.

To solve this problem, we need to find angle \( \angle BCA \) in triangle \( \triangle ABC \).

Given:

• \( \angle BAC = 31^\circ \)
• \( \angle BDC = 75^\circ \)

Since \( D \) lies on \( AC \), \( \angle BDC \) is an external angle for triangle \( \triangle ABC \). The external angle theorem states that \( \angle BDC = \angle BAC + \angle BCA \).

Thus, we have:

\[ \angle BDC = \angle BAC + \angle BCA \]

Substitute the known values:

\[ 75^\circ = 31^\circ + \angle BCA \]

Solve for \( \angle BCA \):

\[ \angle BCA = 75^\circ - 31^\circ = 44^\circ \]

Thus, \( \angle BCA = 44^\circ \).