Let ABCD be a quadrilateral.

Question

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

Accepted Answer

Let the position vectors of $A,B,C,$ and $D$ be $\vec{a},\vec{b},\vec{c},$ and $\vec{d}$, respectively.

Then the position vector of $E$ is:

$\vec{E}=\frac{\vec{a}+\vec{c}}{2}$

And the position vector of $F$ is:

$\vec{F}=\frac{\vec{b}+\vec{d}}{2}$

Now, we are given the equation:

$(\overset{	o \leftarrow }{AB}-\overset{	o \leftarrow }{BC})+(\overset{	o \leftarrow }{AD}-\overset{	o \leftarrow }{DC})=k\overset{	o \leftarrow }{FE}$

We can rewrite this equation using the position vectors:

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

Simplifying the equation, we get:

$(2\vec{b}-2\vec{a}-2\vec{c}+2\vec{d})=\frac{k}{2}(2\vec{E}-2\vec{F})$

Now substitute $\vec{E}$ and $\vec{F}$ expressions we found earlier:

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

Simplifying the equation:

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

Since both sides of the equation are equal:

$k=-4$

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