Higher November 2018 Paper 2 Q21
21
\(A\), \(B\), \(R\) and \(P\) are four points on a circle with centre \(O\).
\(A\), \(O\), \(R\) and \(C\) are four points on a different circle.
The two circles intersect at the points \(A\) and \(R\).
\(CPA\), \(CRB\) and \(AOB\) are straight lines.
Prove that angle \(CAB\) = angle \(ABC\). (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| proof | C1 | uses cyclic quad eg if \(CAB = x\) then \(CRO = 180 - x\) (Opposite angles of a cyclic quadrilateral add up to 180°.) |
| C1 | establishes relationship outside a circle eg \(ORB = x\) (Angles on a straight line add up to 180) | |
| C1 | uses properties of a circle eg \(RO = OB\) (both radii) so \(ABC = x\) (Base angles of an isosceles triangle are equal.) | |
| C1 | Complete proof and conclusion |
Additional guidance
Underlined words need to be shown; reasons need to be linked to their method; any reasons not linked do not credit.
Correct method can be implied from angles on the diagram if no ambiguity or contradiction.
Full reasons given without any redundant reasons and correct reasoning throughout.