An angle inside a regular pentagon A student sent me a question and I am stucked. I'd like to use classical geometry, instead of hard trigonometry if possible (can be trigonometric, but I'd like to not use scientific calculator; the only answer I got I need this), but I am in circles. The pentagon $ABCDE$ is regular. The answer is $48^\circ$ (I've constructed in Geogebra). Thank you in advance!

 A: Without loss of generality, let the sides of the pentagon be $1$. Also let $ \mid BF \mid =x$ and $ \mid CF \mid =y$.

Apply the sine rule to $ABF$
\begin{eqnarray*}
\frac{x}{ \sin(54)} =\frac{1}{\sin(102)}.
\end{eqnarray*}
Next apply the cosine rule to $BCF$
\begin{eqnarray*}
y^2=x^2+1-2\cos(84).
\end{eqnarray*}
Cosine rule again, this time on $CDF$
\begin{eqnarray*}
1= 2y^2-2y^2\cos(\theta).
\end{eqnarray*}
Plug that into your casio (other brands of calculator are available) & you get $ \theta= \color{red}{48^{\circ}}$.
With such a neat final answer you certainly get the feeling there could be a much more elegant method ?
A: Consider the point T such that
 TE = SE = SB
 Be the angle SET = 60
 So the triangle SET is equal to time.
 On the other hand
 CTD = TED = ASE
 So the angle specified in the figure is proven (because I didn't have the right to post the photo)
 We know STC = 96 and because ST = CT
 So CST = SCT = 42
 So SCD = SDC = 66 and so on
 CSD = 48
A: Let $G$ be reflection of $A$ about $EF$. Clearly $EG=AE=ED$ and $\angle GED = \angle AED - \angle AEF - \angle FEG = 108^\circ - 24^\circ -24^\circ=60^\circ$. Hence $GED$ is an equilateral triangle. So $GD=DE=DC$ and $\angle GDC=48^\circ$. Hence $\angle DCG=\angle CGD=66^\circ$, $\angle DGE=60^\circ$, and $\angle EGF =\angle FAE = 54^\circ$. So angles $CGD, DGE, EGF$ sum up to $180^\circ$. So $G$ lies on $FC$. So $\angle DCF = 66^\circ$ and by symmetry $\angle FDC=66^\circ$. It follows that $\angle CFD=48^\circ$.
