Let $ABC$ be a triangle. Find the point $D$ on its side $AC$ and the point $E$ on its side $AB$ such that the area of triangle $ADE$ equals to the area of the quadrilateral $DEBC$, and the segment $DE$ has minimum possible length.
1. **Given**: We have a triangle \(ABC\). We need to find points \(D\) on side \(AC\) and \(E\) on side \(AB\) such that the area of triangle \(ADE\) equals the area of quadrilateral \(DEBC\), and the segment \(DE\) has the minimum possible length.

2. **Area Relationship**: 
   - The area of triangle \(ABC\) is given by:
     \[
     \text{Area}(ABC) = \frac{1}{2} AB \cdot AC \cdot \sin A
     \]
   - The area of triangle \(ADE\) is given by:
     \[
     \text{Area}(ADE) = \frac{1}{2} AD \cdot AE \cdot \sin A
     \]
   - Since the area of triangle \(ADE\) equals the area of quadrilateral \(DEBC\), we have:
     \[
     \text{Area}(ADE) = \frac{1}{2} \text{Area}(ABC)
     \]
   - Therefore:
     \[
     \frac{1}{2} AD \cdot AE \cdot \sin A = \frac{1}{2} \left( \frac{1}{2} AB \cdot AC \cdot \sin A \right)
     \]
   - Simplifying, we get:
     \[
     AD \cdot AE \cdot \sin A = \frac{1}{2} AB \cdot AC \cdot \sin A
     \]
   - Dividing both sides by \(\sin A\), we obtain:
     \[
     AD \cdot AE = \frac{1}{2} AB \cdot AC
     \]

3. **Minimizing \(DE\)**:
   - To minimize the length of \(DE\), we need to consider the geometric mean. The minimum length of \(DE\) occurs when \(AD = AE\).
   - Setting \(AD = AE\), we have:
     \[
     AD^2 = \frac{1}{2} AB \cdot AC
     \]
   - Solving for \(AD\), we get:
     \[
     AD = \sqrt{\frac{1}{2} AB \cdot AC}
     \]

4. **Conclusion**:
   - The points \(D\) and \(E\) are such that \(AD = AE = \sqrt{\frac{1}{2} AB \cdot AC}\).
   - This ensures that the area of triangle \(ADE\) is half the area of triangle \(ABC\) and the segment \(DE\) is minimized.

The final answer is \( \boxed{ AD = AE = \sqrt{\frac{1}{2} AB \cdot AC} } \).