The circle theorems

7) The alternate segment theorem

The angle between a tangent and a chord 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.

Circle containing triangle at tangent

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

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

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.

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

Angles in a triangle add up to 180°.
Angle DOB = 180 – DBO – BDO
Angle DOB = 180 – (90 – x) – (90 – 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.


Exercise 1 | Answer Keys
Exercise 2 | Answer Keys
Circle Theoremsexam-styled questions (PDF)

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