If ΔABC ~ ΔEDF and ΔABC is not similar

Solution:

(c) Given,       $\triangle A B C \sim \triangle E D F$

$\therefore$ $\frac{A B}{E D}=\frac{B C}{D F}=\frac{A C}{E F}$

Taking first two terms, we get

$\frac{A B}{E D}=\frac{B C}{D F}$

$\Rightarrow$ $A B \cdot D F=E D \cdot B C$

Or $B C \cdot D E=A B \cdot D F$

So, option (d) is true.

Taking last two terms, we get

$\frac{B C}{D F}=\frac{A C}{E F}$

$\Rightarrow \quad B C \cdot E F=A C \cdot D F$

So, option (a) is also true.

Taking first and last terms, we get

$\frac{A B}{E D}=\frac{A C}{E F}$

$\Rightarrow \quad A B \cdot E F=E D \cdot A C$

Hence, option (b) is true.