### 7) The alternate segment theorem

The angle between a **tangent** and a **chord** is equal to the angle in the alternate segment.

#### Example

Calculate the missing angles *x*, *y* and *z*.

The angle in a semicircle is 90°. Therefore, *y *is 90°.

Angles in a triangle add up to 180°.

Therefore, *z* is 180 – 30 – 90 = 60°

Using the alternate segment theorem: angle *x = z*

Therefore, *x* = 60°

#### Proof

Let angle CDB = *x*.

The angle between a tangent and the **radius** is 90°.

Angle BDO = 90° –* x*

Triangle DOB is an **isosceles** triangle so angle DBO is 90° –* x*.

Angles in a triangle add up to 180°.

Angle DOB = 180 – DBO – BDO

Angle DOB = 180 – (90 – *x*) – (90 – *x*) = 2*x*

The angle at the centre is double the angle at the circumference.

Angle DAB = *x*

Therefore BDC = DAB.

**Exercise 1**** | Answer Keys**

**Exercise 2**

**| Answer Keys**

**Circle Theorems**– exam-styled questions (PDF)