Question

If $M$ and $N$ are the mid-points of the sides $BC$ and $CD$ respectively of a parallelogram $ABCD$, then $AM+AN$ equals

• A. $\frac{4}{3}AC$
• B. $\frac{5}{3}AC$
• C. $\frac{3}{2}AC$
• D. $\frac{6}{5}AC$

Answer 1

Answer: (C)

Since, $M$ and $N$ are the mid-points of $BC$ and $CD$.

$\therefore BM=\frac{1}{2}b$
and $DN=\frac{1}{2}a$
Now, in $\Delta ABM,$
$=a+\frac{1}{2}b$
and in $\Delta ADN$,
$AN=AD+DN$
$=b+\frac{1}{2}a$
$\therefore AM+AN=a+\frac{1}{2}b+b+\frac{1}{2}a$

$=\frac{3}{2}a+\frac{3}{2}b=\frac{3}{2}(a+b)$
$\Rightarrow AM+AN=\frac{3}{2}AC[\because AC=a+b]$