1) a) **Show that $\angle FEM = \angle MAB$.** 1. We are given that $G$ is the symmetric of $E$ with respect to $F$, and $C$ is the symmetric of $A$ with respect to $B$. 2. Since $F$ is the midpoint of segment $EG$, symmetry implies $FE = FG$ and $\angle EFM = \angle GFM$. 3. Similarly, $B$ is the midpoint of segment $AC$, so $AB = BC$ and $\angle MAB = \angle MBC$. 4. Points $F, M, B$ are collinear and points $A, M, E$ are collinear. 5. By symmetry and collinearity, the angle $\angle FEM$ corresponds to $\angle MAB$. b) **Show that triangles $EFM$ and $AMB$ are congruent and write the homologous elements.** 1. From the symmetry, $FE = AB$ and $FM = MB$ because $F$ and $B$ are midpoints. 2. We have shown $\angle FEM = \angle MAB$. 3. By SAS (Side-Angle-Side) criterion, triangles $EFM$ and $AMB$ are congruent. 4. Homologous elements: - $E \leftrightarrow A$ - $F \leftrightarrow M$ - $M \leftrightarrow B$ - $EF \leftrightarrow AM$ - $FM \leftrightarrow MB$ - $\angle FEM \leftrightarrow \angle MAB$ **Final answers:** $$\angle FEM = \angle MAB$$ Triangles $EFM$ and $AMB$ are congruent by SAS. Homologous vertices: $E \leftrightarrow A$, $F \leftrightarrow M$, $M \leftrightarrow B$.