The answer is $75$. : Let $K=(AFE) \cap BC$.We have $\angle{FAE} =30$. We know that $AE=AF$ $\Longrightarrow \angle{AEF} =\angle{AFE} =75 \Longrightarrow \angle{FKC} =\angle{EKB} =75$. Now let $T=(KEB) \cap (FKC) $. It is enough to prove that $B, T$ and $F$ are collinear (in the same way we can prove that $C, T$ and $E$ are collinear). We have : $$\angle{KTB} =\angle{KEB} =60=180-\angle{FTK} $$As desired... $\blacksquare$

Since, $AC=EA \implies \Delta CFA \text{ is isosceles-right triangle } \implies \boxed{\angle ECA=45^{\circ}}$, If $AB=x $ and Let $FT \perp CB$ such, $T \in CB$, Hence, $$FT=\frac{x\sqrt{3}}{2} \text{ and } TB=\frac{3x}{2} \implies \tan \angle CBF =\frac{\tfrac{x\sqrt{3}}{2}}{\tfrac{3x}{2}}=\frac{1}{\sqrt{3}} \implies \boxed{\angle CBF=30^{\circ}}$$ Therefore, the acute angle between the lines $CE$ and $FB$ will be $75^{\circ}$

Extremely trivial. $\angle ACE=45^\circ$, $AB=AC=AF \implies \angle BFC=90^\circ, \angle CBF=30^\circ, \angle ? = 75^\circ$.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20X.pdf p. 38... Sincerely Jean-Louis

Observe that $BC=BD$. Then, we have $\angle BCD=45^\circ$. Observe that $AB=BF$. We have \[\angle AFB=\frac{180^\circ - \angle ABF}{2} = \frac{\angle CBF}{2}=30^\circ\]Then, $\angle HFC=90^\circ$. Because $\angle BCF=60^\circ$, we have \[\angle DCF=\angle BCF-\angle BCD=60^\circ-45^\circ=15^\circ\]And then \[\angle FGC=180^\circ-\angle GFC-\angle GCF=180^\circ-90^\circ -15^\circ = \boxed{75^\circ}\]

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