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State the converse of basic proportionality theorem. Also find $\frac{BF}{FC}$ in the following figure, given that $AB || DC || EF$ and $\frac{AE}{ED} = \frac{2}{3}$. Also, find the length of EF if $AB = 10$ cm and $DC = 15$ cm.
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Correct statement of converse of Basic Proportionality Theorem.
In $\Delta ADC, EG || DC \implies \frac{AE}{ED} = \frac{AG}{GC} = \frac{2}{3}$
In $\Delta ABC, GF || AB \implies \frac{AG}{GC} = \frac{BF}{FC} = \frac{2}{3}$
$\Delta AEG \sim \Delta ADC$
$\implies \frac{AE}{AD} = \frac{AG}{AC} = \frac{EG}{DC} \implies \frac{2}{5} = \frac{EG}{DC} \implies EG = \frac{2}{5} \times 15 = 6$ cm
Similarly, $\Delta CFG \sim \Delta CBA$ and $\frac{FC}{BF} = \frac{3}{2} \implies \frac{FC}{BC} = \frac{GF}{AB} = \frac{3}{5} \implies GF = \frac{3}{5} \times 10 = 6$ cm
$EF = EG + GF = 6 + 6 = 12$ cm
In $\Delta ADC, EG || DC \implies \frac{AE}{ED} = \frac{AG}{GC} = \frac{2}{3}$
In $\Delta ABC, GF || AB \implies \frac{AG}{GC} = \frac{BF}{FC} = \frac{2}{3}$
$\Delta AEG \sim \Delta ADC$
$\implies \frac{AE}{AD} = \frac{AG}{AC} = \frac{EG}{DC} \implies \frac{2}{5} = \frac{EG}{DC} \implies EG = \frac{2}{5} \times 15 = 6$ cm
Similarly, $\Delta CFG \sim \Delta CBA$ and $\frac{FC}{BF} = \frac{3}{2} \implies \frac{FC}{BC} = \frac{GF}{AB} = \frac{3}{5} \implies GF = \frac{3}{5} \times 10 = 6$ cm
$EF = EG + GF = 6 + 6 = 12$ cm