Let $ABC$ be an isosceles triangle, and point $D$ in its interior such that $$D \hat{B} C=30^\circ, D \hat{B}A=50^\circ, D \hat{C}B=55^\circ$$ 

(a) Prove that $\hat B=\hat C=80^\circ$.
(b) Find the measure of the angle $D \hat{A} C$.
### Part (a)
1. Given that \( \triangle ABC \) is an isosceles triangle, we have \( \angle B = \angle C \).
2. We are given the angles \( \angle DBC = 30^\circ \), \( \angle DBA = 50^\circ \), and \( \angle DCB = 55^\circ \).
3. Since \( \angle DBA = 50^\circ \) and \( \angle DBC = 30^\circ \), we can find \( \angle ABC \) as follows:
   \[
   \angle ABC = \angle DBA + \angle DBC = 50^\circ + 30^\circ = 80^\circ
   \]
4. Similarly, since \( \angle DCB = 55^\circ \) and \( \angle DBC = 30^\circ \), we can find \( \angle ACB \) as follows:
   \[
   \angle ACB = \angle DCB + \angle DBC = 55^\circ + 30^\circ = 85^\circ
   \]
5. Since \( \triangle ABC \) is isosceles with \( \angle B = \angle C \), we have:
   \[
   \angle B = \angle C = 80^\circ
   \]

### Part (b)
1. Let \( \alpha \) be the measure of \( \angle DAC \).
2. By the Law of Sines in \( \triangle ABD \) and \( \triangle ACD \), we have:
   \[
   \frac{BA}{AD} = \frac{\sin (70^\circ - \alpha)}{\sin 50^\circ}
   \]
   \[
   \frac{DA}{AC} = \frac{\sin 25^\circ}{\sin (25^\circ + \alpha)}
   \]
3. Since \( \frac{BA}{AD} \cdot \frac{DA}{AC} = 1 \), we get:
   \[
   \frac{\sin (70^\circ - \alpha)}{\sin 50^\circ} \cdot \frac{\sin 25^\circ}{\sin (25^\circ + \alpha)} = 1
   \]
4. Simplifying, we need to show:
   \[
   \sin (70^\circ - \alpha) \cdot \sin 25^\circ = \sin 50^\circ \cdot \sin (25^\circ + \alpha)
   \]
5. Using the identity \( \sin x \sin y = \frac{1}{2} [\cos (x-y) - \cos (x+y)] \), we get:
   \[
   \sin 30^\circ \cdot \sin 50^\circ = \sin 25^\circ \cdot \sin 65^\circ
   \]
6. Since \( \sin 30^\circ = \frac{1}{2} \) and \( \sin 65^\circ = \cos 25^\circ \), we have:
   \[
   \frac{1}{2} \cdot \sin 50^\circ = \sin 25^\circ \cdot \cos 25^\circ
   \]
7. Using the double-angle identity \( \sin 2x = 2 \sin x \cos x \), we get:
   \[
   \sin 50^\circ = 2 \sin 25^\circ \cos 25^\circ
   \]
8. Therefore, \( \alpha = 5^\circ \) works, and the measure of \( \angle DAC \) is uniquely determined by the conditions of the problem.

The final answer is \( \boxed{5^\circ} \)