We are given that $ABCD$ is a rhombus and $\angle DBC = 55^\circ$. We need to find $\angle DCA$.

Key Properties of Rhombus:

1. All sides of a rhombus are equal.
2. The diagonals of a rhombus bisect each other at right angles (90°).
3. The diagonals bisect the angles of the rhombus.

Step-by-Step Solution:

1. Since $ABCD$ is a rhombus, the diagonals bisect each other at right angles, and they bisect the angles at the vertices.

2. We know that $\angle DBC = 55^\circ$. This angle is part of $\triangle DBC$, and since $\triangle DBC$ is an isosceles triangle (because sides $BD = BC$, being sides of the rhombus), the angle $\angle DCB$ will also be $55^\circ$.

3. Now, in $\triangle DBC$, the sum of the angles must be $180^\circ$. So, we can find $\angle BDC$ as follows: $\angle BDC = 180^\circ - \angle DBC - \angle DCB = 180^\circ - 55^\circ - 55^\circ = 70^\circ.$

4. The diagonals of a rhombus bisect the opposite angles, so $\angle DCA$ is half of $\angle BDC$. Therefore: $\angle DCA = \frac{1}{2} \times \angle BDC = \frac{1}{2} \times 70^\circ = 35^\circ.$

Thus, $\angle DCA = 35^\circ$.

Would you like more details, or do you have any questions?

Here are 5 related questions for further exploration:

1. What are the properties of a rhombus that distinguish it from other quadrilaterals?
2. How do the diagonals of a rhombus affect its internal angles?
3. Can you prove that the diagonals of a rhombus bisect the angles?
4. If $\angle DBC$ were changed, how would that affect $\angle DCA$?
5. How would the solution change if the rhombus were instead a square?

Tip: In rhombuses, the diagonals always divide the rhombus into four right-angled triangles.