
Here $AI$ is the internal bisector of $A$ and $AL$ is the altitude through $A$.
Now We know that if $AM$ is the median through $A$ then $AM=BM=MC=\frac12 BC$
In $\triangle ALI$,
$LI^2=4^2-3^2$
$\Rightarrow LI=\sqrt 7$
therefore $\tan \theta = \frac{\sqrt 7}{3}$
From the figure we can see that $\angle LAB=\frac{\pi}{4}+\theta$ and $\angle LAC=\frac{\pi}{4}-\theta$
Now from $\triangle ALB$ and $\triangle ALC$ the value of $BL$ and $LC$ can easily be obtained.
and Median $=\frac12 BC = \frac 12 [BL+LC]$
DONE!!!!