If `A B C D` is quadrilateral and `Ea n dF` are the mid-points of `A Ca n dB D` respectively, prove that ` vec A B+ vec A D` +` vec C B` +` vec C D` =4 ` vec E Fdot`

Explanation

`therefore` Position vector of `E=frac{1}{2}(vec{a}+vec{c})`


`F` is the mid point of `B D`


`therefore` Position vector of `F=frac{1}{2}(vec{b}+vec{d})`


`therefore vec{E F}=` position vector of `F=` position vector of `hat{t}`



`=frac{1}{2}(vec{b}+vec{d})-frac{1}{2}(vec{a}+vec{c}) `


`=frac{1}{2}(vec{b}+vec{d}-vec{a}-vec{c}) `


`text { Now } vec{A B}+vec{A D}+vec{C B}+vec{C D}=(vec{b} cdot vec{a})+(vec{d}-vec{c})+(vec{d}-vec{c}) `


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