Question

If AD, BE and CF are the medians of $\Delta$ABC, then which one of the following statements is correct?

• A. $(AD+BE+CF)=(AB+BC+CD)$
• B. $(AD+BE+CF)>\frac{3}{4}(AB+BC+CA)$
• C. $(AD+BE+CF)<\frac{3}{4}(AB+BC+CA)$
• D. $(AD+BE+CF)=\frac{1}{2}(AB+BC+CA)$

Answer 1

Answer: (B)

$(AD+BE+CF)>\frac{3}{4}(AB+BC+CA)$

In $△BCG$

$BG+GC>BC$

Now $G$ divides $BE$ in $2:1$

$\therefore BG=\frac{2}{3}BE$

$||ly$ $GC=\frac{2}{3}CF$

$\frac{2}{3}BE+\frac{2}{3}CF>BCBE+CF>\frac{3}{2}BC........\left(i\right)$

Similarly ,,

$CF+AD>\frac{3}{2}CA......\left(ii\right)AD+BE>\frac{3}{2}AB......\left(iii\right)$

Adding $\left(i\right),\left(ii\right)$ and $\left(iii\right)$

$2(AD+BE+CF)>\frac{3}{2}(AB+BC+CA)AD+BC+CF>\frac{3}{4}(AB+BC+CA)$