Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.
Given $\ln \triangle A B C, A D$ is a median.
Construction
Produce $A D$ to a point $E$ such that $A D=D E$ and join $C E$.
To prove $A C+A B>2 A D$
Proof In $\triangle A B D$ and $\triangle E C D$,
$A D=D E$ [by construction]
$B D=C D$ [given $A D$ is the median]
and $\angle A D B=\angle C D E$ [vertically opposite angle]
$\therefore$ $\triangle \mathrm{ABD} \equiv \triangle \mathrm{ECD}$ [by SAS congruence rule]
$\Rightarrow$ $A B=C E$ [by CPCT] ...(i)
Now, in $\triangle A E C$,
$A C+E C>A E$ [sum of two sides of a triangle is greater than the third side]
$\therefore \quad A C+A B>2 A D$ [from Eq. (i) and also taken that $A D=D E]$
Hence proved.
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