In this problem, we are given quadrilateral BCDE as a square, and the angle $\angle DFE = 42^\circ$. We are tasked with finding BD, which is one of the diagonals of the square.

Steps to Solve:

1. Properties of a square:

   • All sides of a square are equal.
   • Diagonals of a square are congruent and bisect each other at right angles ($90^\circ$).
   • Each diagonal divides the square into two isosceles right triangles.

2. Understand $BD$: Since $BCDE$ is a square:

   • $BD$ is one diagonal.
   • The diagonals of a square can be found using the Pythagorean theorem.

   If the side length of the square is $s$, then: $BD = s\sqrt{2}.$

3. Angle $DFE$: The given angle $42^\circ$ at $F$ inside the square does not directly influence the length of the diagonals because the diagonals’ properties are already determined by the square’s geometry.

Key Insight:

To find $BD$, we only need the side length of the square because the diagonals are determined using: $BD = s\sqrt{2}.$

If more specific information about the side length $BC$ (or any other side) is provided, we can substitute that to calculate $BD$.

Next Steps:

• Do you have the side length of the square? If so, I can calculate $BD$ for you.
• Let me know if you'd like further clarification or help interpreting the problem setup.

Related Questions:

1. What are the properties of diagonals in a square?
2. How do you derive the formula for the diagonal of a square?
3. What are the angles formed where the diagonals intersect in a square?
4. How can you use the Pythagorean theorem for a square’s geometry?
5. How are the diagonals of a square related to its symmetry?

Tip: Remember, in a square, the diagonals are always equal and divide the square into four equal right triangles.