O-LevelMathematicsGeometry4024/21 · May/June 2016 · Q48 Marks
[Figure 4.1] ANB, BLC and CMA are straight lines. NM is parallel to BC and LN is parallel to CA.
- (a)Prove that triangle ANM is similar to triangle NBL. Give a reason for each statement you make.[3]
- (b)(i)AN : NB = 2 : 3. Find NM : BC.[2]
- (b)(ii)Find area ANM : area NBL.[1]
- (b)(iii)Find area ANM : area NMCL.[2]
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Mark scheme
Marking points from the official mark scheme (May/June 2016). The mark or mark type is shown in brackets.
- [3] (a) Complete proof
- [3] (a) B2 for 2 pairs of equal angles 1 pair with reason.
- [3] (a) B1 for 1 pair of equal angles.
- [2] (b)(i) 2:5
- [2] (b)(i) B1 for NM : BL = 2:3 oe or NM = LC
- [1] (b)(ii) 4:9
- [2] (b)(iii) 1:3
- [2] (b)(iii) B1 for such as ΔANM/ΔABC = 4/25 or ANBL/ΔABC = 9/25
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