To prove that \( DC \) is perpendicular to \( AE \) using only Euclidean geometry, let's break down the proof step by step.

### Given:
- \( \triangle DBE \) is a right-angled isosceles triangle with \( \angle DBE = 90^\circ \) and \( DB = BE \).
- \( \triangle ABC \) is also a right-angled isosceles triangle with \( \angle ABC = 90^\circ \) and \( AB = BC \).
- We are required to prove that \( DC \) is perpendicular to \( AE \).

### Step 1: Consider the properties of the triangles
Since \( \triangle DBE \) and \( \triangle ABC \) are isosceles and right-angled:
- In \( \triangle DBE \), \( \angle BDE = \angle BED = 45^\circ \).
- In \( \triangle ABC \), \( \angle BAC = \angle BCA = 45^\circ \).

### Step 2: Investigate triangle relationships and congruences
Let's look at the triangles \( \triangle BDC \) and \( \triangle AEB \).

#### 1. \( \triangle BDC \):
- \( \angle BDC \) is made up of \( \angle BDA \) and \( \angle CDA \).
- \( \angle BDA = \angle BED = 45^\circ \) (since \( \angle BED \) and \( \angle BDA \) are vertical angles).
- \( \angle CDA \) is part of \( \triangle ABC \), where \( \angle BCA = 45^\circ \).
- Therefore, \( \angle BDC = 45^\circ + 45^\circ = 90^\circ \).

#### 2. \( \triangle AEB \):
- Similarly, \( \angle AEB = \angle AEC + \angle CEB \).
- \( \angle CEB \) is part of \( \triangle ABC \), where \( \angle BAC = 45^\circ \).
- \( \angle AEC \) is in \( \triangle DBE \), where \( \angle DBE = 90^\circ \).
- \( \angle AEB = 45^\circ + 45^\circ = 90^\circ \).

### Step 3: Conclusion using perpendicularity
Since \( \angle BDC = 90^\circ \) and \( \angle AEB = 90^\circ \), \( DC \) and \( AE \) must be perpendicular to each other by definition.

### Summary
By analyzing the angle relationships in the two right-angled isosceles triangles and using their properties, we can conclude that \( DC \) is perpendicular to \( AE \).

Would you like any further clarification or have any additional questions?

### Related Questions:
1. How do angle bisectors in right-angled isosceles triangles relate to perpendicular lines?
2. Can this proof be extended if \( \triangle ABC \) and \( \triangle DBE \) are not isosceles?
3. What happens if \( \triangle ABC \) and \( \triangle DBE \) share more than one point?
4. How does the concept of vertical angles assist in proving perpendicularity in Euclidean geometry?
5. What role does the Pythagorean Theorem play in verifying perpendicularity in right triangles?

### Tip:
In Euclidean geometry, recognizing key angle relationships, particularly in right-angled triangles, often leads directly to conclusions about perpendicularity and parallelism.

Learn how to prove perpendicularity in Euclidean geometry using right-angled isosceles triangles. This detailed solution examines the properties of triangles and angle relationships to demonstrate perpendicularity. Suitable for advanced high school students.

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