What is the length of side AB of the triangle? Point X is 3m from A, 4m from B and 5m from C. Point X is inside the triangle formed by ABC. If AB = BC and angle B is right angle. Find the length of side AB.
I have come up with the following figure and aware that this is an isosceles right triangle, unfortunately i'm not sure what to do next.



Reflect point $X$ across the sides of $\triangle ABC$. Observe,
$$\begin{align*}
\angle DAF&=\angle DAX+\angle FAX=2(\angle BAX+\angle CAX)=90^{\circ}\;\implies DF=3\sqrt{2}\\
\angle DCE&=\angle DCX+\angle ECX=2(\angle ACX+\angle BCX)=90^{\circ}\;\implies DE=5\sqrt{2}\\
\angle FBE&=\angle FBX+\angle EBX=2(\angle ABX+\angle CBX)=180^{\circ}\implies F,\;B,\;E \;\;\text{are collinear.}
\end{align*}
$$
Using the cosine rule, in $\triangle EDF,$
$$\begin{align*}
EF^2&=DE^2+DF^2-2\cdot DE\cdot DF\cdot\cos\angle EFD \\
64&=18+50-60\cdot\cos\angle EFD \\
\cos \angle EFD&=\frac{1}{15} \implies \sin\angle EFD=\frac{4\sqrt{14}}{15}
\end{align*} $$
Using the cosine rule, in $\triangle ADC,$
$$\begin{align*}
AC^2 &=AD^2+CD^2-2\cdot AD\cdot CD\cdot \cos\angle ADC\\
 &=9+25-30\cos(90^{\circ}+\angle EFD)\\
&=34+30\cdot\sin\angle EFD\\
&=34+8\sqrt{14} \\ \\
\therefore\; AB&=\sqrt{\frac{AC^2}{2}}=\boxed{\sqrt{17+4\sqrt{14}}\approx 5.6539}
\end{align*}$$
