Given BAT = 54^.

The angle between the tangent TA and the chord AB is equal to the angle in the alternate segment.
Given $\angle BAT = 54^\circ$.
Therefore, $\angle ACB = \angle BAT$.

$$\angle ACB = 54^\circ$$

We are given that chord BC is parallel to tangent TA ($BC \parallel TA$).
Considering AB as a transversal line intersecting the parallel lines TA and BC, the alternate interior angles are equal.
Therefore, $\angle ABC = \angle BAT$.

$$\angle ABC = 54^\circ$$

The sum of angles in triangle ABC is $180^\circ$.

$$\angle BAC + \angle ABC + \angle ACB = 180^\circ$$

Substitute the known values:

$$\angle BAC + 54^\circ + 54^\circ = 180^\circ$$ $$\angle BAC + 108^\circ = 180^\circ$$ $$\angle BAC = 180^\circ - 108^\circ$$ $$\angle BAC = 72^\circ$$

The value of $\angle BAC$ is $\boxed{\text{72}^\circ}$.