In a triangle $ABC$ with $ \angle A = 36^o$ and $AB = AC$, the bisector of the angle at $C$ meets the oposite side at $D$. Compute the angles of $\triangle BCD$. Express the length of side $BC$ in terms of the length $b$ of side $AC$ without using trigonometric functions.
1. **Identify the given information and draw the triangle:**
   - Given: \( \angle A = 36^\circ \)
   - \( AB = AC \) (Isosceles triangle)
   - The angle bisector of \( \angle C \) meets \( AB \) at \( D \).

2. **Determine the angles of the triangle:**
   - Since \( \triangle ABC \) is isosceles with \( AB = AC \), the base angles are equal.
   - Let \( \angle B = \angle C = x \).
   - Using the angle sum property of a triangle, we have:
     \[
     \angle A + \angle B + \angle C = 180^\circ
     \]
     \[
     36^\circ + x + x = 180^\circ
     \]
     \[
     2x = 144^\circ
     \]
     \[
     x = 72^\circ
     \]
   - Therefore, \( \angle B = \angle C = 72^\circ \).

3. **Determine the angles of \( \triangle BCD \):**
   - The angle bisector of \( \angle C \) divides \( \angle C \) into two equal parts.
   - Therefore, \( \angle DCB = \frac{72^\circ}{2} = 36^\circ \).
   - Since \( \angle BCD \) is an external angle for \( \triangle BCD \), we have:
     \[
     \angle BCD = \angle B + \angle DCB = 72^\circ + 36^\circ = 108^\circ
     \]
   - The remaining angle \( \angle CBD \) can be found using the angle sum property of a triangle:
     \[
     \angle BCD + \angle CBD + \angle DCB = 180^\circ
     \]
     \[
     108^\circ + \angle CBD + 36^\circ = 180^\circ
     \]
     \[
     \angle CBD = 180^\circ - 144^\circ = 36^\circ
     \]

4. **Express the length of side \( BC \) in terms of \( b \):**
   - Since \( \angle A = 36^\circ \) and \( AB = AC \), \( \triangle ABC \) can be inscribed in a circle with radius \( b \).
   - The side \( BC \) is the side length of a regular decagon inscribed in a circle of radius \( b \).
   - The side length \( s \) of a regular decagon inscribed in a circle of radius \( R \) is given by:
     \[
     s = R \left( \frac{\sqrt{5} - 1}{2} \right)
     \]
   - Therefore, for \( R = b \):
     \[
     BC = b \left( \frac{\sqrt{5} - 1}{2} \right)
     \]

The final answer is \( \boxed{ BC = \frac{b}{2} (\sqrt{5} - 1) } \)