Answer: 54
Analysis: As shown in the figure, construct square $\mathrm{APBM}$, and let $\mathrm{Q}$ be the reflection of $\mathrm{F}$ over $\mathrm{AD}$. Connect $\mathrm{DQ}$, $\mathrm{PQ}$, and $\mathrm{PM}$. $P M=A B=M F=F D=D Q=Q P$ (since $\mathrm{APBM}$ is a square, $P M=A B$), $\angle Q D F=2 \angle A D F=2 \times\left(45^{\circ}+9^{\circ}\right)=108^{\circ}$. Therefore, PQDFMPQDFM is a regular pentagon, and by the same reasoning, $\angle F M G=360-108 \times 2-90=54$ (degrees)
4. As shown in the figure, triangles $A B C$ and $A D E$ are isosceles right triangles with $\angle A=90$. Point $M$ is the midpoint of $B C$. It is known that $A B=A C=D F=F M=E G=G M, \angle F D E=\angle G E D=9^{\circ}$, and points $F$ and $G$ are outside triangle ADE. What is the degree measure of $\angle \mathrm{FMG}$?