Angles in a Semicircle

Welcome to your first lesson on Circle Theorems. This course is designed to build your understanding from the ground up, preparing you for the full range of problems you'll encounter in your IGCSE Extended tier exam.

Today, we will begin with a foundational theorem that is both elegant and essential for many other geometric problems. Our focus will be on defining and applying the theorem that the angle in a semicircle is a right angle. By the end of this lesson, you will be able to identify this configuration, use it to find unknown angles, and, crucially, state your reasoning formally, just as required in your exams.

1. Defining the Angle in a Semicircle Theorem

Let's start by formally defining the theorem and understanding the specific terminology. The key to mastering circle theorems is to first recognize the geometric setup in a diagram.

Angle in a Semicircle | Cambridge (CIE) IGCSE Maths Revision Notes 2023

This revision note from Save My Exams clearly defines the theorem and explains how to identify it. It also points out the exact phrasing you'll need for your exam answers.

Please read the section "Circle Theorem: The angle in a semicircle is 90°". Focus on the bullet points that describe the theorem and the key elements to look for in a diagram, such as the side being a diameter and all vertices being on the circumference. Pay close attention to the required exam wording highlighted in bold.

To build a strong intuition for this property, it's helpful to see it in a more dynamic way. The following video demonstrates how the right angle is formed no matter where the vertex is placed on the semicircle's arc.

Angle in a Semicircle (1 of 6: Standard proof)

In this video, Eddie Woo provides a clear visual introduction to the theorem.

Watch the segment from the beginning of the explanation. Observe how he constructs the angle and how the triangle moves while the angle at the circumference remains consistently 90 degrees.

2. Proving the Theorem

A key part of mathematics is not just accepting a rule but understanding why it works. Since you are preparing for formal proofs later in your studies, let's look at the logical argument that proves this theorem. The proof relies on your existing knowledge of isosceles triangles and the sum of angles in a triangle.

The next segment of Eddie Woo's video provides a concise and elegant proof.

Angle in a Semicircle (1 of 6: Standard proof)

Here, Eddie Woo walks through a standard proof for the theorem. This is a great example of the kind of deductive reasoning you will build on throughout this course.

Please watch the proof carefully from this section. Pay attention to how he constructs a central radius to split the main triangle into two isosceles triangles and then uses algebra to show that the angle at the circumference must be 90°.

The core steps of the proof are:

Draw a radius from the centre to the vertex on the circumference, creating two new triangles.
Recognize that both triangles are isosceles because two of their sides are radii.
Use the property that base angles of an isosceles triangle are equal to label the angles (e.g., as $α$ and $β$ ).
Consider the large triangle and sum all its internal angles to 180°.
An algebraic simplification, $2 α + 2 β = 18 0^{\circ}$ , leads directly to the conclusion that $α + β = 9 0^{\circ}$ .

3. Applying the Theorem in Problems

Now that you know what the theorem is and why it's true, let's focus on applying it to solve problems. This is the skill you'll need most for your exam.

A typical problem involves finding a missing angle in a triangle inscribed in a semicircle.

This image shows a typical problem. A triangle PQR is inscribed in a circle where PQ is the diameter. The angle at R is identified as 90° using the theorem, allowing the unknown angle at P to be calculated using the sum of angles in a triangle.

The logic shown in the image is a standard two-step process:

Identify the right angle: Since PQ is the diameter, the angle opposite it on the circumference, $∠ P R Q$ , is 90°. The reason given is "Angle in a semicircle".
Use other geometry facts: The angles in triangle PQR must sum to 180°. Therefore, $∠ QP R = 18 0^{\circ} - (9 0^{\circ} + 4 0^{\circ}) = 5 0^{\circ}$ .

Let's look at another worked example to solidify the process.

Angle in a semi circle

The resource from Third Space Learning provides several examples. We will focus on the first one, which is a classic application.

Read through "Example 1: standard diagram". Follow the three steps outlined: locating the key parts, using other angle facts, and applying the theorem to find the missing angle.

Your Turn: Practice Problems

It's time to try applying the theorem yourself. Below is a worksheet with several problems.

This worksheet contains six problems. Questions 1-3 ask to find a shaded angle, while Questions 4-6 ask to find a missing length.

Take a few minutes to solve Question 1 and Question 2. For each problem, identify the right angle first and then use the sum of angles in a triangle to find the shaded angle. Remember to think about the formal reason you would state in an exam.

Solution for Q1:

The line passing through the center is the diameter. The angle opposite this diameter is 90°.
The shaded angle = $18 0^{\circ} - (9 0^{\circ} + 3 0^{\circ}) = 6 0^{\circ}$ .
Reasoning: Angle in a semicircle is 90°; sum of angles in a triangle is 180°.

Solution for Q2:

The line passing through the center is the diameter. The angle opposite it is 90°.
The small triangle containing the 10° angle and the shaded angle is a right-angled triangle.
The shaded angle = $18 0^{\circ} - (9 0^{\circ} + 1 0^{\circ}) = 8 0^{\circ}$ .
Reasoning: Angle in a semicircle is 90°; sum of angles in a triangle is 180°.

4. Common Misconceptions

When learning a new theorem, it's useful to be aware of common mistakes. This helps you avoid them in your own work.

Angle in a semi circle

This short section highlights frequent errors students make when applying this theorem.

Please read the four bullet points under the "Common misconceptions" heading. The most critical point is the one about mistaking a chord for a diameter. The theorem only applies if the line segment forming the side of the triangle passes through the center of the circle.

Conclusion

In this lesson, we established the first of your circle theorems. You learned how to define and identify the "angle in a semicircle" theorem, understood the logical proof behind it, and applied it to solve for unknown angles in geometric problems.

Here are the key takeaways:

Definition: The angle subtended by a diameter at any point on the circumference is a right angle (90°).
Identification: Look for a triangle inscribed in a circle where one of its sides is the diameter.
Formal Reason: For your IGCSE exam, the required statement is "The angle in a semicircle is 90°".
Application: This theorem is frequently combined with the fact that the sum of angles in a triangle is 180° to find missing angles.

In our next lesson, we will explore a more general theorem: "The angle at the center is double the angle at the circumference." You will see that the angle in a semicircle theorem we learned today is, in fact, a special case of this upcoming theorem.