B2. $\frac{3}{2}$ holds

Triangles $A D F, B D A$ and $B A F$ are similar (AA). Therefore,

$$
\frac{b}{a}=\frac{|B A|}{|D A|}=\frac{|A F|}{|D F|}=\frac{|B F|}{|A F|}
$$

In other words,

$$
\left(\frac{b}{a}\right)^{2}=\frac{|A F|}{|D F|} \cdot \frac{|B F|}{|A F|}=\frac{|B F|}{|D F|}=\frac{9}{4} .
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_88e9b1bd823632ad9c59g-2.jpg?height=268&width=346&top_left_y=2419&top_left_x=1523)

Thus, we find $\frac{b}{a}=\frac{3}{2}$.
B2. A rectangle $A B C D$ has sides $a$ and $b$, where $a<b$. The perpendiculars from $A$ and $C$ to the diagonal $B D$ divide that diagonal into three segments with lengths 4, 5, and 4.

Calculate $\frac{b}{a}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_8b40d2c2e9518068878ag-2.jpg?height=277&width=351&top_left_y=1569&top_left_x=1521)