Let $$ABCD$$ be a quadrilateral. If $$E$$ and $$F$$ are the mid points of the diagonals $$AC$$ and $$BD$$ respectively and $$\left(\vec{AB} - \vec{BC}\right) + \left(\vec{AD} - \vec{DC}\right) = k\vec{FE}$$, then $$k$$ is equal to

Let's use position vectors. Let positions of A, B, C, D be $$\vec{a}, \vec{b}, \vec{c}, \vec{d}$$.

E is midpoint of AC: $$\vec{e} = \frac{\vec{a}+\vec{c}}{2}$$

F is midpoint of BD: $$\vec{f} = \frac{\vec{b}+\vec{d}}{2}$$

$$\vec{FE} = \vec{e} - \vec{f} = \frac{\vec{a}+\vec{c}}{2} - \frac{\vec{b}+\vec{d}}{2} = \frac{\vec{a}+\vec{c}-\vec{b}-\vec{d}}{2}$$

Now compute $$(\vec{AB} - \vec{BC}) + (\vec{AD} - \vec{DC})$$:

$$\vec{AB} = \vec{b}-\vec{a}$$, $$\vec{BC} = \vec{c}-\vec{b}$$, $$\vec{AD} = \vec{d}-\vec{a}$$, $$\vec{DC} = \vec{c}-\vec{d}$$

$$(\vec{AB}-\vec{BC}) + (\vec{AD}-\vec{DC})$$

$$= (\vec{b}-\vec{a}-\vec{c}+\vec{b}) + (\vec{d}-\vec{a}-\vec{c}+\vec{d})$$

$$= 2\vec{b} - 2\vec{a} - 2\vec{c} + 2\vec{d}$$

$$= 2(\vec{b}+\vec{d}-\vec{a}-\vec{c})$$

$$= -2(\vec{a}+\vec{c}-\vec{b}-\vec{d})$$

$$= -4 \times \frac{\vec{a}+\vec{c}-\vec{b}-\vec{d}}{2} = -4\vec{FE}$$

So $$k = -4$$.

This matches option 4.