Let $\mathrm{ABCD}$ be a quadrilateral. If $\mathrm{E}$ and $\mathrm{F}$ are the mid points of the diagonals $\mathrm{AC}$ and $\mathrm{BD}$ respectively and $(\overrightarrow{A B}-\overrightarrow{B C})+(\overrightarrow{A D}-\overrightarrow{D C})=k \overrightarrow{F E}$, then $k$ is equal to :

<p>Let the position vectors of $A, B, C,$ and $D$ be $\vec{a}, \vec{b}, \vec{c},$ and $\vec{d}$, respectively.</p> <p>Then the position vector of $E$ is:</p> <p>$\vec{E} = \frac{\vec{a} + \vec{c}}{2}$</p> <p>And the position vector of $F$ is:</p> <p>$\vec{F} = \frac{\vec{b} + \vec{d}}{2}$</p> <p>Now, we are given the equation:</p> <p>$$ (\overrightarrow{AB} - \overrightarrow{BC}) + (\overrightarrow{AD} - \overrightarrow{DC}) = k\overrightarrow{FE} $$</p> <p>We can rewrite this equation using the position vectors:</p> <p>$$ (\vec{b} - \vec{a} - (\vec{c} - \vec{b})) + (\vec{d} - \vec{a} - (\vec{c} - \vec{d})) = k(\vec{E} - \vec{F}) $$</p> <p>Simplifying the equation, we get:</p> <p>$(2\vec{b} - 2\vec{a} - 2\vec{c} + 2\vec{d}) = \frac{k}{2}(2\vec{E} - 2\vec{F})$</p> <p>Now substitute $\vec{E}$ and $\vec{F}$ expressions we found earlier:</p> <p>$$ (2\vec{b} - 2\vec{a} - 2\vec{c} + 2\vec{d}) = \frac{k}{2}\left(2\left(\frac{\vec{a} + \vec{c}}{2}\right) - 2\left(\frac{\vec{b} + \vec{d}}{2}\right)\right) $$</p> <p>Simplifying the equation:</p> <p>$$ (2\vec{b} - 2\vec{a} - 2\vec{c} + 2\vec{d}) = -\frac{k}{2}(\vec{b} + \vec{d} - \vec{a} - \vec{c}) $$</p> <p>Since both sides of the equation are equal:</p> <p>$k = -4$</p>