The alternate segment theorem - Higher - Circle theorems - Higher - AQA - GCSE Maths Revision - AQA

The alternate segment theorem - Higher

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

Circle contaning triangle with 2 pairs of identical angles, inside and outside the triangle

Example

Calculate the missing angles \(x\), \(y\) and \(z\).

The angle in a semicircle is 90°.

\(y = 90°\)

Angles in a triangle add up to 180°.

\(z = 180^\circ - 30^\circ - 90^\circ = 60^\circ\)

Using the alternate segment theorem:

angle \(x = z\)

\(x = 60^\circ\)

Proof

Let angle CDB = \(x\).

Circle on tangent, EDC, with triangle (ADB) inside circle and external angle x labelled

The angle between a tangent and the is 90°.

Angle BDO = \(90^\circ - x\)

Triangle DOB is an triangle so angle DBO is \(90^\circ - x\).

Internal angles of triangle (ODB) labelled, 90-x

Angles in a triangle add up to 180°.

Angle DOB = \(180^\circ - \text{BDO} - \text{DBO}\)

Angle DOB = \(180^\circ - (90^\circ - x) - (90^\circ - x) = 2x\)

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

Circle on tangent, EDC, with triangle (ADB) inside circle. Internal angles of triangle (ODB) labelled, 90-x and 2x

Angle DAB = \(x\)

Therefore BDC = DAB.

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Circle theorems - Higher - AQAThe alternate segment theorem - Higher