GCSE Maths Geometry and Measure

Circle Theorems

Angle in a Semicircle

# Angle in a Semicircle

Here we will learn about angles in a semicircle, including their application, proof, and using them to solve more difficult problems.

There are also circle theorem worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

## What are angles in a semicircle?

Angles in a semicircle are \pmb{90} degrees:

Angles in a semicircle are created when you join the two ends of the diameter to one point on the arc using chords.

In the diagram above, AB is the diameter of a circle that divides the circle into two. The two points A and B are joined to another point C on the circumference using two chords. This triangle is a right-angle triangle with the 90 degree angle touching the arc. This is the same for any point that is placed on the arc.

If the point is placed on another point in a semicircle, the result would be the same (the two chords would meet at the circumference at \pmb{90} degrees). The angle in a semicircle is 90 degrees.

### What are angles in a semicircle? ### Key parts of a circle needed for this theorem

• The diameter of the circle is the width of a circle, through the centre. The diameter is twice the length of the radius of a circle. In a semicircle, the diameter is the straight edge.
• A chord is a straight line that meets the circumference in two places. The longest chord in a circle is the diameter.
• A semicircle is half of a circle.

Subtended angles (in a semicircle):

An angle within a semicircle is created by two chords meeting at the two ends of the diameter. The diagrams below show the angle subtended by diameter AC from point B for two different semicircles:

### Proving that angles in the same segment are equal

To be able to prove this theorem, you do not need to know any other circle theorem. You just need to be confident with angles in a triangle.

## How to use the angle in a semicircle theorem

In order to use the fact that angles in a semicircle equal 90° :

1. Locate the key parts of the circle for the theorem.
2. Use other angle facts to determine angles within the triangle.
3. Use the angles in a semicircle theorem to state the other missing angle.

### How to use the angle in a semicircle theorem ## Angle in a semicircle examples

### Example 1: standard diagram

ABCD is an arrowhead where C is the centre of the circle and A, B, and D lie on the circumference. Calculate the size of angle BAD :

1. Locate the key parts of the circle for the theorem.

Here we have:

• The angle ABC = 67°
• AB is a diameter
• AC and BC are chords
• The angle CAB = θ

2Use other angle facts to determine angles within the triangle.

We already know that ABC=67° so we do not need to use any other angle fact to determine this angle for this example.

3Use the angles in a semicircle theorem to state the other missing angle.

As the angle in a semicircle is equal to 90° , angle ACB = 90° . We can therefore use the fact that angles in a triangle total 180° to calculate the size of angle BAC :

BAC = 180 - (90 + 67)

BAC=23°

ABCD is a cyclic quadrilateral. ACD is an isosceles triangle. Calculate the size of angle ACB .

Here we have:

• The angle BAD = 107°
• AC is a diameter
• AB and BC are chords
• The angle ACB = θ

Triangle ACD is interesting because it is an isosceles triangle but also a right angle triangle with the hypotenuse AC .

Using the theorem that the angle in a semicircle is 90° , we can say that angle ADC is equal to 90° .

As the triangle is isosceles, the other two angles are equal to (180 - 90) ÷ 2 = 45° . This is important because we now know that angle DAC is equal to 45° , and so we can calculate the size of angle BAC which is in the triangle containing θ.

Angle BAC is therefore equal to:

BAC = 107 - 45

BAC = 62°

We now have the following information added to the diagram:

As the angle in a semicircle is equal to 90° , angle ABC = 90° . We can therefore use the fact that angles in a triangle total 180° to calculate the size of the inscribed angle BCA :

BCA = 180 - (90 + 62)

BAC = 28°

### Example 3: angles in the same segment

Points A, B, C, and D lie on the circumference of the circle. AD is a diameter of the circle. BD and AC intersect at the point E . Calculate the size of the angle CDB .

Here we have:

• The angle CAD = 36
• AB, BD, AC and CD are chords
• The angle BEA = 48°
• The angle CAD = 36°
• The angle BDC = θ

In order to find the angle θ, we need to know the size of the angle EDA . BED is a straight line and so, as angles on a straight line total 180° , we can calculate angle AED :

AED = 180 - 48

AED = 132°

We can now calculate angle ADE as angles in a triangle total 180° :

ADE = 180 - (132 + 36)

As the angle in a semicircle is equal to 90° , angle ACD = 90° , we can, therefore, use the fact that angles in a triangle total 180° to calculate the size of angle CDA , and hence angle CDE :

CDA = 180 - (90 + 36)

CDA = 54°

CDE = 54 - 12

CDE = 42°

### Example 4: angles in parallel lines

AC and BD are diameters of the circle with centre O . BC and AD are parallel. Calculate the size of the angle OAB .

Here we have:

• The angle COD = 58°
• BD and AC are diameters
• OA, OB, OC, and OD are radii
• The angle BAC = θ

AOD is an isosceles triangle as OA and OD are radii. This means that the other two angles are the same and so:

OAD = (180 - 122) ÷ 2

As the angle in a semicircle is equal to 90° , angle BAD = 90° . Angle OAB is therefore equal to:

OAB = 90 - 29

OAB = 61°

### Example 5: the alternate segment theorem

ABCD are points on a circle with centre O . EF is a tangent to the circle through the point C . BD and EF are parallel. Angle DCE = 63° .

Calculate the size of angle BAC .

Here we have:

• The angle DCE = 63°
• AC is a diameter
• The angle BAC = θ

The alternate segment theorem states that angles in the alternate segment are equal. The angle CBD is, therefore, equal to 63° . Also, as the lines BD and EF are parallel, the line AC is perpendicular to the line BD . So the angles at their intersection G are equal to 90° .

BCG is a triangle. As we know two of the three angles, we can now calculate the size of angle BCG , which will help us determine angle CAB as the angles are congruent.

BCG = 180 - (90 + 63)

BCG = 27°

As the angle in a semicircle is equal to 90° , angle ABC = 90° . Angle BAC is therefore equal to:

BAC = 180 - (90 + 27)

BAC = 63°

### Example 6: angle at the centre

A, B, C, D and E are points on the circle with centre O . Points A and O lie on a line of symmetry. Calculate the size of the angle DAE .

Here we have:

• The angle DOC = 90°
• BE is a diameter
• The angle DAE = θ

The angle at the centre is twice the angle at the circumference and so angle DAC = 45° .

As the shape is symmetrical, the angle OAD is half the angle DAC . So OAD = 22.5° .

As the angle in a semicircle is 90° , the angle BAE = 90° . Due to the shape being symmetrical, angle OAE is half of the angle BAE . Angle OAE = 45° .

We can now calculate the angle DAE :

DAE = 45 - 22.5

DAE = 22.5°

### Common misconceptions

The angles that are either end of the diameter total 180° as if the triangle were a cyclic quadrilateral. They should total 90° as the angle in a semicircle is 90° .

• Isosceles triangles

The triangle is assumed to be isosceles and so the angles at either end of the diameter are the same.

• A diameter or a chord?

The angle at the circumference is assumed to be 90° when the associated chord does not intersect the centre of the circle and so the diagram does not show a semicircle.

• Incorrect angle

The angle that is established as 90° , is labelled as an angle by the diameter, and not the subtended angle at the circumference.

### Practice angle in a semicircle questions

1. The points A, B, and C are on the circumference of a circle. Calculate the size of angle BCA . 127^{\circ} 37^{\circ} 53^{\circ} 90^{\circ} • ABC = 90^{\circ} (angles in a semicircle)
• BCA = 180 – (90+53) = 37^{\circ}

2. The points A, B, and C are on the circumference of a circle. Calculate the size of angle BCA . 56^{\circ} 101^{\circ} 79^{\circ} 50.5^{\circ} • BAC = BCD = 90^{\circ} (angles in a semicircle)
• ABC = (180 – 90)\div2 = 45^{\circ} (isosceles triangle)
• CBD = 79 – 45 = 34^{\circ}
• CDB = 180 – (90+34) = 56^{\circ}

3. The points A, B, C, and D are on the circumference of a circle. AC and BD are chords that intersect at the point E . Calculate the size of angle DAE . 63^{\circ} 35^{\circ} 55^{\circ} 27^{\circ} • AED = 35^{\circ} (vertically opposite angles are equal)
• ADE = 90^{\circ} (angles in a semicircle)
• DAE = 180 – (90+35)=55^{\circ}

4. The points A, B, C, and D are points on the circumference of a circle. AC and BC intersect at the centre O at an angle of 152^{\circ} . Calculate the size of angle ABO . 76^{\circ} 28^{\circ} 14^{\circ} 90^{\circ} • BOC = 152^{\circ} (vertically opposite angles are equal)
• OBC = (180-152)\div2 = 14^{\circ} (isosceles triangle)
• ABO = 90 – 14 = 76^{\circ}

5. The points A, B, C and D are on the circumference of a circle with centre O . The chords AC and BD intersect at the point G . AC is parallel to the tangent EF which passes through the point B . Angle FBC = 70^{\circ} . Calculate the size of angle BDC . 90^{\circ} 40^{\circ} 20^{\circ} 70^{\circ} • CBG = 90 – 70 = 20^{\circ} (tangent meets radius at 90^{\circ} )
• BCD = 90^{\circ} (angles in a semicircle)
• BDC = 180 – (90+20) = 70^{\circ} (angles in a triangle)

6. The points A, B, C, D and E are on the circumference of a circle with centre O . AB = AE , and AD = AC. Angle ACD = 26^{\circ} . Calculate the size of angle DAE . 24^{\circ} 42^{\circ} 21^{\circ} 45^{\circ} • CAD = 180 – (66+66) = 48^{\circ} (isosceles triangle)
• BAE = 90^{\circ} (angles in a semicircle)
• DAE = (90-48)\div2 = 21^{\circ}

### Angle in a semicircle GCSE questions

1.

(a) The circle with centre O has a diameter AD . Points A, B, C, and D lie on the circumference of the circle and AC intersects with BD at the point E . Calculate the size of angle ABE marked x . Show your working. (b) Hence calculate the size of angle BAE , marked y . (10 marks)

(a)

DCE = 90^{\circ}

(1)

The angle in a semicircle is 90^{\circ}

(1)

DEC = 180 – (90+40) = 50^{\circ}

(1)

Angles in a triangle

(1)

BEA = 50^{\circ}

(1)

Vertically opposite angles are equal

(1)

(b)

ABE = 90^{\circ}

(1)

The angle in a semicircle is 90^{\circ}

(1)

BAE = 180 – (90+50) = 40^{\circ}

(1)

Angles in a triangle

(1)

2. ABC is a triangle inscribed into a circle. Calculate the size of angle ABC . (4 marks)

BAC = 90^{\circ}

(1)

The angle in a semicircle is 90^{\circ}

(1)

ABC = 180 – (90 + 15) = 75^{\circ}

(1)

Angles in a triangle total 180^{\circ}

(1)

3.

(a) By calculating the size of angle ACD , find the size of angle \theta . State any reasons with your answer. (b) What type of triangle is BCD ?

(5 marks)

(a)

ACD = (180 – 120)\div2 = 30^{\circ}

(1)

ACB = 90^{\circ}

(1)

The angle in a semicircle is 90^{\circ}

(1)

BCD = 90 – 30 = 60^{\circ}

(1)

(b)

Equilateral

(1)

## Learning checklist

You have now learned how to:

• Apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results

## Still stuck?

Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.

Find out more about our GCSE maths revision programme.