Question

Use vector methods to show that the diagonals of a parallelogram bisect each other.

Short answer

It is proved that the diagonals of a parallelogram bisect each other.

Step-by-step solution

Given

Let us consider a parallelogram $ABCD$, $A$be at origin and $\overrightarrow{AB}=\overrightarrow{a},\overrightarrow{AD}=\overrightarrow{b}$

We have to prove that $H$is the mid-point of $AC\mathrm{and}BD$.

Proof

Let us consider,

$\overrightarrow{AH}=x\overrightarrow{AC}\mathrm{and}\overrightarrow{HB}=y\overrightarrow{DB}...\left(1\right)$

From above figure,

$$\begin{gathered} \overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AB}+\overrightarrow{AD}=\overrightarrow{a}+\overrightarrow{b} \\ \overrightarrow{DB}=\overrightarrow{DA}+\overrightarrow{AB}=\overrightarrow{AB}-\overrightarrow{AD}=\overrightarrow{a}-\overrightarrow{b} \\ \mathrm{From}\mathrm{equation}\left(1\right), \\ \overrightarrow{AH}=x(\overrightarrow{a}+\overrightarrow{b}) \\ \overrightarrow{HB}=y(\overrightarrow{a}-\overrightarrow{b}) \\ \mathrm{Now},\overrightarrow{AB}=\overrightarrow{AH}+\overrightarrow{HB} \\ \Rightarrow \overrightarrow{a}=x(\overrightarrow{a}+\overrightarrow{b})+y(\overrightarrow{a}-\overrightarrow{b}) \\ \Rightarrow \overrightarrow{a}=x\overrightarrow{a}+x\overrightarrow{b}+y\overrightarrow{a}-y\overrightarrow{b} \\ \Rightarrow \overrightarrow{a}=(x+y)\overrightarrow{a}+(x-y)\overrightarrow{b} \end{gathered}$$

Equating the coefficient,

$x+y=1\mathrm{and}x-y=0$

After solving,

$x=\frac{1}{2}\mathrm{and}y=\frac{1}{2}$

Therefore,

$\overrightarrow{AH}=\frac{1}{2}\overrightarrow{AC}\mathrm{and}\overrightarrow{HB}=\frac{1}{2}\overrightarrow{DB}$

Hence, proved.