https://r2.erweima.ai/mathbot/image/1726061037407ktlzcuyt.jpg

The problem shown in the image asks to prove a geometric relation in triangle \(ABC\), where the triangle is an isosceles right triangle with \( \angle BAC = 90^\circ \). The detailed problem is as follows:

Given:
- \( \triangle ABC \) is a right isosceles triangle with \( \angle BAC = 90^\circ \).
- \( D \) is the midpoint of side \( AB \).
- Line \( AE \perp CD \) at point \( E \).
- Line \( CD \) intersects \( BC \) at point \( F \).
- Prove: \( CD = AF + DF \).

This proof involves geometric properties of right triangles and perpendicular bisectors. Here's how we can approach the problem:

1. **Isosceles Triangle Properties**:
   Since \( \triangle ABC \) is an isosceles right triangle, we know:
   - \( AB = AC \)
   - \( \angle ABC = \angle ACB = 45^\circ \)

2. **Midpoint and Perpendicular**:
   Since \( D \) is the midpoint of \( AB \), segment \( AD = DB \). Also, \( AE \perp CD \) introduces another right angle at \( E \).

3. **Utilizing Similar Triangles**:
   Triangles \( AEF \) and \( CDF \) may be similar based on the angles and perpendicular relationships. This similarity can help establish the ratio between the sides of these triangles.

### Steps to prove \( CD = AF + DF \):
- Investigate the triangle similarity and proportionality between the segments.
- Use the properties of isosceles triangles and the right angle at \( E \) to apply the Pythagorean theorem in appropriate triangles.

Would you like me to proceed with a detailed geometric proof, or do you need clarification on specific steps?

如图，△ABC为等腰直角三角形，其中∠BAC=90°，D为AB中点。AE ⊥ CD，交CD于点E，交BC于点F。连接DF。求证：CD=AF+DF。

In this problem, we explore a geometric proof involving an isosceles right triangle ABC, where ∠BAC = 90°, and AE is perpendicular to CD. Using properties of isosceles triangles, midpoints, and triangle similarity, we prove that CD = AF + DF.

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