Let's analyze the given problem:

1. Given Information:

   • Quadrilateral $ABCD$ has angles $\angle A = 160^\circ$ and $\angle B = 80^\circ$.
   • The sides $AB = BC = AD$.

2. Understanding the Shape: Since $AB = AD$, triangles $\triangle ABD$ and $\triangle BCD$ are isosceles triangles. Also, $AB = BC$, indicating symmetry about line $BD$.

3. Calculating $\angle D$: Using the properties of quadrilaterals: $\angle A + \angle B + \angle C + \angle D = 360^\circ$ Plugging in the values for $\angle A$ and $\angle B$: $160^\circ + 80^\circ + \angle C + \angle D = 360^\circ$ $\angle C + \angle D = 360^\circ - 240^\circ = 120^\circ$

4. Relation between $\angle C$ and $\angle D$: Since $\triangle BCD$ is isosceles, $\angle B = \angle C$.

5. Therefore, the solution yields ( C=80^\circ