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In the given figure, $AB || DE$ and $BD || EF$. Prove that $DC^2 = CF \times AC$.
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$EF || BD \Rightarrow \frac{CF}{DC} = \frac{CE}{CB}$ (i)
$DE || AB \Rightarrow \frac{DC}{AC} = \frac{CE}{CB}$ (ii)
Using (i) & (ii), $\frac{CF}{DC} = \frac{DC}{AC} \Rightarrow DC^2 = CF \times AC$
$DE || AB \Rightarrow \frac{DC}{AC} = \frac{CE}{CB}$ (ii)
Using (i) & (ii), $\frac{CF}{DC} = \frac{DC}{AC} \Rightarrow DC^2 = CF \times AC$