We can start by finding the relationship between $\angle AEF$ and $\angle BAC$. Since $\overline{DA} \parallel \overline{EF}$, we have $\angle AEF = \angle ACD$, and since $\overline{AB}$ bisects $\angle DAC$, we have $\angle BAC = \angle BAD = \frac{1}{2}\angle DAC$.
From the given information, we have $\angle AEF = 3 \cdot \angle BAC - 15^\circ$. Substituting the expressions we found above, we get $\angle ACD = 3 \cdot \frac{1}{2}\angle DAC - 15^\circ$, which simplifies to $\angle ACD = \frac{3}{2}\angle DAC - 15^\circ$.
Since $\angle ACD$ and $\angle DAC$ are part of a straight angle, we have $\angle ACD + \angle DAC = 180^\circ$. Substituting the expression we found for $\angle ACD$, we have $\frac{3}{2}\angle DAC - 15^\circ + \angle DAC = 180^\circ$, which simplifies to $\frac{5}{2}\angle DAC = 195^\circ$. Solving for $\angle DAC$, we get $\angle DAC = \boxed{78^\circ}$.
In the diagram below, point $A$ is on line segment $\overline{CE}$, and $\overline{AB}$ bisects $\angle DAC.$ If $\overline{DA} \parallel \overline{EF}$ and $\angle AEF = 3 \cdot \angle BAC - 15^\circ$, then what is $\angle DAC$ in degrees?
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