Given : $\vec{a}\times\vec{b}$ = $\vec{c}\times\vec{d}$ and $\vec{a}\times\vec{c}$ = $\vec{b}\times\vec{d}$ To show $\vec{a}-\vec{d}$ is parallel to $\vec{b}-\vec{c}$ i.e $(\vec{a}-\vec{d})$ × $(\vec{b}-\vec{c})$ = 0 Consider $(\vec{a}-\vec{d})$ × $(\vec{b}-\vec{c})$ = $\vec{a}\times(\vec{b}-\vec{c})-\vec{d}\times(\vec{b}-\vec{c})$ = $\vec{a}\times\vec{b}$ - $\vec{a}\times\vec{c}$ - $\vec{d}\times\vec{b}$ + $\vec{d}\times\vec{c}$ $\vec{c}\times\vec{d}$ - $\vec{b}\times\vec{d}$ - $\vec{d}\times\vec{b}$ + $\vec{d}\times\vec{c}$ [Since $\vec{a}\times\vec{b}$ = $\vec{c}\times\vec{d}$ and $\vec{a}\times\vec{c}$ = $\vec{b}\times\vec{d}$] $\vec{c}\times\vec{d}$ - $\vec{b}\times\vec{d}$ + $\vec{b}\times\vec{d}$ - $\vec{c}\times\vec{d}$ [Since $\vec{d}\times\vec{c}$ = $-\vec{c}\times\vec{d}$ and $\vec{d}\times\vec{b}$ = - $\vec{b}\times\vec{d}$] = 0 Therefore $\vec{a}-\vec{d}$ is parallel to $\vec{b}-\vec{c}$