# Angle in a semicircle is 90 degrees

We can draw a triangle where the base is a diameter of a circle, and the third point of the triangle lies on the circumference of the circle:

The angle at the circumference (called the angles in a semicircle) is 90°.

This theorem is covered in this video on circle theorems:

## Proof

We can prove this as follows.

We label the vertices of the triangle **A**, **B** and **C**, and label the centre of the circle **O**. We are trying to prove that the angle at **B** is 90°.

We first draw an extra line from **O** to **B** like this:

Notice that the lines **OA**, **OB** and **OC** are all radii of the circle, and therefore all of equal length.

Looking at the triangle **AOB**, this is an isosceles triangle (from the rule 2 radii form an isosceles triangle). So the two angles at the circumference are equal (we will call them *a*):

By the same logic, the two angles at the circumference in triangle **COB** are equal, and we will call them *b*:

Looking at the original triangle **ABC**:

The angle at **A** is *a*, the angle at **B** is *a + b*, and the angle at **C** is *b*. Since the three angles of a triangle add up to 180° we have:

```
a + (a + b) + b = 180°
```

So:

```
2(a + b) = 180°
```

Which means:

```
a + b = 90°
```

Since the angle at **B** is *a + b*, this proves that the angle at **B** is 90°.

## See also

- Tangent and radius of a circle meet at 90°
- Two radii form an isosceles triangle
- Perpendicular bisector of a chord
- Angle at the centre of a circle is twice the angle at the circumference

## Join the GraphicMaths Newletter

Sign up using this form to receive an email when new content is added: