$\triangle ABC$ is isosceles with $\overline{AB} \cong \overline{CB}$.

$D$, $E$, $H$ are midpoints of $\overline{AB}$, $\overline{CB}$, $\overline{AC}$. $F$, $G$ are midpoints of $\overline{DH}$, $\overline{EH}$.

By the midpoint triangle theorem, $\triangle DEH$ has $\frac{1}{4}$ the area of $\triangle ABC$.

$\triangle FGH$ has $\frac{1}{4}$ the area of $\triangle DEH$.

$\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle FGH} = \frac{1}{1/16} = 16:1$