Finding ratio of area of triangle and area of trapezium

Question

Can anyone help me with this? –>
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I already answered a, b, c i) and c ii) but how am I supposed to do c iii)?
I'm quite confused and I don't know how to solve this particular question.

Answer 1

Hint:

$\text{area}(NMCL)=\text{area}(ABC)-\text{area}(ANM)-\text{area}(NBL)$ and each of the l.h.s. area is a fraction of $\text{area}(ANM)$.

However, it's a parallelogram, not a trapezium.

Variant: $$\frac{\text{area}(ANM)}{\text{area}(NMCL)}=\frac{\text{area}(ANM)}{\text{area}(NBL)}\cdot\frac{\text{area}(NBL)}{\text{area}(NMCL)}$$ Observe that triangle $NBL$ and parallelogram $NMCL$ have the same height, hence the last ratio is $$\frac{\text{area}(NBL)}{\text{area}(NMCL)}=\frac{\frac12 BL}{NB}.$$