Question

$D,E,F$ are the mid points of $BC,CA,andAB$ of $\Delta ABC$. If $AD$ and $BE$ intersect in $G$, then $AG+BG+CG$ is equal to

• A. $AD=BE=CF$
• B. $\frac{2}{3}(AD+BE+CF)$
• C. $\frac{3}{2}(AD+BE+CF)$
• D. $\frac{1}{3}(AD+BE+CF)$

Answer 1

The correct option is D $\frac{1}{3}(AD+BE+CF)$

(d) Since, $D,E,F$ are the mid points of $BC,CA$ and $AB$ respectively; $AD,BE$ and $CF$ are the medians of $\Delta ABC$. $G$ is the centroid of $\Delta ABC$.
$\therefore G$ divides $AD,BE$ and $CF$ in the ratio $2:1$, i.e.,
$AG:GD=2:1,BG:GE=2:1CG:GF=2:1$
$\Rightarrow AG=\frac{1}{3}AD,BG=\frac{1}{3}BE,CG=\frac{1}{3}CF$
$\Rightarrow AG+BG+CG=\frac{1}{3}(AD+BE+CF)$.