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To solve this problem, let's address each part step-by-step:

### Given Information and Setup
1. **Diagram details**: \( D \) is the center of the semi-circle on \( AB \) with \( ED \perp AB \). Thus, \( D \) is the midpoint of \( AB \).
2. **Triangle Relations**: \( \triangle BAC \) and \( \triangle BED \) are important triangles here.

### Solution Steps

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#### 2.1 Prove that \( \triangle BAC \parallel \triangle BED \)

To prove that \( \triangle BAC \parallel \triangle BED \), we need to show that they are similar. One approach is to use angle similarity:

1. Since \( D \) is the center of the semi-circle, \( \angle BAC = 90^\circ \) (a right angle in a semi-circle).
2. \( \angle BED = 90^\circ \) as given, because \( ED \perp AB \).
3. Both triangles share \( \angle ACB \) as a common angle.

Since both triangles have two equal angles (\( 90^\circ \) and \( \angle ACB \)), they are similar by the **AA (Angle-Angle) similarity criterion**.

Thus, \( \triangle BAC \parallel \triangle BED \).

---

#### 2.2 Deduce that \( \frac{AB}{BE} = \frac{BC}{BD} \)

Since \( \triangle BAC \sim \triangle BED \) by similarity, the corresponding sides of similar triangles are proportional. Therefore:

\[
\frac{AB}{BE} = \frac{BC}{BD}
\]

---

#### 2.3 Prove that \( \frac{BE \times BC}{BD} = \sqrt{AC^2 + BC^2} \)

To prove this, consider the Pythagorean relationship in \( \triangle BAC \) (since \( \angle BAC = 90^\circ \)):

1. From \( \triangle BAC \), we have:
   \[
   AC^2 + BC^2 = AB^2
   \]
2. Using the proportion found in 2.2:
   \[
   AB = \frac{BE \times BC}{BD}
   \]
3. Substitute \( AB \) in terms of \( AC \) and \( BC \) using the relationship:
   \[
   \frac{BE \times BC}{BD} = \sqrt{AC^2 + BC^2}
   \]

---

#### 2.4 If \( BE = 25 \) units and \( AB = 40 \) units, determine the length of \( CE \).

Given \( BE = 25 \) units and \( AB = 40 \) units:

1. Since \( D \) is the midpoint of \( AB \), \( BD = \frac{AB}{2} = \frac{40}{2} = 20 \) units.
2. Use the proportion from part 2.2:
   \[
   \frac{AB}{BE} = \frac{BC}{BD}
   \]
   Substitute the known values to find \( BC \):
   \[
   \frac{40}{25} = \frac{BC}{20}
   \]
3. Solving for \( BC \):
   \[
   BC = \frac{40 \times 20}{25} = 32 \, \text{units}
   \]
4. Now, using the Pythagorean theorem in \( \triangle CEB \) (since \( \angle BED = 90^\circ \)):
   \[
   CE = \sqrt{BE^2 + BC^2} = \sqrt{25^2 + 32^2} = \sqrt{625 + 1024} = \sqrt{1649}
   \]
   Thus, \( CE \approx 40.62 \) units.

---

### Summary of Answers
1. **2.1**: \( \triangle BAC \parallel \triangle BED \).
2. **2.2**: \( \frac{AB}{BE} = \frac{BC}{BD} \).
3. **2.3**: \( \frac{BE \times BC}{BD} = \sqrt{AC^2 + BC^2} \).
4. **2.4**: \( CE \approx 40.62 \) units.

Would you like further details or have any specific questions?

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### Follow-up Questions
1. How would you prove the similarity between two other triangles in a different context?
2. Can the proportionality approach used here be applied to any pair of similar triangles?
3. What is the significance of having \( D \) as the midpoint in this problem?
4. How would the solution change if \( AB \) were not the diameter?
5. Can you explain why \( \angle BAC \) is always \( 90^\circ \) when \( D \) is the center of the semi-circle?

### Tip
Remember that in similar triangles, corresponding angles are equal, and the ratios of corresponding sides are proportional. This is a powerful property for solving many geometric problems.

In the given diagram, D is the center of the semi-circle ACB with ED perpendicular to AB. Prove that:

2.1 ΔBAC || ΔBED
2.2 Hence deduce that AB = (BE * BC) / BD
2.3 Prove that (BE * BC) / BD = √(AC² + BC²)
2.4 If BE = 25 units and AB = 40 units, determine the length of CE.

Solve a geometry problem involving a semi-circle with a center at D and a perpendicular line ED to AB. Prove similarity of triangles ΔBAC and ΔBED, deduce a proportionality relationship, and apply the Pythagorean theorem to determine lengths. Suitable for high school students in Grades 10-12.

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