Higher November 2017 Paper 3 Q20

\(A\), \(B\) and \(C\) are points on the circumference of a circle, centre \(O\).
\(AOB\) is a diameter of the circle.

Prove that angle \(ACB\) is \(90^\circ\)
You must not use any circle theorems in your proof. (4)

Mark scheme

Answer	Mark	Notes
Proof	C1	draws \(OC\) and considers angles in an isosceles triangle (algebraic notation may be used, eg two angles labelled \(x\))
	C1	finds sum of angles in triangle \(ABC\), eg \(x + x + y + y = 180\), or sum of angles at \(O\), eg \(180 - 2x + 180 - 2y\)
	C1	complete method leading to \(ACB = 90\)
	C1	complete proof with all reasons given, eg base angles of an isosceles triangle are equal, angles in a triangle add up to \(180^\circ\), angles on a straight line add up to \(180^\circ\)