Given:

• \( AD = AC \)
• \( D \) is midpoint of \( BC \)
• \( \angle DAC = 96^\circ \)

Since \( AD = AC \), triangle \( \triangle ADC \) is isosceles.

Let \( \angle DBC = \angle x \), then \( \angle DAC = x \) and the external angle becomes:

\[ \angle DAC + \angle CAB = 96^\circ \Rightarrow 2x = 96^\circ \Rightarrow x = 32^\circ \]

Now, use this to find angle \( \angle C = 2x = 64^\circ \)

✅ Final Answer: \( \boxed{64^\circ} \)