<s>[INST] Find the distance $ID$ in the triangle below For reference: In the triangle $ABC. \angle A = 2\angle C, IG|| AC$, where $I$ is the incenter and $G$ the centroid. If $IG =x$, calculate the distance $ID$ if $BD$ is an interior angle bisector. (Answer: $3x\sqrt2$)
My progress:

I was able to see these relationships but I couldn't finish
$\frac{BI}{DI} = \frac{a+c}{b}\\
IG \parallel AC \implies b =\frac{a+c}{2} \\
ID = \frac{BD}{3} = 2BI$
Th. Angle Bissector
$\frac{AD}{c}=\frac{CD}{a}\\
BD^2 = a.c - AD.CD\\
\triangle BIG \sim \triangle BDM \implies\\
\frac{BI}{BD} = \frac{x}{DM}\rightarrow \frac{2}{3} = \frac{x}{DM} \therefore DM = \frac{3x}{2}$
Th. Median:
$a^2+c^2 = 2BM^2+\frac{b^2}{2}$
 [/INST] Question says $IG = a$. To avoid confusion, I will instead use $IG = x$ and standard notation for the side lengths of the triangle.
$IG \parallel AC \implies  r = \dfrac h 3$ where $h$ is altitude from $B$ to $AC$
So using $A = r \cdot s, $ where $r$ is inradius, $s$ is sub-perimeter and $A$ is area,
$\displaystyle \frac h 3 = \frac{2 A}{a + b + c} = \frac{2 \cdot b \cdot h / 2}{a+b+c} \implies 2b = a+c$
Now using $\triangle AIB \sim \triangle CDB$
$\displaystyle \frac{BI}{c} = \frac{BD}{a}$
Given $BI = \dfrac{2}{3} BD, c = \dfrac{2a}{3}$
So we have, $ \displaystyle a = \frac{6b}{5}, c = \frac{4b}{5}$
As $ \displaystyle DM = \frac{3x}{2}, \dfrac{a}{c} = \frac{3}{2} = \frac{CD}{AD} = \frac{b + 3x}{b - 3x} \implies b = 15x$
Using formula for the length of angle bisector,
$ \displaystyle BD^2 = ac \left(1 - \frac{b^2}{(a+c)^2}\right) = \frac{3ac}{4} = 162 x^2$
$ \displaystyle \implies BD = 9 x \sqrt2  ~ $ and $ ~ ID = \dfrac{BD}{3} = 3 x \sqrt2$
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