## Angle at the Centre of a Circle Proof

**Here we want to prove that the angle ACB is double angle ADB**

**ACB = 2(ADB)**

First: Join point D to point C and extend on to Radius to create point X

To prove: Angle ACB = 2 Angle ADB

Proof:

Similarly Angle BCX = 2 Angle CDB

Angle ACX + Angle BCX = 2 Angle CDA + 2 Angle CDB

Therefore: Angle ACB = 2 Angle ADB

To prove: Angle ACB = 2 Angle ADB

Proof:

- AC = CB because they are the same radius.
- Angle CAB = Angle CBD because it is an isosceles triangle.
- Angle ACX = Angle DAC and angle CDA because 2 opposite interior angles are equal to the exterior angle of a triangle.
- Angle ACX =Angle CDA and Angle CDA
- Therefore Angle ACX = 2 Angle CDA

Similarly Angle BCX = 2 Angle CDB

Angle ACX + Angle BCX = 2 Angle CDA + 2 Angle CDB

Therefore: Angle ACB = 2 Angle ADB

**Interactive of Proof**