The diagram you've uploaded presents a triangle with a point $D$ outside of the triangle. Here is a step-by-step solution to the problem.

Understanding the Diagram:

• Triangle $\triangle BCD$ is depicted, where the following information is provided:
  • $\angle BCD = 70^\circ$ (interior angle)
  • $\angle BDA = 130^\circ$ (exterior angle)
  • The lines $BD$ and $BC$ are marked with measurements, though they do not affect the angle calculations.

Key Concept:

From the properties of exterior and interior angles, we know that the sum of the interior angles in any triangle is 180°. Additionally, an exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles.

Calculations:

1. Calculate $\angle BDC$ (interior angle): Since $\angle BDA = 130^\circ$ is the exterior angle, and the exterior angle equals the sum of the two non-adjacent interior angles in the triangle, we have: $\angle BDA = \angle BCD + \angle BDC$ $130^\circ = 70^\circ + \angle BDC$ Solving for $\angle BDC$: $\angle BDC = 130^\circ - 70^\circ = 60^\circ$

2. Compare $m \angle BCD - m \angle DBC$: We know that:

   • $m \angle BCD = 70^\circ$
   • $m \angle DBC = 60^\circ$

   So, $m \angle BCD - m \angle DBC = 70^\circ - 60^\circ = 10^\circ$

Conclusion:

The correct option is: $\boxed{A. \, m \angle BCD - m \angle DBC = 10^\circ}$

Would you like further clarification or details on any part of the solution?

Follow-up Questions:

1. How do we apply the exterior angle theorem in other scenarios?
2. What are the other properties of triangle angle sums?
3. Can the exterior angle theorem be used for polygons with more than three sides?
4. How would the problem change if we altered the exterior angle to a different value?
5. How can we calculate the angles if the lengths of sides are provided instead?

Tip:

Always remember that the sum of interior angles in any triangle is $180^\circ$, and an exterior angle is equal to the sum of the two opposite interior angles.