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Q.
If $M$ and $N$ are the mid-points of the sides $B C$ and $C D$ respectively of a parallelogram $A B C D$, then $AM + AN$ equals
TS EAMCET 2015
Solution:
Since, $M$ and $N$ are the mid-points of $B C$ and $C D$.
$\therefore B M=\frac{1}{2} b $
and $ D N=\frac{1}{2} a$
Now, in $\Delta A B M,$
$= a +\frac{1}{2} b$
and in $\Delta A D N$,
$ A N =A D+D N $
$= 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} A C [\because AC = a + b ] $
