
Here is some information about regular pentagons.
$ABCDE$ is a regular pentagon.
We first prove that triangles $AEX$ and $ADC$ are similar and
that the ratio $AD/AE$ is equal to the golden ratio ${1+\sqrt
5\over 2}$.

As $ABCDE$ is a regular pentagon, triangles $AEX$ and $ADC$
have angles $36^o, 72^o, 72^o$ and so they are similar isosceles
triangles. Label the side length of the pentagon $s$ so
$AE=AX=CD=CX=s$ and the chord length $c$ so $AD=EC = c$ and
$EX=cs$. From the similar triangles $${cs\over s}={s\over c}$$ so
writing $AD/AE=c/s = x$ gives $$x  1 = {1\over x}$$ which is the
quadratic equation $x^2x1=0$ and hence, as it must be positive,
the ratio $x=AD/AE$ is the golden ratio ${1+\sqrt 5\over 2}$.
Now we can find the exact values of $\cos 72^o$ and $\cos
36^o$.
Drawing a perpendicular from $E$ to $AD$ gives a right angled
triangle from which $$\cos 36^o = {c\over 2s} ={\sqrt 5+1\over
4}.$$
Drawing a perpendicular from $A$ to $DC$ gives a right angled
triangle from which $$\cos 72^o = {s\over 2c} ={1\over2\phi}={\sqrt
51\over 4}.$$
In explaining the construction of a regular pentagon
coordinates are useful.
If the pentagon is inscribed in a unit circle with $A$ the
point $(0,1)$ you can find the exact coordinates of $B$ and $C$
using the exact values of $\cos 72^o$ and $\cos 36^o$ which we have
just found.
If the pentagon is inscribed in a unit circle with $A$ the
point $(0,1)$ then $B=(\sin 72^o, \cos 72^o)$ and $C=(\sin 36^o,
\cos 36^o)$.
Hint: When you explain the construction focus on the $'y'$
coordinates of $B, C, D$ and $E$.