The problem involves determining the measure of the angle $\angle ADF$ in the given geometric diagram. Here's a breakdown of the problem:

1. $ABCD$ is a square.
2. $\triangle EFB$ is an isosceles triangle with its vertex at $F$.
3. We are tasked with finding $\angle ADF$.

Step-by-step solution:

Step 1: Properties of the square

In a square, all sides are equal and all angles are $90^\circ$. Thus:

• $AB = BC = CD = DA$.
• $\angle DAB = \angle ABC = \angle BCD = \angle CDA = 90^\circ$.

The diagonals of the square, $AC$ and $BD$, bisect each other at right angles.

Step 2: Placement of $E$ and triangle $\triangle EFB$

Since $E$ lies on diagonal $AC$, and $\triangle EFB$ is isosceles at $F$, the sides $EF = BF$. The orientation of $E$ along the diagonal affects the angle $\angle ADF$.

Step 3: Analyze the angle $\angle ADF$

To determine $\angle ADF$, note:

• The diagonal $AC$ bisects $\angle DAB$, so $\angle DAC = 45^\circ$.
• Adding $\angle EDF$ to $\angle DAC$ yields $\angle ADF$.

The key is the relationship between $EF$, $BF$, and the geometry of $\triangle EFB$. Let’s calculate $\angle EDF$ by analyzing the isosceles triangle $EFB$.

Let me confirm details from the user before proceeding further. Do you know the exact location of point $E$ along the diagonal, or should I solve with general assumptions?