In the following figure,
$CD=5$ cm, $BE=10$ cm, $AE=12$ cm,
$\angle DAB=\angle DCB$, and $\angle DAE=\angle DBC=90^{\circ }$
Points AFCD create a rhombus.


The length of BF (in cm) is

The problem asks for the length of the segment $\text{BF}$ based on the geometric constraints involving the rhombus $\text{AFCD}$ and the right-angled triangle $\Delta \text{DAE}$.

1. Analyze the Rhombus AFCD

The figure $\text{AFCD}$ is a rhombus. The diagonals ($\text{AC}$ and $\text{DF}$) intersect at point $\text{B}$.

- All sides are equal: $\text{AF}=\text{FC}=\text{CD}=\text{DA}$. Given $\text{CD}=5\text{ cm}$, so $\text{DA}=5\text{ cm}$.
- Diagonals bisect each other: $\text{DB}=\text{BF}$. (Let $\text{DB}=\text{BF}=x$).

Our goal is to find $x=\text{BF}$.

**2. Analyze the Right-Angled Triangle $\Delta \text{DAE}$ **

We are given $\angle \text{DAE}=90^{\circ }$. We use the Pythagorean theorem on $\Delta \text{DAE}$ to find the length of $\text{DE}$.

- $\text{DA}=5\text{ cm}$ (Side of rhombus).
- $\text{AE}=12\text{ cm}$ (Given).

$${\text{DE}}^2={\text{DA}}^2+{\text{AE}}^2$$

$${\text{DE}}^2=5^2+12^2=25+144=169$$

$$\text{DE}=\sqrt{169}=13\text{ cm}$$

3. Relate Segment Lengths on the Line DE

From the diagram, the points $\text{D, B, F, E}$ are collinear in that order. The total length $\text{DE}$ can be broken down into segments.

$$\text{DE}=\text{DB}+\text{BF}+\text{FE}$$

We know $\text{DB}=\text{BF}=x$.

$$\text{DE}=x+x+\text{FE}=2x+\text{FE}$$

We are also given $\text{BE}=10\text{ cm}$. From the collinear points, $\text{BE}$ is:

$$\text{BE}=\text{BF}+\text{FE}$$

$$10=x+\text{FE}\Rightarrow \text{FE}=10-x$$

Substitute $\text{FE}$ back into the equation for $\text{DE}$:

$$\text{DE}=x+\text{BE}$$

$$13=x+10$$

$$x=13-10=3\text{ cm}$$

Conclusion

Since $\text{BF}=x$, the length of $\text{BF}$ is $3\text{ cm}$.