Question

$M$ and $N$ are the midpoints of the diagonals $AC$ and $BD$ respectively of quadrilateral $ABCD$, then $\overline{AB}+\overline{AD}+\overline{CB}+\overline{CD}=$................

• A. $2\overline{MN}$
• B. $2\overline{NM}$
• C. $4\overline{MN}$
• D. $4\overline{NM}$

Answer 1

The correct option is C $4\overline{MN}$

Let $a,b,c,d,m,n$ are the position vectors of $A,B,C,D,M$ and $N$ respectively

$M$ and $N$ are the midpoints of t diagonals $AC$ and $BD$ respectively ..... [Given]

$\therefore m=\frac{a+c}{2}$ and $n=\frac{b+d}{2}$

$⟹2m=a+c$ and $2n=b+d$ ........ $\left(i\right)$

Now, consider $\overline{AB}+\overline{AD}+\overline{CB}+\overline{CD}$

$=(b-a)+(d-a)+(b-c)+(d-c)$

$=2b-2a-2c+2d$

$=2(b+d)-2(a+c)$

$=2\left(2n\right)-2\left(2m\right)$

$=4n-4m=4(n-m)=4\overline{MN}$

Hence, $\overline{AB}+\overline{AD}+\overline{CB}+\overline{CD}=4\overline{MN}$