Question

If D, E are the midpoints $\overline{AB},\overline{AC}$ of $△ABC$, then $\overline{BE}+\overline{DC}=$

• A. $\overline{BC}$
• B. $\frac{1}{2}\overline{BC}$
• C. $2\overline{BC}$
• D. $\frac{3}{2}\overline{BC}$

Answer 1

The correct option is D $\frac{3}{2}\overline{BC}$

Let position vector of $\overrightarrow{A}=\overrightarrow{a}$

Let position vector of $\overrightarrow{B}=\overrightarrow{b}$

Let position vector of $\overrightarrow{C}=\overrightarrow{c}$

Let position vector of $\overrightarrow{D}=\overrightarrow{d}$

Let position vector of $\overrightarrow{E}=\overrightarrow{e}$

$\Rightarrow \overrightarrow{d}=\frac{\overrightarrow{a}+\overrightarrow{b}}{2}\to \left(1\right)$

$\Rightarrow \overrightarrow{e}=\frac{\overrightarrow{a}+\overrightarrow{c}}{2}\to \left(2\right)$

$\overrightarrow{BE}+\overrightarrow{DC}=(\overrightarrow{e}-\overrightarrow{b})+(\overrightarrow{c}-\overrightarrow{d})=(\overrightarrow{e}-\overrightarrow{b})+(\overrightarrow{c}-\overrightarrow{b})\to \left(3\right)$

Subtracting (1) from (2),

$\overrightarrow{e}-\overrightarrow{d}=\frac{\overrightarrow{c}-\overrightarrow{b}}{2}$

Adding $(\overrightarrow{c}-\overrightarrow{b})$ on both sides

$\Rightarrow (\overrightarrow{e}-\overrightarrow{d})+(\overrightarrow{c}-\overrightarrow{b})=(\overrightarrow{c}-\overrightarrow{b})+\frac{\overrightarrow{c}-\overrightarrow{b}}{2}=\frac{3}{2}(\overrightarrow{c}-\overrightarrow{b})$

$\Rightarrow \left(3\right)=\overrightarrow{BE}+\overrightarrow{DC}=\frac{3}{2}\overrightarrow{BC}$