In $\triangle A B C$, if $D$ and $E$ are the mid-points of the sides $B C$ and $C A$ respectively, then…
- $3 \mathrm{AB}$
- $\frac{3}{2} \mathrm{AB}$
- $2 \mathrm{AB}$
- 3BC
Solution


On adding Eqs. (i) and (ii), we get $ \begin{aligned} \mathbf{A D}+\mathbf{E B} & =\frac{3}{2} \mathbf{A C}+\frac{3}{2} \mathbf{C B} \\ & =\frac{3}{2}(\mathbf{A C}+\mathbf{C B})=\frac{3}{2} \mathbf{A B} \\ \Rightarrow 2(\mathbf{A D}+\mathbf{E B}) & =3 \mathbf{A B} \end{aligned} $
Asked in: AP EAMCET 2018 (24 Apr Shift 1)