What's the maximum integer value of the segment (IG) joining the incenter to the centroid in the triangle below? For reference:

Given the obtuse triangle $ABC$, obtuse at $B$,
where $IG \parallel BC$, $I$ being the incenter and $G$ the centroid of this triangle and the perimeter of the triangle is $144$.
Calculate the maximum integer value of $IG$.

My progress:
My draw and the relationships I found
$IG \parallel BC \implies a = \dfrac{b+c}{2}\\
\dfrac{AI}{IJ} = \dfrac{b+c}{2p}=\dfrac{b+c}{144}\\
c^2+a^2=2BM^2+\dfrac{b^2}{2}$
From angle bissector theorem ($\triangle ABC-AN::$),
$\dfrac{BJ}{CJ}=\dfrac{c}{b}\\
AJ^2  =bc-CJ\cdot BJ \\
\triangle ADK \sim \triangle ACB \implies:\\
\dfrac{AM}{CM}=\dfrac{AK}{BK}=\dfrac{b}{c}$
but I don't see where to fit $IG$..


$ \displaystyle \frac{AI}{IJ} = \frac{b+c}{a} = 2 ~ $ (given $IG \parallel BC$)
$b+c = 2a \implies a = 48 ~ $ (using $a+b+c = 144$)
$ \displaystyle IG = \frac23 \cdot JN = \frac23 \left(BN - BJ \right)$
$ \displaystyle = \frac23 \left(\frac a2 - \frac{ac}{b+c} \right) = \frac{a (b-c)}{3 (b+c)} = \frac{b-c}{6}$
(given $a = 48, b + c = 96$)
Using Triangle inequality, we must have $b - c \lt 48$ so the next biggest value of $(b-c)$ that gives an integer value for $IG$ is $42$ and hence max integer value of $IG$ is $7$
