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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?

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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.

A, B, C, D, and E are five points on the circumference of a circle. EB is parallel to DC, EAC = 72°, AEB = 25°, and X is the intersection of AC and EB. Find (i) angle EBC, (ii) angle CXB, and (iii) angle EDC.

Explore how to solve for angles EBC, CXB, and EDC in a cyclic quadrilateral with parallel lines and points on a circle. This question covers concepts like the alternate segment theorem and properties of cyclic quadrilaterals, suitable for Grades 9-10.

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