Angle in a Semicircle

Class 9Circles

Thales' Theorem states: the angle subtended by a diameter of a circle at any point on the circle is 90° (a right angle). This is one of the oldest and most celebrated results in geometry.

The theorem is a special case of the Central Angle Theorem. Since a diameter subtends an angle of 180° at the centre, the inscribed angle at any point on the circle = 180°/2 = 90°.

The converse is equally important: if the angle subtended by a line segment at a point is 90°, then the point lies on the circle with that line segment as diameter. This is used to prove that a point lies on a specific circle.

What is Angle in a Semicircle?

Theorem (Angle in a Semicircle):

The angle in a semicircle is a right angle (90°).

If AB is a diameter of a circle with centre O, and C is any point on the circle (other than A or B), then:

∠ACB = 90°

Converse:

If ∠ACB = 90°, then AB is a diameter of the circle passing through A, B, and C.

Key terms:

Semicircle: Half of a circle; the arc cut off by a diameter.
Thales' Theorem: Another name for the angle in a semicircle theorem, attributed to Thales of Miletus (624–546 BC).

Important:

The angle is 90° regardless of where C is on the semicircle.
Moving C along the semicircle changes the shape of triangle ACB but the angle at C remains 90°.
This holds for both semicircles defined by the diameter.

Angle in a Semicircle Formula

Key Results:

1. Angle in a semicircle:

∠ACB = 90° (when AB is a diameter)

2. Relationship to the Central Angle Theorem:

∠AOB = 180° (AB is a straight line through O)
∠ACB = ∠AOB / 2 = 180° / 2 = 90°

3. Converse — finding the diameter:

If ∠ACB = 90° and A, B, C lie on a circle, then AB must be a diameter.

4. Midpoint of the hypotenuse:

In a right triangle, the midpoint of the hypotenuse is equidistant from all three vertices.
This distance equals half the hypotenuse = the radius of the circumscribed circle.
The circumcentre of a right triangle lies at the midpoint of the hypotenuse.

5. In right ▵ACB with ∠C = 90°:

AB (hypotenuse) = diameter = 2r
OC = OA = OB = r (O is midpoint of AB)

Derivation and Proof

Proof: Angle in a Semicircle is 90°

Given: AB is a diameter of a circle with centre O. C is a point on the circle.

To prove: ∠ACB = 90°

Method 1: Using the Central Angle Theorem

AB is a diameter, so A, O, B are collinear.
∠AOB = 180° (straight angle).
By the Central Angle Theorem: ∠ACB = ∠AOB / 2 = 180/2 = 90°. ■

Method 2: Using isosceles triangles

Construction: Join OC. Since O is the centre, OA = OB = OC = r (radii).

▵OAC is isosceles (OA = OC). Let ∠OAC = ∠OCA = α.
▵OBC is isosceles (OB = OC). Let ∠OBC = ∠OCB = β.
In ▵ACB:
∠A + ∠B + ∠C = 180°
α + β + (α + β) = 180°
2(α + β) = 180°
α + β = 90°
∠ACB = α + β = 90° ■

Proof of the Converse:

Given: ∠ACB = 90°; A, B, C lie on a circle.

To prove: AB is a diameter.

Proof (by contradiction):

Suppose AB is NOT a diameter. Then the centre O is not on AB.
By the Central Angle Theorem: ∠AOB = 2 ∠ACB = 2 × 90 = 180°.
∠AOB = 180° means A, O, B are collinear, i.e., AB passes through O.
But a chord passing through the centre IS a diameter.
This contradicts our assumption. Therefore AB is a diameter. ■

Types and Properties

Applications and Extensions of the Theorem:

1. Constructing a right angle

Draw a circle with AB as diameter.
Any point C on the circle gives ∠ACB = 90°.
This is a standard compass-and-straightedge construction for a right angle.

2. Circumcircle of a right triangle

The circumscribed circle of a right triangle has the hypotenuse as its diameter.
The circumcentre is the midpoint of the hypotenuse.
The circumradius = half the hypotenuse.

3. Finding the diameter

If a triangle inscribed in a circle has a 90° angle, the side opposite the 90° angle is a diameter.
This identifies the diameter without needing to know the centre.

4. Tangent perpendicular to radius

A tangent at any point on the circle is perpendicular to the radius at that point.
This can be viewed as a limiting case of the angle in a semicircle.

5. Locus interpretation

The locus of all points that subtend a right angle at a given line segment AB is the circle with AB as diameter (excluding A and B themselves).

Solved Examples

Example 1: Example 1: Basic application

Problem: AB is a diameter of a circle. C is a point on the circle such that ∠BAC = 35°. Find ∠ABC.

Solution:

Given: AB = diameter; ∠BAC = 35°

By the angle in a semicircle theorem:

∠ACB = 90°

In ▵ACB:

∠BAC + ∠ABC + ∠ACB = 180°
35 + ∠ABC + 90 = 180
∠ABC = 180 − 125 = 55°

Answer: ∠ABC = 55°.

Example 2: Example 2: Finding angle at centre

Problem: PQ is a diameter. R is on the circle. ∠RPQ = 50°. Find ∠PQR and ∠PRQ.

Solution:

∠PRQ = 90° (angle in semicircle)

In ▵PRQ:

50 + ∠PQR + 90 = 180
∠PQR = 180 − 140 = 40°

Answer: ∠PRQ = 90°; ∠PQR = 40°.

Example 3: Example 3: Identifying the diameter

Problem: In a circle, three points A, B, C lie on the circumference. ∠ACB = 90°. Identify the diameter.

Solution:

By the converse of the angle in a semicircle theorem:

If ∠ACB = 90°, then the side opposite to the 90° angle is the diameter.
The side opposite ∠ACB is AB.

Answer: AB is the diameter.

Example 4: Example 4: Finding the circumradius of a right triangle

Problem: A right triangle has legs 6 cm and 8 cm. Find the radius of its circumscribed circle.

Solution:

Step 1: Find the hypotenuse

Hypotenuse = √(6² + 8²) = √(36 + 64) = √100 = 10 cm

Step 2: The hypotenuse is the diameter (by the angle in a semicircle theorem)

Diameter = 10 cm
Radius = 10/2 = 5 cm

Answer: The circumradius is 5 cm.

Example 5: Example 5: Midpoint of hypotenuse property

Problem: In right ▵ABC (∠C = 90°), hypotenuse AB = 20 cm. O is the midpoint of AB. Find OC.

Solution:

Since ∠C = 90°, the circumscribed circle has AB as diameter and O as centre.

OA = OB = OC = radius = AB/2 = 20/2 = 10 cm

The midpoint of the hypotenuse is equidistant from all three vertices.

Answer: OC = 10 cm.

Example 6: Example 6: Algebraic problem

Problem: AB is a diameter. C is on the circle. ∠CAB = (2x + 15)° and ∠CBA = (3x)°. Find x.

Solution:

∠ACB = 90° (angle in semicircle)

In ▵ACB:

(2x + 15) + (3x) + 90 = 180
5x + 105 = 180
5x = 75
x = 15

Angles:

∠CAB = 2(15) + 15 = 45°
∠CBA = 3(15) = 45°

This makes ▵ACB an isosceles right triangle.

Answer: x = 15.

Example 7: Example 7: Two points on the semicircle

Problem: AB is a diameter. C and D are two points on the same semicircle. Find ∠ACB and ∠ADB.

Solution:

Both C and D lie on the semicircle defined by diameter AB.

∠ACB = 90° (angle in semicircle)
∠ADB = 90° (angle in semicircle)

Every point on the semicircle subtends a right angle at the diameter.

Answer: Both angles are 90°.

Example 8: Example 8: Proving a right angle

Problem: O is the centre of a circle. AB is a chord (not a diameter). M is the midpoint of AB. Prove that ∠AMO = 90° using the angle in a semicircle concept.

Solution:

This is more directly proved by the perpendicular-from-centre theorem. However, we can approach it alternatively:

Direct method:

The perpendicular from centre O to chord AB bisects AB at M.
Therefore ∠OMA = 90° (by Theorem 10.3).

Angle-in-semicircle perspective:

If we draw a circle with OM as a chord, M being the midpoint and O the centre relates to the perpendicular bisector property, not directly the semicircle theorem.

Answer: ∠AMO = 90° (by the perpendicular from centre to chord theorem).

Example 9: Example 9: Construction using the theorem

Problem: Construct a right angle at point P on a given line segment AB.

Solution (construction):

Draw AB as the given line segment.
Find the midpoint O of AB (using compass: arcs from A and B).
With O as centre and OA as radius, draw a circle (semicircle suffices).
Mark any point C on the semicircle (not at A or B).
Join CA and CB.
∠ACB = 90° (angle in semicircle).

If P is one of the endpoints, say P = A, then ∠CAB gives an acute angle. To get 90° at A, use a different construction.

Answer: The angle ∠ACB at any point C on the semicircle is a right angle.

Example 10: Example 10: Real-world application

Problem: A bridge has a semicircular arch with a span (diameter) of 20 m. A vertical support is placed at a point on the arch such that it makes a 90° angle with the line joining the two endpoints of the diameter. Find the height of the support at the midpoint of the diameter.

Solution:

Given: Diameter = 20 m, so radius = 10 m.

The highest point of the semicircular arch is directly above the centre of the diameter.

Height at the midpoint = radius = 10 m

At this point, C is directly above O (midpoint of AB), and:

∠ACB = 90° (angle in semicircle ✔)
CA = CB (C is on the perpendicular bisector of AB)

Answer: The height at the midpoint is 10 m.

Real-World Applications

Applications of the Angle in a Semicircle:

Construction of right angles: Drawing a semicircle on a line segment and picking any point on it gives a perfect 90° angle. Used in drafting and engineering.
Circumscribed circle of right triangles: The hypotenuse is always the diameter. The circumcentre is the midpoint of the hypotenuse. This is used to find circumradii.
Architecture: Semicircular arches guarantee that the angle subtended at any point on the arch equals 90°, distributing load symmetrically.
Optics: In circular mirror systems, the angle of incidence at the semicircle boundary is always 90° relative to the diameter, enabling precise reflection calculations.
Locus problems: The set of all points that see a given line segment at a right angle forms a circle with that segment as diameter.
Proving perpendicularity: In coordinate geometry, if three points form a right angle, they lie on a circle whose diameter is the side opposite the right angle.

Key Points to Remember

The angle in a semicircle is always 90° (right angle). This is Thales' Theorem.
This is a special case of the Central Angle Theorem: diameter gives central angle = 180°, so inscribed angle = 90°.
The converse: if ∠ACB = 90° and A, B, C lie on a circle, then AB is a diameter.
The circumcentre of a right triangle is the midpoint of the hypotenuse.
The circumradius of a right triangle = half the hypotenuse.
The midpoint of the hypotenuse is equidistant from all three vertices.
Moving the point C along the semicircle changes the triangle shape but ∠C remains 90°.
This theorem is used to construct right angles using compass and straightedge.
The locus of points subtending 90° at a line segment is the circle with that segment as diameter.
In NCERT Class 9, this is a corollary of Theorem 10.8 (Central Angle Theorem).

Practice Problems

AB is a diameter of a circle. C is on the circle such that ∠BAC = 28°. Find ∠ABC.
PQ is a diameter. R is on the circle. If PR = 5 cm and QR = 12 cm, find PQ.
In a circle, ∠ACB = 90° and AB = 26 cm. Find the radius of the circle.
AB is a diameter. C and D are on opposite semicircles. Find ∠ACB + ∠ADB.
A right triangle has a hypotenuse of 30 cm. Find the circumradius and the distance from the circumcentre to each vertex.
Prove that the midpoint of the hypotenuse of a right triangle is equidistant from all three vertices.
Using ruler and compass, construct a right angle at a point on a given line using the angle in a semicircle.
In right △ABC (∠C = 90°), the midpoint M of hypotenuse AB is such that MC = 8 cm. Find AB.

Frequently Asked Questions

Q1. What is the angle in a semicircle?

The angle subtended by a diameter at any point on the circle is exactly 90° (a right angle). This is called the angle in a semicircle or Thales' Theorem.

Q2. Who discovered this theorem?

It is attributed to Thales of Miletus (624–546 BC), a Greek mathematician and philosopher. He is said to have been the first to prove that the angle in a semicircle is 90°.

Q3. Why is the angle in a semicircle 90°?

A diameter subtends a straight angle (180°) at the centre. By the Central Angle Theorem, the inscribed angle = 180°/2 = 90°.

Q4. Does this work for any point on the semicircle?

Yes. Any point on the semicircle (except the endpoints of the diameter) gives an angle of exactly 90°. This is true for both semicircles defined by the diameter.

Q5. What is the converse of this theorem?

If three points A, B, C lie on a circle and ∠ACB = 90°, then AB must be a diameter. This helps identify the diameter when the centre is not known.

Q6. Where is the circumcentre of a right triangle?

The circumcentre of a right triangle is at the midpoint of the hypotenuse. The circumradius equals half the hypotenuse.

Q7. How is this theorem used to construct a right angle?

Draw a segment AB. Find its midpoint O. Draw a semicircle with centre O and radius OA. Any point C on the semicircle gives ∠ACB = 90°.

Q8. Can the angle in a semicircle ever be different from 90°?

No. For a true semicircle (arc cut by a diameter), the angle is always exactly 90°. If the angle differs, the chord is not a diameter.

Q9. What is the relationship between this theorem and Pythagoras?

In a semicircle with diameter AB and point C on the circle, ∠ACB = 90° means triangle ACB is right-angled. By Pythagoras: AC² + BC² = AB². Both theorems apply simultaneously.

Q10. Is this theorem in NCERT Class 9?

Yes. It appears as a direct consequence of Theorem 10.8 in Chapter 10 (Circles). NCERT states it as: 'Angle in a semicircle is a right angle.'

Angle in a Semicircle

What is Angle in a Semicircle?

Angle in a Semicircle Formula

Derivation and Proof

Types and Properties

Solved Examples

Example 1: Example 1: Basic application

Example 2: Example 2: Finding angle at centre

Example 3: Example 3: Identifying the diameter

Example 4: Example 4: Finding the circumradius of a right triangle

Example 5: Example 5: Midpoint of hypotenuse property

Example 6: Example 6: Algebraic problem

Example 7: Example 7: Two points on the semicircle

Example 8: Example 8: Proving a right angle

Example 9: Example 9: Construction using the theorem

Example 10: Example 10: Real-world application

Real-World Applications

Key Points to Remember

Practice Problems

Frequently Asked Questions

Q1. What is the angle in a semicircle?

Q2. Who discovered this theorem?

Q3. Why is the angle in a semicircle 90°?

Q4. Does this work for any point on the semicircle?

Q5. What is the converse of this theorem?

Q6. Where is the circumcentre of a right triangle?

Q7. How is this theorem used to construct a right angle?

Q8. Can the angle in a semicircle ever be different from 90°?

Q9. What is the relationship between this theorem and Pythagoras?

Q10. Is this theorem in NCERT Class 9?

Related Topics

More Maths Topics

Quick Links