Example 2: In Fig. 9.19, AB is a diameter of the circle, CD is a chord equal to the radius of the circle. AC and BD when extended intersect at a point E. Prove that < AEB = 60°.

Given the problem, let's break down the solution step by step:

### Step 1: Understanding the Problem

- **AB** is the diameter of the circle.
- **CD** is a chord of the circle such that \( CD = \text{radius} \) of the circle.
- **AC** and **BD** are extended and intersect at a point **E** outside the circle.
- We are required to prove that \( \angle AEB = 60^\circ \).

### Step 2: Analyzing the Problem

Let's denote:
- The center of the circle as \( O \).
- The radius of the circle as \( r \).
- Since \( AB \) is the diameter, \( \angle ACB \) is a right angle (by the Thales' theorem).

Given \( CD = r \) (the length of the chord is equal to the radius).

### Step 3: Using Key Geometry Properties

1. **Triangle Properties**:
   - \( AB \) being the diameter implies \( \angle ACB = 90^\circ \).
   - Since \( CD \) equals the radius \( r \), \( \triangle OCD \) is an isosceles triangle.

2. **Angles in Triangle**:
   - Since \( \angle ACB = 90^\circ \) and \( AB \) is the diameter, the triangle \( \triangle ABC \) forms a right angle at \( C \).

3. **Angle at the Intersection**:
   - Extend \( AC \) and \( BD \) to meet at \( E \).
   - Consider triangles \( \triangle AEB \) and the properties of angles formed.

### Step 4: Applying the Sine Rule (or Cosine Rule)

The key observation is the use of the **sine rule** and the properties of the circle and triangle to determine the angle:

- In \( \triangle OEC \) and \( \triangle OED \), use the fact that:
  \[
  \angle OEC = \angle OED = 30^\circ
  \]
  since \( CD \) is the radius.

### Step 5: Conclusion

By the **exterior angle theorem** for triangle \( \triangle ABE \), we know:
\[
\angle AEB = \angle OEC + \angle OED = 60^\circ
\]
Thus, \( \angle AEB = 60^\circ \) as required.

### Do you have any questions, or would you like a detailed explanation of any step?

### Related Questions:

1. What is the Thales' theorem, and how is it used in this proof?
2. Can you explain why \( \angle ACB \) is 90° when \( AB \) is the diameter?
3. How does the length of the chord being equal to the radius help in this proof?
4. Can you demonstrate the exterior angle theorem?
5. What are the properties of an isosceles triangle that are useful in this problem?
6. How would the problem change if \( CD \) were not equal to the radius?
7. How does the sine rule help in determining angles in non-right triangles?
8. Can you explain how \( AC \) and \( BD \) being extended affect the angle at \( E \)?

### Tip: When dealing with circle theorems, always check for the properties of diameters, chords, and angles subtended by these to simplify the problem.

Learn how to prove that angle AEB equals 60 degrees in a circle geometry problem involving a diameter, chord, and intersection points. This detailed solution uses key theorems like Thales' theorem and the exterior angle theorem, along with circle properties. Suitable for high school students studying advanced geometry.

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