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 $IG$..


As you find out $2a = b+c$ so $a= 48$ and $b+c= 96$.
By angle bisector theorem we have $$BJ = {ac\over b+c}= {c\over 2}$$
Then $$IG = {2\over 3} JN = {2\over 3} ({a\over 2}-{c\over 2})$$
So $$IG = {b-48\over 3}$$ Since $b< a+c \implies b<72 $ so $$IG < {24\over 3} =8\implies IG\leq 7$$
Value $7$ is achieved at $b = 71$ and $ c = 25$ so $IG_{\max} =7$.
