# An angle inside a regular pentagon

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 $$angle CFD=48^circ$$.

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 $$BCF$$
$$begin{eqnarray*} y^2=x^2+1-2cos(84). end{eqnarray*}$$
Cosine rule again, this time on $$CDF$$
$$begin{eqnarray*} 1= 2y^2-2y^2cos(theta). end{eqnarray*}$$
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 ?

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