Let's analyze the problem step by step:

Question Details:

The figure shows several straight lines and points labeled as $A, B, C, D, E, F, G$. The problem involves proving:

1. $\angle ECD = x$ and $\angle ECA = y$.
2. $F, G, D,$ and $C$ are concyclic.

The given conditions include:

• $BA = AE$.
• $\angle DAF = x$ and $\angle AEB = y$.

Approach:

Part (a) $\angle ECD = x$ and $\angle ECA = y$:

1. Note that $BA = AE$ implies symmetry and equal subtended angles in the configuration. Since $\angle DAF = x$, we can trace the arcs and corresponding angles subtended at point $C$.
2. Use properties of isosceles triangles and cyclic quadrilaterals to connect the relationships.

Part (b) $F, G, D,$ and $C$ are concyclic:

1. Use the property that if four points are concyclic, the sum of opposite angles equals $180^\circ$.
2. Show that $\angle FGD + \angle FCD = 180^\circ$ or $\angle FCG + \angle FDG = 180^\circ$.
3. Relate these angles to the given $x$ and $y$ values using cyclic properties and subtended arcs.

Let me know if you'd like me to solve this in detail, or clarify specific aspects!