ A C B = 9 0^{\circ}$ (angle in a semicircle).
Do this: Mark every diameter and its right angles before starting.
Usually costs: Usually the first step of a 3- or 4-mark part.
Mistake: No reasons
Examiners give marks for the theorem used.
Wrong working
$∠ B D C = 2 8^{\circ}$ .
Right working
$∠ B D C = ∠ B A C = 2 8^{\circ}$ (angles in the same segment).
Do this: Name the chord as well as the theorem.
Usually costs: The 2024 report (Q7(i)) notes many candidates did not give reasons.

Exam tips

2024 Q7(i) (3 marks) and 2025 Q1(vi) (1 mark) were both angle-chasing questions in circles; the examiners ask candidates to give reasons and to find angles by more than one route.
The proof of 'angle in a semicircle is a right angle' can be asked; 'angles in the same segment are equal' is without proof.
In a right triangle inscribed in a circle, the hypotenuse is a diameter: use this for radius questions.
When chords cross inside the circle, list both pairs of equal angles before calculating.

Practice questions on Angle in a Semicircle and Angles in the Same Segment

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
$A B$ is a diameter of a circle and $C$ is a point on it with $∠ A B C = 6 1^{\circ}$ . Then $∠ B A C$ is:
(a) $2 9^{\circ}$
(b) $6 1^{\circ}$
(c) $9 0^{\circ}$
(d) $11 9^{\circ}$
Hint
$∠ A C B = 9 0^{\circ}$ .
Answer
(a)
Q2 · 2 marks · Short answerBoard level
$C$ and $D$ lie on a circle on the same side of chord $A B$ . $∠ A C B = (3 x + 8)^{\circ}$ and $∠ A D B = (5 x - 20)^{\circ}$ . Find $x$ and the angle.
Hint
Angles in the same segment are equal.
Answer
$3 x + 8 = 5 x - 20$ , so $x = 14$ ; each angle is $5 0^{\circ}$ .
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If $C$ and $D$ lie on opposite sides of chord $A B$ of a circle, then $∠ A C B = ∠ A D B$ .
Reason (R): Angles in the same segment of a circle are equal.
(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
Opposite sides of the chord means different segments.
Answer
(b): on opposite sides the angles are supplementary ( $A C B D$ is cyclic).
Q4 · 3 marks · Short answerHarder board level
$A B$ is a diameter of a circle. $D$ is a point on the circle with $∠ A B D = 3 5^{\circ}$ , and $C$ is a point on the circle on the same side of $A D$ as $B$ . Find $∠ A D B$ , $∠ B A D$ and $∠ A C D$ , giving reasons.
Hint
Chord $A D$ subtends equal angles at $B$ and $C$ .
Answer
$∠ A D B = 9 0^{\circ}$ (angle in a semicircle); $∠ B A D = 5 5^{\circ}$ ; $∠ A C D = ∠ A B D = 3 5^{\circ}$ (same segment).
Q5 · 3 marks · Short answerHarder board level
A triangle has sides 7 cm, 24 cm and 25 cm. Find the radius of the circle through its three vertices, and the distance of its centre from the 24 cm side.
Hint
$7^{2} + 2 4^{2} = 2 5^{2}$ ; where is the centre of a right triangle's circumcircle?
Answer
The triangle is right-angled, so the hypotenuse is a diameter: radius 12.5 cm. The centre is the mid-point of the hypotenuse, $\frac{7}{2} = 3.5$ cm from the 24 cm side.
Q6 · 3 marks · Short answerHarder board level
Chords $A C$ and $B D$ of a circle meet at $E$ . $∠ A B D = 3 y^{\circ}$ , $∠ A C D = (2 y + 15)^{\circ}$ and $∠ B A C = 4 0^{\circ}$ . Find $y$ , $∠ B D C$ and $∠ A E B$ .
Hint
$∠ A B D$ and $∠ A C D$ stand on chord $A D$ .
Answer
$3 y = 2 y + 15$ , $y = 15$ ( $∠ A B D = 4 5^{\circ}$ ); $∠ B D C = ∠ B A C = 4 0^{\circ}$ ; $∠ A E B = 18 0^{\circ} - 4 5^{\circ} - 4 0^{\circ} = 9 5^{\circ}$ .
Q7 · 4 marks · Long answerHOTS / competency
Two circles meet at $P$ and $Q$ . $P A$ is a diameter of the first circle and $P B$ a diameter of the second.
(a) Prove that $A$ , $Q$ and $B$ are collinear.
(b) If $P A = 10$ cm, $P B = 8$ cm and $P Q = 6$ cm, find $A B$ .
Hint
Join $P Q$ , $A Q$ and $B Q$ ; find $∠ A QP$ and $∠ B QP$ .
Answer
(a) $∠ A QP = 9 0^{\circ}$ and $∠ B QP = 9 0^{\circ}$ (angles in semicircles), so $∠ A QB = 18 0^{\circ}$ . (b) $A Q = 100 - 36 = 8$ cm, $QB = 64 - 36 = 27$ cm, so $A B = 8 + 27 \approx 13.29$ cm.

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.Angle in a semicircle?Show answer
$9 0^{\circ}$
2.Angles in the same segment?Show answer
Equal.
3. $A B$ a diameter, $∠ C A B = 4 0^{\circ}$ . $∠ C B A$ ?Show answer
$5 0^{\circ}$
4.Which of the two properties is without proof?Show answer
Angles in the same segment are equal.
5.Where is the circumcentre of a right-angled triangle?Show answer
At the mid-point of the hypotenuse.

Angle in a Semicircle and Angles in the Same Segment: frequently asked questions

Why is the angle in a semicircle 90°?

The diameter makes a straight angle of 180° at the centre, and the angle at the circumference is half the angle at the centre, so it is 90°.

What does 'angles in the same segment' mean?

A chord divides a circle into two segments. Angles made by the chord at points in the same segment are equal; at points in opposite segments they add up to 180°.

Do I need to prove that angles in the same segment are equal?

No. The CISCE syllabus marks it 'without proof'. The angle in a semicircle should be proved if asked.