Higher November 2017 Unit 2 Q10
10. The line \(GH\) is a tangent to the circle at point \(Y\).
The line \(EF\) is parallel to the line \(GH\).
The vertices of triangle \(EFY\) lie on the circle.
\(E\hat{Y}G\) = 60°.
Diagram not drawn to scale
Prove that \(EFY\) is an equilateral triangle.
Give a reason for each step to justify your proof. [3]
| Scheme | Marks | Comments |
|---|---|---|
| For a correct proof: i.e. each angle within the triangle is correctly evaluated as 60° AND with correct reasons. No assumptions can be made at any stage of the proof for the evaluation of any angles. | E1 E1 E1 | If any other angle is used within the proof e.g. angle FYH, then a correct reason must again be stated (there is no E1 for this angle as it is working towards one of the angles within the triangle). Award E2 for two angles within the triangle correctly evaluated as 60° AND with correct reasons. Award E1 for one angle within the triangle correctly evaluated as 60° AND with correct reason(s). *Do not accept 'Z' angles for alternate angles. |
Examples
<EFY = 60° AND Alternate Segment Theorem; E1
<FEY = 60° AND Alternate angles; E1
(<EYF = 60° AND) angles in a triangle (therefore equilateral) OR
(<EYF = 60° AND) therefore equilateral E1
<FEY = 60° AND Alternate angles E1
<FYH = 60° AND Alternate Segment Theorem
<FYE = (180°–60°–60°)=60° AND straight line E1
(<EFY = 60° AND) angles in a triangle (therefore equilateral) OR
(<EFY = 60° AND) therefore equilateral E1
<EFY = 60° AND Alternate Segment Theorem E1
<FYH = 60° AND Alternate angles
<FYE = (180°–60°–60°)=60° AND straight line E1
(<FEY = 60° AND) angles in a triangle (therefore equilateral) OR
(<FEY = 60° AND) therefore equilateral E1
<EFY = 60° AND Alternate Segment Theorem E1
<FYE = 60° AND Interior angles E1
(<FEY = (180°–60°–60°)=60° AND) angles in a triangle (therefore equilateral) OR
(<FEY = 60° AND) therefore equilateral E1