If $\overrightarrow{a} \times \overrightarrow{b} = \overrightarrow{c} \times \overrightarrow{d}$ and $\overrightarrow{a} \times \overrightarrow{c} = \overrightarrow{b} \times \overrightarrow{d}$, show that $\overrightarrow{a} - \overrightarrow{d} \parallel \overrightarrow{b} - \overrightarrow{c}$
Given: $\;$ $\overrightarrow{a} \times \overrightarrow{b} = \overrightarrow{c} \times \overrightarrow{d}$ $\;\;\; \cdots \; (1)$; $\;\;\;$
$\overrightarrow{a} \times \overrightarrow{c} = \overrightarrow{b} \times \overrightarrow{d}$ $\;\;\; \cdots \; (2)$
To prove that $\;$ $\overrightarrow{a} - \overrightarrow{d} \parallel \overrightarrow{b} - \overrightarrow{c}$
i.e. $\;$ To prove that $\;$ $\left(\overrightarrow{a} - \overrightarrow{d}\right) \times \left(\overrightarrow{b} - \overrightarrow{c}\right) = \overrightarrow{0}$
Now, $\;$ $\left(\overrightarrow{a} - \overrightarrow{d}\right) \times \left(\overrightarrow{b} - \overrightarrow{c}\right)$
$= \overrightarrow{a} \times \left(\overrightarrow{b} - \overrightarrow{c}\right) - \overrightarrow{d} \times \left(\overrightarrow{b} - \overrightarrow{c}\right)$
$= \left(\overrightarrow{a} \times \overrightarrow{b}\right) - \left(\overrightarrow{a} \times \overrightarrow{c}\right) - \left(\overrightarrow{d} \times \overrightarrow{b}\right) + \left(\overrightarrow{d} \times \overrightarrow{c}\right)$
$= \left(\overrightarrow{c} \times \overrightarrow{d}\right) - \left(\overrightarrow{b} \times \overrightarrow{d}\right) - \left(\overrightarrow{d} \times \overrightarrow{b}\right) + \left(\overrightarrow{d} \times \overrightarrow{c}\right)$ $\;\;\;$ [from $(1)$ and $(2)$]
$= \left(\overrightarrow{c} \times \overrightarrow{d}\right) - \left(\overrightarrow{b} \times \overrightarrow{d}\right) + \left(\overrightarrow{b} \times \overrightarrow{d}\right) - \left(\overrightarrow{c} \times \overrightarrow{d}\right)$
$= \overrightarrow{0}$
$\implies$ $\overrightarrow{a} - \overrightarrow{d} \parallel \overrightarrow{b} - \overrightarrow{c}$ $\;\;\;$ Hence proved.