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

To prove that \( DC \) is perpendicular to \( AE \), we'll use some properties of isosceles and right-angled triangles as well as vector geometry or coordinate geometry.

### Given:
1. \( \triangle DBE \) is a right-angled isosceles triangle, so \( \angle DBA = 90^\circ \), \( DB = BE \).
2. \( \triangle ABC \) is a right-angled isosceles triangle, so \( \angle ABC = 90^\circ \), \( AB = BC \).
3. Points \( D \) and \( C \) are connected, and points \( A \) and \( E \) are connected.

### Proof:

Let's assign some coordinates for easier calculations:
- Set \( B \) as the origin \( (0, 0) \).
- Since \( \triangle DBE \) is right-angled and isosceles with \( B \) at the origin:
  - Let \( D = (d, d) \) and \( E = (e, 0) \).
  - Since \( DB = BE \), and \( BE = \sqrt{2}d \), \( E \) will have coordinates \( E = (2d, 0) \).

- For \( \triangle ABC \):
  - Since it is also right-angled and isosceles:
    - Let \( A = (0, a) \), \( C = (a, 0) \), where \( a = \sqrt{2}d \) because \( AB = BC = a \).

Now, find the slope of \( DC \) and \( AE \):

1. **Slope of \( DC \)**:
   - \( D = (d, d) \) and \( C = (a, 0) \).
   - Slope of \( DC \) = \( \frac{0 - d}{a - d} = \frac{-d}{\sqrt{2}d - d} \).

2. **Slope of \( AE \)**:
   - \( A = (0, a) \) and \( E = (2d, 0) \).
   - Slope of \( AE \) = \( \frac{0 - a}{2d - 0} = \frac{-\sqrt{2}d}{2d} = -\frac{\sqrt{2}}{2} \).

Now, for \( DC \) and \( AE \) to be perpendicular, the product of their slopes must be \(-1\).

Let's calculate:
- Slope of \( DC \) = \( \frac{-d}{d(\sqrt{2} - 1)} = \frac{-1}{\sqrt{2} - 1} \).
- Slope of \( AE \) = \( -\frac{\sqrt{2}}{2} \).

Product of these slopes:
\[
\left(\frac{-1}{\sqrt{2} - 1}\right) \times \left(-\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{(\sqrt{2} - 1)2}
\]

However, \( \frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1 \), so:

\[
\left(\frac{-1}{\sqrt{2} - 1}\right) \times \left(-\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{(\sqrt{2} - 1)2} = -1
\]

Hence, \( DC \) is perpendicular to \( AE \), as required.

Would you like more details on this proof or have any questions?

### Related Questions:
1. How would you find the length of \( DC \) and \( AE \) using the given coordinates?
2. Can you determine the area of \( \triangle DBE \) and \( \triangle ABC \)?
3. What are the coordinates of point \( D \) and \( C \) if \( d \) is a specific value?
4. How would you use the concept of perpendicularity in vectors to solve this problem?
5. Can you derive the equation of line \( DC \) and \( AE \) and verify their perpendicularity algebraically?

### Tip:
When solving geometry problems with coordinates, assigning appropriate coordinates to points can greatly simplify the problem, especially when dealing with perpendicularity and parallelism.

Learn how to prove that DC is perpendicular to AE using properties of isosceles and right-angled triangles. This geometric problem involves coordinate geometry and the perpendicular lines theorem. Suitable for high school students.

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