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In the given figure, $AB \parallel DE$ and $AC \parallel DF$. Show that $\triangle ABC \sim \triangle DEF$. If $BC = 10$ cm, $EB = CF = 5$ cm and $AB = 7$ cm, then find the length $DE$.
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Sol. $AB \parallel DE \Rightarrow \angle DEF = \angle ABC$ (I Mark)
$AC \parallel DF \Rightarrow \angle DFE = \angle ACB$
Hence $\triangle ABC \sim \triangle DEF$
$\frac{BC}{EF} = \frac{AB}{DE} \Rightarrow \frac{10}{20} = \frac{7}{DE}$ (II Mark)
$DE = 14$ cm (III Mark)
$AC \parallel DF \Rightarrow \angle DFE = \angle ACB$
Hence $\triangle ABC \sim \triangle DEF$
$\frac{BC}{EF} = \frac{AB}{DE} \Rightarrow \frac{10}{20} = \frac{7}{DE}$ (II Mark)
$DE = 14$ cm (III Mark)