First, note that $\angle EAF + \angle ECF = \frac{\angle BAD}{2} + 90^{\circ} + \frac{\angle BCD}{2} = 180^{\circ}$, so $A,E,C,F$ are concyclic. Let $CD$ intersect $(AECF)$ at point $H \ne C$. Since $\angle ADH = \angle ABC = \angle ABE$ and $\angle AHD = \angle AHC = \angle AEB$, we have $\triangle ADH \sim \triangle ABE$. Thus, $\frac{1}{2} = \frac{AD}{AB} = \frac{AH}{AE} = \frac{AH}{2AM}$ and $AH = AM$. Also, $\angle FAH = \angle DAH = \angle BAE = \angle EAD = \angle MAF$. From $AH = AM$ and $\angle HAF = \angle MAF$, we have $\triangle HAF \cong \triangle MAF$. Now, $\angle ABD = \angle ACD = \angle ACH = \angle AFH = \angle AFM = \angle AFG$, so $\triangle ABD \sim \triangle AFG$ (since $\angle DAB = \angle GAF$). Since $2 = \frac{AB}{AD} = \frac{AF}{AG} = \frac{MF}{MG}$ from $\triangle ABD \sim \triangle AFG$ and the angle bisector theorem, we get the conclusion that $MF = 2MG$.

It is similar to P6 in Grade 11.Or you can use Menalaus.