Higher November 2024 Paper 2 Q18
18 \(A\), \(B\) and \(C\) are three points on a circle, centre \(O\).
\(BA = BC\)
Prove that \(OB\) bisects angle \(ABC\). (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| Proof | M1 | begins proof to show that triangles \(ABO\) and \(CBO\) or triangles \(ABD\) and \(CBD\) are congruent by giving one pair of equal sides or equal angles with reason |
| M1 | for different pair of equal sides or angles with reason | |
| C1 | for full proof that triangles \(ABO\) and \(CBO\) are congruent, SSS, or triangles \(ABD\) and \(CBD\) are congruent, RHS, and therefore angle \(ABO\) = angle \(CBO\) \(AB = CB\) (given) \(BO\) (or \(BD\)) is common \(AO = CO\) radii of circle angle \(BAD\) = angle \(BCD\) angles in a semicircle are 90 (\(BO = AO = CO\) radii of circle) counts as two sides with reasons | |
| OR | ||
| M1 | draws \(OA\), \(OC\) and \(AC\) and labels angle \(OAC = x\) and angle \(OCA = x\) with reason given, \(AO = CO\) radii of circle and base angles of an isosceles triangle are equal or \(BAC = BCA\) since \(ABC\) is isosceles | |
| M1 | shows \(OAC = OCA\) and shows \(BAC = BCA\) and uses these to show \(OAB = OCB\) with all reasons given | |
| C1 | for full proof concluding with angle \(ABO = y\) and angle \(CBO = y\) with reason given, eg \(OA = OB = OC\) radii of circle and \(OBC\) and \(OAB\) are isosceles |
Additional guidance
Where \(D\) is point such that \(BOD\) is diameter