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Problem Analysis:

The image contains a geometric figure with labeled points and angles. The task is to find the measure of angle \( \angle BAC \). Here's the breakdown of the given information:

- \( \triangle ABC \) is an isosceles triangle with \( AB = AC \).
- \( D \) is a point on \( BC \) such that \( BD = DC \), making \( D \) the midpoint of \( BC \).
- \( E \) is a point on \( AC \) such that \( AE = EC \), making \( E \) the midpoint of \( AC \).
- \( F \) is a point on \( AB \) such that \( AF = FB \), making \( F \) the midpoint of \( AB \).
- The measure of \( \angle BFD \) is given as \( 120^\circ \).

We need to determine the measure of \( \angle BAC \).

---

Step-by-Step Solution:

#### Step 1: Understand the Geometry
Since \( \triangle ABC \) is isosceles with \( AB = AC \), the base angles \( \angle ABC \) and \( \angle ACB \) are equal. Let:
\[
\angle ABC = \angle ACB = x
\]
The sum of the angles in a triangle is \( 180^\circ \), so:
\[
\angle BAC + \angle ABC + \angle ACB = 180^\circ
\]
Substituting \( \angle ABC = \angle ACB = x \):
\[
\angle BAC + x + x = 180^\circ \implies \angle BAC + 2x = 180^\circ \implies \angle BAC = 180^\circ - 2x
\]

#### Step 2: Analyze the Midpoints
- \( D \) is the midpoint of \( BC \), so \( BD = DC \).
- \( E \) is the midpoint of \( AC \), so \( AE = EC \).
- \( F \) is the midpoint of \( AB \), so \( AF = FB \).

Since \( D \), \( E \), and \( F \) are midpoints, the segments \( AD \), \( BE \), and \( CF \) are medians of \( \triangle ABC \). The medians of a triangle intersect at the centroid, which divides each median into a ratio of \( 2:1 \).

#### Step 3: Use the Given Angle \( \angle BFD = 120^\circ \)
The key insight here is that \( F \) and \( D \) are midpoints, and the angle \( \angle BFD \) is given. Since \( F \) is the midpoint of \( AB \) and \( D \) is the midpoint of \( BC \), the segment \( FD \) is part of the medial triangle formed by connecting the midpoints of the sides of \( \triangle ABC \).

In any triangle, the medial triangle (formed by connecting the midpoints of the sides) is similar to the original triangle and has half the side lengths. Importantly, the angles of the medial triangle are the same as the angles of the original triangle.

Given \( \angle BFD = 120^\circ \), this angle corresponds to an angle in the medial triangle. In the context of \( \triangle ABC \), this angle is related to the vertex angle \( \angle BAC \).

#### Step 4: Relate \( \angle BFD \) to \( \angle BAC \)
In an isosceles triangle, the angle formed by the medians from the base vertices (like \( \angle BFD \)) is supplementary to the vertex angle \( \angle BAC \). Specifically, if \( \angle BFD = 120^\circ \), then:
\[
\angle BAC = 180^\circ - \angle BFD = 180^\circ - 120^\circ = 60^\circ
\]

#### Step 5: Verify the Solution
If \( \angle BAC = 60^\circ \), then the base angles \( \angle ABC \) and \( \angle ACB \) are:
\[
\angle ABC = \angle ACB = \frac{180^\circ - \angle BAC}{2} = \frac{180^\circ - 60^\circ}{2} = 60^\circ
\]
This makes \( \triangle ABC \) an equilateral triangle, where all angles are \( 60^\circ \). In an equilateral triangle, the medians form angles of \( 120^\circ \) between them, which matches the given \( \angle BFD = 120^\circ \).

Thus, the solution is consistent.

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Final Answer:

\[
\boxed{60^\circ}
\]