To solve this problem, let's analyze the given information step by step.

Given:

• Points $D$, $E$, and $F$ are midpoints of the sides of $\triangle ABC$.
• $DF = 34$, $AC = 48$, and $BC = 78$.

Using Midpoint Theorem:

1. Midsegment Theorem: The segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.
   • $DF$ is parallel to $AB$ and $DF = \frac{1}{2} \cdot AB$.
   • $DE$ is parallel to $BC$ and $DE = \frac{1}{2} \cdot BC$.
   • $EF$ is parallel to $AC$ and $EF = \frac{1}{2} \cdot AC$.

Calculations:

1. Finding $AB$: Since $DF = 34$ and $DF = \frac{1}{2} \cdot AB$, $AB = 2 \cdot DF = 2 \cdot 34 = 68.$

2. Finding $DE$: Since $DE = \frac{1}{2} \cdot BC$ and $BC = 78$, $DE = \frac{1}{2} \cdot 78 = 39.$

3. Finding $BE$: Since $BE$ is half of $BC$ because $E$ is the midpoint of $BC$, $BE = \frac{1}{2} \cdot BC = \frac{1}{2} \cdot 78 = 39.$

Final Answers:

• $DE = 39$,
• $AB = 68$,
• $BE = 39$.

Would you like a detailed explanation of the midsegment theorem or any other part? Here are some related questions for further exploration:

1. How does the Midpoint Theorem simplify solving triangle problems?
2. What happens when all midsegments of a triangle are drawn?
3. Could these midsegments form another triangle within $\triangle ABC$?
4. What are the proportional relationships between the original triangle and the inner triangle?
5. How could these calculations apply in a coordinate geometry context?

Tip: Always check midpoint and parallel segment relationships for similar triangles when using the Midpoint Theorem!