Two angle bisector

Let $AC=x$ and $\angle C = \theta$ . Then we have:

$\begin{aligned} BC: \quad x\cos \theta & = 8\sqrt 2 \cos \frac \theta 2 \\ AB: \quad x \sin \theta & = 9 \cos \left(45^\circ - \frac \theta 2\right) \\ & = \frac 9{\sqrt 2}\left(\cos \frac \theta 2 + \sin \frac \theta 2 \right) \\ \frac {AB}{BC}: \quad \tan \theta & = \frac {9\left(\cos \frac \theta 2 + \sin \frac \theta 2 \right)}{16 \cos \frac \theta 2} \\ & = \frac 9{16} \left(1 + \blue{\tan \frac \theta 2} \right) & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ \frac {32t}{1-t^2} & = 9(1+t) \\ 32t & = 9(1+t-t^2 - t^3) \\ 9t^3 + 9t^2+23t-9 & = 0 \\ \implies t & = \tan \frac \theta 2 = \frac 13 & \small \blue{\implies \cos \frac \theta 2 = \frac 3{\sqrt{10}}} \\ BC: \quad x \cdot \frac 45 & = 8\sqrt 2 \cdot \frac 3{\sqrt{10}} & \small \blue{\implies \cos \theta = 2 \left(\frac 3{\sqrt{10}}\right)^2 - 1 = \frac 45} \\ \implies x & = 3\sqrt{20} \approx \boxed{13.4} \end{aligned}$

We can write $C=90\degree-A,h\cos A=9\cos \dfrac {A}{2}, h\sin A=8\sqrt 2\cos \left (45\degree -\dfrac {A}{2}\right )$ , where $h=|\overline {AC}|$

$\implies 8\left (\cos \dfrac{A}{2}+\sin \dfrac {A}{2}\right ) =h\sin A$

$\implies h^2=\dfrac {64(1+\sin A) }{\sin^2 A}=\dfrac {81(1+\cos A) }{2\cos^2 A}$

$\implies \sin A=\dfrac {81+81\cos A-209\cos^2 A-81\cos^3 A}{128\cos^2 A}$

Squaring, converting into cosine function only and solving we get

$\cos A=\dfrac 35,\sin A=\dfrac 45,h^2=180$ .

So, $h=|\overline {AC}|=\sqrt {180}\approx \boxed {13.4164}$ .