Converse of Triangle Midsegment Theorem: If a line segment connecting two midpoints of a triangle's sides is parallel to the third side and half its length, it is the midsegment.

Problem Statement:

The task is to solve a problem using the Converse of the Triangle Midsegment Theorem. The theorem states:

> If a line segment connecting two midpoints of the opposite sides of a triangle is parallel to the third side and is half the length of the third side, then it is the midsegment of the triangle.

In the given triangle \( \Delta ABC \):
- \( D \) and \( E \) are the midpoints of sides \( AB \) and \( AC \), respectively.
- \( DE \parallel BC \).

We need to show that \( DE \) is the midsegment of \( \Delta ABC \).

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Solution:

#### Step 1: Understand the given information
- \( D \) is the midpoint of \( AB \), so \( AD = DB \).
- \( E \) is the midpoint of \( AC \), so \( AE = EC \).
- \( DE \parallel BC \).

#### Step 2: Apply the Converse of the Triangle Midsegment Theorem
The Converse of the Triangle Midsegment Theorem states:
- If a line segment connects the midpoints of two sides of a triangle and is parallel to the third side, then it is the midsegment of the triangle.

From the problem:
- \( D \) and \( E \) are the midpoints of \( AB \) and \( AC \), respectively.
- \( DE \parallel BC \).

These conditions directly satisfy the hypothesis of the Converse of the Triangle Midsegment Theorem.

#### Step 3: Conclusion
By the Converse of the Triangle Midsegment Theorem, since \( D \) and \( E \) are midpoints of \( AB \) and \( AC \), respectively, and \( DE \parallel BC \), it follows that \( DE \) is the midsegment of \( \Delta ABC \).

Additionally, the midsegment theorem also implies that:
\[ DE = \frac{1}{2} BC. \]

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Final Answer:

\[
\boxed{DE \text{ is the midsegment of } \Delta ABC}
\]