# L and M are the mid-points of sides BC & DC of a paralellogram ABCD. Prove that $\overline{AL}+\overline{AM}=\frac{3}{2}\overline{AC}$

Toolbox:
• Section formula: If C is mid point of AB then position vector of C,$\overrightarrow {OC}=\frac{\overrightarrow {OA}+\overrightarrow {OB}}{2}$
• Triangular law of addition:$\overrightarrow {AB}+\overrightarrow {BC}=\overrightarrow {AC}$
Given ABCD is a parallelogram and L, M are mid points of BC and DC reapectively.
In $\Delta\:ALC$, using triangular law of addition we get
$\overrightarrow {AL}+\overrightarrow {LC}=\overrightarrow {AC}$
But since L is the mid point of BC, $\overrightarrow {LC}=\frac{1}{2}\overrightarrow {BC}$
$\Rightarrow\:\overrightarrow {AL}+\frac{1}{2}\overrightarrow {BC}=\overrightarrow {AC}$
$\Rightarrow\:\overrightarrow {AL}=\overrightarrow {AC}-\frac{1}{2}\overrightarrow {BC}$..................(i)
In $\Delta\:AMC$, using triangular law of addition we get
$\overrightarrow {AM}+\overrightarrow {MC}=\overrightarrow {AC}$
But since M is the mid point of DC, $\overrightarrow {MC}=\frac{1}{2}\overrightarrow {DC}$
$\Rightarrow\:\overrightarrow {AM}+\frac{1}{2}\overrightarrow {DC}=\overrightarrow {AC}$
$\Rightarrow\:\overrightarrow {AM}=\overrightarrow {AC}-\frac{1}{2}\overrightarrow {DC}$..................(ii)
$\Rightarrow\:\overrightarrow {AL}+\overrightarrow {AM}=\overrightarrow {AC}-\frac{1}{2}\overrightarrow {BC}+\overrightarrow {AC}-\frac{1}{2}\overrightarrow {DC}$
$\Rightarrow\:\overrightarrow {AL}+\overrightarrow {AM}=2\overrightarrow {AC}-\frac{1}{2}(\overrightarrow {BC}+\overrightarrow {DC})$
But since ABCD is aparallelogram, $\overrightarrow {DC}=\overrightarrow {AB}$
$\Rightarrow\:\overrightarrow {AL}+\overrightarrow {AM}=2\overrightarrow {AC}-\frac{1}{2}(\overrightarrow {BC}+\overrightarrow {AB})$
But we know from triangular law of addition in $\Delta ABC$,
$\overrightarrow {AB}+\overrightarrow {BC}=\overrightarrow {AC}$
$\Rightarrow\:\overrightarrow {AL}+\overrightarrow {AM}=2\overrightarrow {AC}-\frac{1}{2}\overrightarrow {AC}$
$\Rightarrow\:\overrightarrow {AL}+\overrightarrow {AM}=\frac{3}{2}\overrightarrow {AC}$
Hence proved.