Let $ABC$ be an equilateral triangle whose angle bisectors of  $B$ and $C$ intersect at $D$. Perpendicular bisectors of $BD$ and $CD$ intersect $BC$ at points $E$ and $Z$ respectively.
a) Prove that $BE=EZ=ZC$.
b) Find the ratio of the areas of the triangles $BDE$ to $ABC$
### Part (a)
1. **Identify the properties of the equilateral triangle:**
   - Since $ABC$ is an equilateral triangle, all its sides are equal, i.e., $AB = BC = CA$.
   - All its angles are equal, i.e., $\angle A = \angle B = \angle C = 60^\circ$.

2. **Angle bisectors in an equilateral triangle:**
   - The angle bisectors of an equilateral triangle are also the medians and the altitudes.
   - Therefore, $D$ is the centroid of $\triangle ABC$.

3. **Properties of the centroid:**
   - The centroid divides each median in the ratio $2:1$.
   - Hence, $BD = DC$ and $AD = \frac{2}{3} \cdot AM$ where $M$ is the midpoint of $BC$.

4. **Perpendicular bisectors of $BD$ and $CD$:**
   - Let $E$ be the point where the perpendicular bisector of $BD$ intersects $BC$.
   - Let $Z$ be the point where the perpendicular bisector of $CD$ intersects $BC$.

5. **Prove $BE = EZ = ZC$:**
   - Since $E$ and $Z$ are on the perpendicular bisectors of $BD$ and $CD$ respectively, $E$ and $Z$ are equidistant from $B$ and $D$, and $C$ and $D$ respectively.
   - Therefore, $BE = ED$ and $CZ = ZD$.
   - Since $D$ is the centroid, $BD = DC$.
   - Hence, $BE = EZ = ZC$.

### Part (b)
1. **Identify the areas of $\triangle BDE$ and $\triangle ABC$:**
   - Since $D$ is the centroid, it divides $\triangle ABC$ into three smaller triangles of equal area.
   - Therefore, the area of $\triangle ABD = \frac{1}{3}$ of the area of $\triangle ABC$.

2. **Area of $\triangle BDE$:**
   - Since $E$ is on the perpendicular bisector of $BD$, $\triangle BDE$ is an isosceles triangle with $BE = ED$.
   - The area of $\triangle BDE$ is $\frac{1}{3}$ of the area of $\triangle ABD$.

3. **Ratio of the areas:**
   - The area of $\triangle BDE$ is $\frac{1}{3} \times \frac{1}{3}$ of the area of $\triangle ABC$.
   - Therefore, the ratio of the areas of $\triangle BDE$ to $\triangle ABC$ is $\frac{1}{9}$.

\[
\boxed{\frac{1}{9}}
\]