1. Identify the given elements and relationships:
- Triangle $ABC$ is a right-angled triangle with $\angle C = 90^\circ$.
- $D$ is the midpoint of the altitude from $C$ to $AB$.
- The reflections of line $AB$ about $AD$ and $BD$ meet at point $F$.

2. Establish the properties of the reflections:
- Reflecting $AB$ about $AD$ and $BD$ will create two new lines that intersect at point $F$.
- Since $D$ is the midpoint of the altitude from $C$, it lies on the altitude $CL$ where $L$ is the foot of the altitude from $C$ to $AB$.

3. Use similarity and geometric properties:
- Note that $\triangle ALC \sim \triangle CLB$ because both are right triangles sharing the angle at $C$.
- This similarity implies that $LC^2 = LA \cdot LB$.

4. Relate the inradius and semi-perimeter:
- Let $a = BF$, $b = AF$, and $c = AB$.
- Let $r$ be the inradius and $s$ be the semi-perimeter of $\triangle AFB$.
- Let $h$ be the length of the altitude from $F$ to $AB$.

5. Calculate the area relationships:
- The area of $\triangle ABC$ is given by:
$$S_{ABC} = \frac{1}{2} \cdot AC \cdot BC$$
- The area of $\triangle ABF$ can be expressed using the altitude $h$:
$$S_{ABF} = \frac{1}{2} \cdot AB \cdot h$$

6. Use the relationship between the inradius and the semi-perimeter:
- The inradius $r$ of $\triangle AFB$ is related to the area and semi-perimeter by:
$$r = \frac{S_{ABF}}{s}$$
- Given that $4r^2 = (s-a)(s-b)$, we can derive:
$$\frac{4(s-c)}{s} = 1 \implies s = \frac{4}{3}c$$

7. Calculate the ratio of the areas:
- Using the relationship between the semi-perimeter and the sides:
$$\frac{S_{ABF}}{S_{ABC}} = \frac{h}{2r} = \frac{\frac{2\Delta}{c}}{\frac{2\Delta}{s}} = \frac{s}{c} = \frac{4}{3}$$

The final answer is $\boxed{\frac{4}{3}}$.