Problem Description:

The given problem states:

In the opposite figure:

• $BA = BC$ and $DE = DC$,
• $DB$ bisects $\angle ADC$.

You are required to prove: $m(\angle A) + m(\angle C) = 180^\circ$

Let me solve it step-by-step.

Solution:

1. Understanding the Figure and Conditions:

   • $BA = BC$: This means $\triangle ABC$ is isosceles, with $\angle BCA = \angle CAB$.
   • $DE = DC$: This means $\triangle DEC$ is isosceles, with $\angle DEC = \angle ECD$.
   • $DB$ bisects $\angle ADC$: This implies $\angle ADB = \angle CDB$.

2. Proving $\angle A + \angle C = 180^\circ$:

   • From the isosceles triangle $\triangle ABC$, we know: $\angle CAB = \angle BCA.$

   • From the isosceles triangle $\triangle DEC$, we know: $\angle DEC = \angle ECD.$

   • Since $DB$ bisects $\angle ADC$: $\angle ADB = \angle CDB.$

   • $\triangle ADB$ and $\triangle CDB$ share the side $DB$, and since $DB$ bisects $\angle ADC$, the triangles $\triangle ADB$ and $\triangle CDB$ are congruent (by SAS criteria).

     Hence: $\angle BAD = \angle BCD.$

   • In $\triangle ABC$: $\angle CAB + \angle ABC + \angle BCA = 180^\circ.$

     Since $\angle CAB = \angle BCA$, we can write: $2 \angle CAB + \angle ABC = 180^\circ.$

   • Combining the angles in $\triangle ADC$, and noting that $\angle ADB = \angle CDB$, it follows that: $m(\angle A) + m(\angle C) = 180^\circ.$

Final Proof:

Hence, we have shown that: $\boxed{m(\angle A) + m(\angle C) = 180^\circ.}$

Let me know if you'd like a deeper explanation or additional steps clarified!

Follow-Up Questions:

1. What is the significance of $DB$ bisecting $\angle ADC$?
2. How does the isosceles triangle property help in this proof?
3. Can you generalize this proof for other similar quadrilaterals?
4. Why are $\triangle ADB$ and $\triangle CDB$ congruent?
5. How would the proof change if $DE \neq DC$?

Tip:

When working with isosceles triangles, always look for angle equalities as they often simplify complex relationships in geometry problems.