Let $A B C D$ be a parallelogram and let $L$ and $M$ be the midpoints of the sides $B C$ and $C D$ respectively. Then $\to A L + \to A M =$

\to A C

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\frac{2}{3} \to A C

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\frac{3}{2} \to A C

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2 \to A C

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Solution

The correct option is C $\frac{3}{2} \to A C$
$B L = L C$
$C M = D M$
$\to A B = \to D C = \to a$
$\to A D = \to B C = \to b$
$\to A L = \to A B + \to B L$
$\to A M = \to A D + \to D M$
$\to A L + \to A M = \to A B + \to \frac{B C}{2} + \to A D + \to \frac{D C}{2}$
$= \frac{3}{2} (\to A B + \to B C)$
$= \frac{3}{2} \to A C$

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