Consider the diagram below where  {jatex options:inline}\alpha = 36^o{/jatex}.

Triangles ABC and CBE are both right angles, have a side in common and a corresponding equal angle  {jatex options:inline}\alpha{/jatex}. They are congruent triangles.
Let  {jatex options:inline}AC=AD=x{/jatex}  then  {jatex options:inline}AE=ED=x/2{/jatex}.
Using trigonometry on triangle  {jatex options:inline}ABE{/jatex}  gives  {jatex options:inline}AB=\frac{x/2}{cos \alpha}{/jatex}  then  {jatex options:inline}CB=\frac{x/2}{cos \alpha}{/jatex}  too.
From triangle  {jatex options:inline}BCD{/jatex}  with  {jatex options:inline}\alpha=36^o{/jatex},  {jatex options:inline}\angle DBC=180^o-3 \times 36=72^o=2 \alpha{/jatex}.
Then triangle  {jatex options:inline}BCD{/jatex}  is isosceles and  {jatex options:inline}CD=\frac{x/2}{cos \alpha}{/jatex}.
Also,  {jatex options:inline}CD=x-\frac{x/2}{cos \alpha}{/jatex}.

Now use the Cosine Rule on triangle  {jatex options:inline}BCD{/jatex}. {jatex options:inline}(x-\frac{x/2}{cos \alpha})^2=(\frac{x/2}{cos \alpha})^2+(\frac{x/2}{cos \alpha})^2-2(\frac{x/2}{cos \alpha})(\frac{x/2}{cos \alpha})^2 cos \alpha{/jatex}
Expanding the brackets, Dividing by  {jatex options:inline}x^2{/jatex}  and multiplying by  {jatex options:inline}4cos^2 \alpha{/jatex}  gives
{jatex options:inline}4cos^2 \alpha-4cos \alpha +1=2-2cos \alpha \rightarrow 4cos^2 \alpha-2 cos \alpha-1=0{/jatex} The from the quadratic formula  {jatex options:inline}cos \alpha=\frac{--2 \pm \sqrt{20}}{2 \times 4}= \frac{1 + \sqrt{5}}{4}{/jatex}.
We ignore the minus option because this returns a negative value for  {jatex options:inline}cos \alpha{/jatex}  but  {jatex options:inline}cos \alpha \gt 0{/jatex}.