Let's go through each part of the question using properties of cyclic quadrilaterals and angles in circles.

Given:

• $EAC = 72^\circ$
• $AEB = 25^\circ$
• $EB \parallel DC$

We need to find:

1. $\angle EBC$
2. $\angle CXB$
3. $\angle EDC$

Solution:

1. Finding $\angle EBC$:

   Since $EB \parallel DC$ and $EAC = 72^\circ$, we can apply the alternate segment theorem. This tells us that: $\angle EBC = \angle EAC = 72^\circ$

   Answer for (i): $\angle EBC = 72^\circ$

2. Finding $\angle CXB$:

   Since $AEB = 25^\circ$, this angle subtends arc $AB$ in the circle. Therefore, $\angle CXB$, which is an angle subtended by the same arc $AB$ on the opposite side, is equal to $\angle AEB$: $\angle CXB = \angle AEB = 25^\circ$

   Answer for (ii): $\angle CXB = 25^\circ$

3. Finding $\angle EDC$:

   Because $EDC$ is opposite $EAC$ in the cyclic quadrilateral $AEDC$, we use the property that opposite angles in a cyclic quadrilateral add up to $180^\circ$: $\angle EDC = 180^\circ - \angle EAC = 180^\circ - 72^\circ = 108^\circ$

   Answer for (iii): $\angle EDC = 108^\circ$

Final Answers:

1. $\angle EBC = 72^\circ$
2. $\angle CXB = 25^\circ$
3. $\angle EDC = 108^\circ$

Would you like further clarification on any of these steps?

Here are some related questions to expand your understanding:

1. Why do opposite angles in a cyclic quadrilateral add up to $180^\circ$?
2. How can the alternate segment theorem be applied in different situations?
3. What are the properties of angles subtended by the same arc in a circle?
4. How would the solution change if $EB$ were not parallel to $DC$?
5. What other theorems are commonly used in cyclic quadrilateral problems?

Tip: Always look for parallel lines in cyclic quadrilateral problems; they often allow the use of the alternate segment theorem for angle calculations.