Question 35

ABCD is a parallelogram. The point E lies on diagonal AC such that EC is one-fifth of AE. Given that DE and AB meet at F, when produced further. The ratio $$CD \colon FB$$ equals

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Since AF lies along AB, which is parallel to DC, triangles EAF and ECD are similar, so $$\frac{AF}{CD} = \frac{AE}{EC} = 5$$ and $$AF = 5CD$$. As $$AB = CD$$, we get $$FB = AF - AB = 5CD - CD = 4CD$$. Hence $$CD \colon FB = 1 \colon 4$$.

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