Higher November 2023 Paper 3 Q19
19 \(AC\) is a diameter of a circle, centre \(E\).
\(E\) is the midpoint of \(BD\).
Not drawn accurately
Prove that triangle \(ABE\) is congruent to triangle \(CDE\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(AE = CE\) | M1 | oe |
| angle \(AEB\) = angle \(CED\) | M1 | oe |
| \(BE = DE\) | M1 | oe |
| \(AE = CE\) and radii and angle \(AEB\) = angle \(CED\) and (vertically) opposite and \(BE = DE\) and \(E\) is the midpoint and SAS | A1 | oe allow \(BE = DE\) and given |
Additional guidance
Up to M3 may be awarded for correct, unambiguous working shown on the diagram
Angles must be correctly identified, do not accept angle \(E\) for angle \(AEB\)
Do not award A mark if any incorrect statement is seen