The angle inscribed in a semi-circle is always a right angle.

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OA and OC are both radii of the circle so they are the same length.

∴ Δ AOC is an isoceles triangle so has two equal angles (x in the diagram).

OB is also a radius so is the same length as OC

∴ Δ OBC is an isoceles triangle so has two equal angles (y in the diagram).

We know that the angles in a Δ add up to 180° so considering Δ ABC:

x + (x + y) + y = 180°

2x + 2y = 180°

2(x+ y) = 180°

∴ x+y = 90°

and since ∠ ACB = x+y

∠ ACB = 90°

The Angle at the Centre Theorem states that:

The angle at the centre of a circle is always twice that at the circumference subtended by the same arc.

The angle in a semi-circle theorem can be considered as a special case of this where the angle at the centre is 180° (i.e. a straight line). The angle at the circumference will therefore be \(\frac{180°}{2} = 90°\)