In the given figure, AB || DE and BD || EF. Prove that $DC^2$ = CF x АС

 In the given figure, AB || DE and BD || EF. Prove that $DC^2$ = CF x АС

Since $AB \parallel DE$ and $BD \parallel EF$, triangles formed by these parallel lines with transversals are similar by the basic proportionality theorem (Thales' theorem).

Using similarity properties in $\triangle DCF$ and $\triangle ACD$:

Since $\triangle DCF \sim \triangle ACD$, we have the property of similar triangles:

$ \frac{DC}{AC} = \frac{CF}{DC} $

Cross multiplying:

$ DC^2 = CF \times AC $

$\boxed{DC^2 = CF \times AC}$

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