The problem provided involves a geometric figure where certain angles and relationships between the segments are given. Here's the breakdown of the problem and how to solve it:

Given:

1. $\angle CAB = \angle DFE$
2. $\angle FBE = 30^\circ$
3. $\angle ABC = \angle FED$

To Prove:

1. $\triangle ACE \cong \triangle FDB$
2. $ME = MB$

Steps to Solve:

1. Proving $\triangle ACE \cong \triangle FDB$:

• Angle $\angle CAB$ is given to be equal to $\angle DFE$.
• Angle $\angle ABC$ is given to be equal to $\angle FED$.
• Since $\angle FBE = 30^\circ$, we can observe that $\angle ACB = \angle DFB$ (vertically opposite angles).

With two angles in one triangle equal to two angles in another triangle, and the included side equal, by ASA (Angle-Side-Angle) Congruence criterion, the triangles $\triangle ACE$ and $\triangle FDB$ are congruent.

2. Proving $ME = MB$:

• Since $\triangle ACE \cong \triangle FDB$, corresponding sides of the triangles are equal.
• In particular, $\overline{ME}$ and $\overline{MB}$ being corresponding medians of the congruent triangles $\triangle ACE$ and $\triangle FDB$, must be equal.

This completes the proof.

Would you like further details or have any questions about this solution?

Here are some related questions:

1. How does the ASA Congruence Theorem work?
2. What are corresponding medians in congruent triangles?
3. How can you prove congruence using other methods like SSS or SAS?
4. What properties of vertical angles are useful in geometric proofs?
5. How do medians relate to triangle congruence?

Tip: Always check if corresponding parts of congruent triangles (CPCTC) can simplify proofs of equal lengths or angles!