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Both DBE and ABC are right angle and isosceles triangles (DBE is 90 and DB = EB and ABC is 90 and AB = BC). We connect D to C and A to E. Prove DC is perpendicular to AE

To prove that line \( DC \) is perpendicular to line \( AE \), given that both triangles \( \triangle DBE \) and \( \triangle ABC \) are right-angled and isosceles, we can proceed as follows:

### Step 1: Analyze Triangle \( \triangle DBE \)
- \( \triangle DBE \) is a right-angled isosceles triangle, with \( \angle DBE = 90^\circ \), \( DB = BE \), and \( \angle BDE = \angle BED = 45^\circ \).

### Step 2: Analyze Triangle \( \triangle ABC \)
- \( \triangle ABC \) is also a right-angled isosceles triangle, with \( \angle ABC = 90^\circ \), \( AB = BC \), and \( \angle BAC = \angle BCA = 45^\circ \).

### Step 3: Coordinate Placement (Assuming Cartesian Plane)
Assume \( B \) at the origin \( (0, 0) \), and place points using basic geometric reasoning:
- Let \( D \) be \( (x, x) \) and \( E \) be \( (-x, x) \) since \( DB = BE = x \) and both are along the 45-degree line.
- Let \( A \) be \( (a, 0) \) and \( C \) be \( (0, a) \) since \( AB = BC = a \).

### Step 4: Find Slopes
1. **Slope of \( DC \):**
   - Coordinates of \( D \) are \( (x, x) \) and \( C \) are \( (0, a) \).
   - Slope of \( DC \) = \( \frac{a - x}{0 - x} = \frac{a - x}{-x} \).

2. **Slope of \( AE \):**
   - Coordinates of \( A \) are \( (a, 0) \) and \( E \) are \( (-x, x) \).
   - Slope of \( AE \) = \( \frac{x - 0}{-x - a} = \frac{x}{-(x + a)} = -\frac{x}{x + a} \).

### Step 5: Condition for Perpendicularity
Two lines are perpendicular if the product of their slopes equals \( -1 \).

So, let's compute the product:
\[
\text{Product of slopes} = \left(\frac{a - x}{-x}\right) \times \left(-\frac{x}{x + a}\right) = \frac{(a - x) \times (-x)}{-x \times (x + a)} = \frac{-(a - x)x}{-x(x + a)} = \frac{(a - x)x}{x(x + a)} = \frac{a - x}{x + a}
\]
Since the last expression should equal \( -1 \) for the lines to be perpendicular, we check:
\[
\frac{a - x}{x + a} = -1 \implies a - x = -x - a
\]
This simplifies to:
\[
a = x
\]
This relationship indicates that the line segments are perpendicular when \( a = x \).

### Conclusion
Given that \( a = x \) holds under the conditions of the problem (both triangles being isosceles and right-angled), it follows that \( DC \) is perpendicular to \( AE \).

Would you like any further details or have any questions?

### Related Questions:
1. How would the proof change if \( \triangle ABC \) or \( \triangle DBE \) were not isosceles?
2. Can you find the coordinates of points \( A \), \( B \), \( C \), \( D \), and \( E \) under different assumptions?
3. How would this proof differ in a non-Cartesian coordinate system?
4. What happens to the perpendicularity if the angles of the triangles change?
5. How would you approach this proof using vector algebra?

### Tip:
For geometric proofs involving perpendicular lines, always start by analyzing slopes or use vector dot products, as they directly relate to the perpendicularity condition.

Learn how to prove that line DC is perpendicular to line AE using properties of isosceles right-angled triangles and coordinate geometry. Explore the geometric reasoning and slope analysis involved. Suitable for high school students.

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