In the figure, $ABC$ and $DAE$ are isosceles triangles ($AB = AC = AD = DE$) and the angles $BAC$ and $ADE$ have measures $36^o$.
a) Using geometric properties, calculate the measure of angle $\angle EDC$.
b) Knowing that $BC = 2$, calculate the length of segment $DC$.
c) Calculate the length of segment $AC$ .
[img]https://1.bp.blogspot.com/-mv43_pSjBxE/XqBMTfNlRKI/AAAAAAAAL2c/5ILlM0n7A2IQleu9T4yNmIY_1DtrxzsJgCK4BGAYYCw/s400/2004%2Bobm%2Bl2.png[/img]
### Part (a)
1. Given that $\triangle ABC$ and $\triangle DAE$ are isosceles triangles with $AB = AC = AD = DE$ and $\angle BAC = \angle ADE = 36^\circ$.
2. Since $\triangle ABC$ is isosceles with $AB = AC$, the base angles $\angle ABC$ and $\angle ACB$ are equal. Let $\angle ABC = \angle ACB = x$.
3. Using the angle sum property of a triangle, we have:
   \[
   \angle BAC + \angle ABC + \angle ACB = 180^\circ
   \]
   Substituting the known values:
   \[
   36^\circ + x + x = 180^\circ \implies 2x = 144^\circ \implies x = 72^\circ
   \]
4. Similarly, in $\triangle DAE$, since it is isosceles with $AD = DE$, the base angles $\angle DAE$ and $\angle DEA$ are equal. Let $\angle DAE = \angle DEA = y$.
5. Using the angle sum property of a triangle, we have:
   \[
   \angle ADE + \angle DAE + \angle DEA = 180^\circ
   \]
   Substituting the known values:
   \[
   36^\circ + y + y = 180^\circ \implies 2y = 144^\circ \implies y = 72^\circ
   \]
6. Now, consider $\triangle ADC$. Since $AD = AC$ and $\angle DAC = 36^\circ$, $\triangle ADC$ is isosceles with $\angle ADC = 72^\circ$.
7. To find $\angle EDC$, note that $\angle EDC = \angle ADC - \angle ADE$:
   \[
   \angle EDC = 72^\circ - 36^\circ = 36^\circ
   \]

### Part (b)
1. From part (a), we know that $\triangle ABC$ and $\triangle DAE$ are congruent by the given conditions.
2. Since $BC = 2$ and $\triangle ABC \cong \triangle ADC$, it follows that $DC = BC = 2$.

### Part (c)
1. Drop a perpendicular from $A$ to $BC$, and call the foot of the perpendicular $Q$. This splits $\triangle ABC$ into two right triangles, $\triangle ABQ$ and $\triangle ACQ$.
2. Since $\angle BAC = 36^\circ$, each of the angles $\angle BAQ$ and $\angle CAQ$ is $18^\circ$.
3. Let $AQ = h$ and $BQ = QC = 1$ (since $BC = 2$ and $Q$ is the midpoint).
4. In $\triangle ABQ$, using the sine function:
   \[
   \sin(18^\circ) = \frac{BQ}{AB} = \frac{1}{AC}
   \]
5. The value of $\sin(18^\circ)$ is known to be:
   \[
   \sin(18^\circ) = \frac{\sqrt{5} - 1}{4}
   \]
6. Therefore:
   \[
   \frac{\sqrt{5} - 1}{4} = \frac{1}{AC} \implies AC = \frac{4}{\sqrt{5} - 1}
   \]
7. Rationalizing the denominator:
   \[
   AC = \frac{4}{\sqrt{5} - 1} \cdot \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{4(\sqrt{5} + 1)}{(\sqrt{5})^2 - 1^2} = \frac{4(\sqrt{5} + 1)}{5 - 1} = \frac{4(\sqrt{5} + 1)}{4} = \sqrt{5} + 1
   \]

The final answer is $\boxed{\sqrt{5} + 1}$