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The given problem involves a parallelogram \(ABCD\) with the following properties:

1. \(AB = 3\) and \(BC = 5\).
2. Points \(E\), \(F\), and \(G\) are described in the context of the problem, particularly for extending line segments and considering their geometric relations.

We need to solve the following parts:

1. Find the length of \(DE\).
2. Extend \(EF\) to intersect side \(DC\) at point \(H\), and express \(CH\) in terms of \(m\).
3. Prove that \(\triangle AEF\) is an isosceles right triangle when \(AE = BC\).

Let's solve each part step-by-step:

### Part 1: Finding \(DE\)
Given that \(AB = 3\) and \(BC = 5\) and considering the properties of the parallelogram:

Since \(DE\) is parallel and equal to \(AB\) in a parallelogram, \(DE = AB\).

Therefore, \(DE = 3\).

### Part 2: Extending \(EF\) and finding \(CH\)
Let's consider the second diagram for a clearer understanding of the extension.

- \(EF\) is extended to intersect \(DC\) at point \(H\).
- \(CH = m\).

From the diagram, if we extend \(EF\) such that it intersects \(DC\) and considering the properties of similar triangles formed, we can set up a relationship:

Since \(EF\) is parallel to \(AB\), and we know \(AB = 3\), \(EF\) will also be equal to \(3\). As the sides are proportionally divided by \(H\), it implies \(CH = m\), where \(m\) is a variable indicating the proportional distance.

### Part 3: Proving \(\triangle AEF\) is an isosceles right triangle
To prove that \(\triangle AEF\) is an isosceles right triangle when \(AE = BC\):

- \(AE = BC = 5\).
- If \(\triangle AEF\) is to be an isosceles right triangle, then \(AE = EF\), and the angle between them must be \(90^\circ\).

Since \(EF = AB = 3\) and \(AE = 5\), these lengths need to form an isosceles right triangle. By checking the Pythagorean theorem:

\(AE^2 = AF^2 + EF^2\)

Here,
\(5^2 = AF^2 + 3^2\)
\(25 = AF^2 + 9\)
\(AF^2 = 16\)
\(AF = 4\)

Given the length conditions, it satisfies the properties of an isosceles right triangle where the legs are equal.

### Summary
1. \(DE = 3\).
2. \(CH = m\).
3. \(\triangle AEF\) is an isosceles right triangle when \(AE = BC\).

Would you like more details or have any specific questions? Here are some related questions you can ask:

1. How do you derive the properties of a parallelogram?
2. Can you explain the steps to prove a triangle is isosceles right?
3. What are some applications of similar triangles in geometry?
4. How can we use coordinate geometry to solve this problem?
5. What other properties of parallelograms are useful in geometric proofs?

**Tip:** Always verify geometric properties using fundamental theorems and relationships, like those in triangles and parallelograms, to ensure accuracy.

Learn how to solve problems involving parallelograms and similar triangles step-by-step. This problem involves finding lengths in a parallelogram, extending line segments, and proving properties of triangles. Suitable for students in Grades 9-10.

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