Question

Let us prove that two medians of triangle are together greater than the third median.

Answer 1

Given: $AD,BE$ and $FC$ are medians.
Extend $AD$ to $H$ such that $(AG=HG)$ then join $BH$ and $CH$.
In $△ABH$,
$F$ is mid-point and $G$ is centroid, so use mid-point theorem,
So, FG $\parallel BH$
Similarly, GC || BH
And BG $\parallel HC\dots \left(2\right)$
From (1) and (2),
We get,
$BGCH$ is a parallelogram,
So, $BH=GC$.....(3) (opp. Sides of a parallelogram are equal)
In $\Delta BGH$,
$BG+GH>BH$ (sum of two sides greater than third side)
$\Rightarrow BG+AG>GC$
Similarly, $BE+CF>AD$ and $AD+CF>BE$
Hence proved.