Question

ABCD is a parallelogram. If L, M be the middle points of BC and CD, express $\overline{AL}$ and $\overline{AM}$ in terms of $\overline{AB}$ and $\overline{AC}$ also show that $\overline{AL}+\overline{AM}=\left(\frac{3}{2}\right)\overline{AC}$.

Answer 1

Given $ABCD$ is a parallelogram and $L,M$ are the mid points of $BC$ and $CD$ respectively.

In $△ALC$

$\overrightarrow{AL}+\overrightarrow{LC}=\overrightarrow{AC}$

Since $L$ is the mid point of $BC$,

$\overrightarrow{LC}=\frac{1}{2}\overrightarrow{BC}$

$⟹\overrightarrow{AL}+\frac{1}{2}\overrightarrow{BC}=\overrightarrow{AC}$

$⟹\overrightarrow{AL}=-\frac{1}{2}\overrightarrow{BC}+\overrightarrow{AC}$

In $△AMC$

$\overrightarrow{AM}+\overrightarrow{MC}=\overrightarrow{AC}$

Since $M$ is the mid point of $DC$,

$\overrightarrow{MC}=\frac{1}{2}\overrightarrow{DC}$

$⟹\overrightarrow{AL}+\frac{1}{2}\overrightarrow{DC}=\overrightarrow{AC}$

$⟹\overrightarrow{AL}=-\frac{1}{2}\overrightarrow{DC}+\overrightarrow{AC}$

Hence $\overrightarrow{AL}+\overrightarrow{AM}=-\frac{1}{2}(\overrightarrow{DC}+\overrightarrow{BC})+2\overrightarrow{AC}$

Since $ABCD$ is a parallelogram, $\overrightarrow{DC}=\overrightarrow{AB}$

$\overrightarrow{AL}+\overrightarrow{AM}=-\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{BC})+2\overrightarrow{AC}$

From triangular law of addition in $△ABC$

$\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$

Hence,

$⟹\overrightarrow{AL}+\overrightarrow{AM}=-\frac{1}{2}\left(\overrightarrow{AC}\right)+2\overrightarrow{AC}$

$⟹\overrightarrow{AL}+\overrightarrow{AM}=\frac{3}{2}\overrightarrow{AC}$

Hence proved.