Let's break down the given information and solve the problem geometrically, aided by a diagram.

Given Information:

• ABCD is a quadrilateral.
• $\angle A = 160^\circ$, $\angle B = 80^\circ$.
• The sides $AB = BC = AD$, so two pairs of equal sides are given.

Diagram:

We can imagine that quadrilateral ABCD has a symmetrical shape due to the equal sides. Based on the equal sides and given angles, this suggests that the shape might be an isosceles trapezium or a kite-like quadrilateral.

Since $AB = BC = AD$, the structure suggests that angles $\angle A$ and $\angle B$ influence the other angles due to symmetry.

Let me describe the diagram setup that reflects this solution:

1. Draw a quadrilateral ABCD.
2. Label points A, B, C, and D in a way that shows symmetry, where AB = BC = AD.
3. Assign $\angle A = 160^\circ$ at vertex A, and $\angle B = 80^\circ$ at vertex B.
4. Notice that angle C will be symmetrical to angle B due to the properties of the quadrilateral, so $\angle C = 80^\circ$ as well.

Solution Outline:

Since AB = BC and AD = AB, the quadrilateral likely forms an isosceles-like symmetry. By geometric symmetry and angle sum in a quadrilateral (which must add up to $360^\circ$):

• $\angle A + \angle B + \angle C + \angle D = 360^\circ$
• $160^\circ + 80^\circ + 80^\circ + \angle D = 360^\circ$
• Solving for $\angle D$: $\angle D = 40^\circ$.

Thus, $\angle C = 80^\circ$, matching the answer.

I can generate a diagram to match this description. Would you like me to generate a visual diagram?

Related Questions:

1. What properties of quadrilaterals make it possible to solve problems like this?
2. How do the angle sums of different polygons relate to their number of sides?
3. What is the significance of symmetry in determining angles in polygons?
4. How does knowing equal side lengths help solve angle problems in quadrilaterals?
5. Can you find the angles of irregular quadrilaterals without symmetry?

Tip: When solving geometry problems, always check if there are symmetrical properties, as they can greatly simplify the problem!