327. The Semi-Circle Secret · why angles on a diameter always hit 90°

Try to break it

Drag C along the semicircle. ∠ACB stays locked at exactly 90° at every position — because the inscribed angle is half the central angle, and the central angle here is the straight diameter (180°). The right angle survives the move because AB is a diameter; cross over and you'd find ∠ACB still 90° on the other semicircle too.

How you build it

Draw a diameter AB and angle at C on the circle.

• Mark a point O — this will be the centre of the circle.
• Draw a circle centred on O — click O, then a point on the rim to set the radius.
• Mark point A on the circle (it snaps onto the rim).
• Mark point B on the circle directly opposite A, so chord AB passes through O. AB is then a diameter.
• Join A to B through O — the diameter.
• Mark point C anywhere else on the circle — on the semi-circle above (or below) the diameter.
• Join A to C.
• Join B to C. Because AB is a diameter, ∠ACB in the semi-circle is exactly 90°.

The proof, step by step

Prove that the angle in a semicircle is a right angle.

• Arc APB subtends ∠AOB at the centre and ∠ACB at point C on the circumference.
• By the central angle theorem, the angle at the centre is double the angle at the circumference: ∠AOB = 2∠ACB.
• Since AB is a diameter, ∠AOB is a straight angle measuring 180°.
• Substituting the value: 2∠ACB = 180°, which gives ∠ACB = 90°.

Worked example

In a circle with centre O, AB is a diameter of length 10 cm. Point C lies on the circumference such that AC = 6 cm. What is the measure of ∠ACB?

By Theorem 7, the angle subtended by a diameter at any point on the circumference is always 90°, regardless of the chord lengths. Thus, ∠ACB = 90°.

• 45°
• 60°
• 90° — correct
• 120°