<s>[INST] What is the value of $ \angle x $ shown below? For reference: AB=CF

My progress:
This is a typical Peruvian exercise where the solutions are provided by auxiliary lines. I tried one way but I didn't find the solution

 [/INST] Construct equilateral triangle $ABE$. Because $\angle AEB=2 \angle BCA$, $E$ is the circumcenter of $ABC$. Thus, $AB=FC=CE=BE=AE$. $\angle EAF=60^\circ-24^\circ=36^\circ, \angle AEC=180^\circ-72^\circ=108^\circ, \angle BEC=108^\circ-60^\circ=48^\circ, $ $\angle ECB=90^\circ-24^\circ=66^\circ, \angle ACE=36^\circ, \angle CFE=72^\circ, \angle FEA=36^\circ \implies AF=FE$  Therefore $\triangle ABF \cong \triangle BFE$ (SSS)Finally, $2\angle AFB=360^\circ-108^\circ, \angle AFB=126^\circ=x+30^\circ, x=96^\circ$

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