Triangles DEF and PQF are similar and the area of triangle PQF is 25 cm². Calculate the area of triangle DEF.

The problem states that triangles ABC and BXY are similar ($\triangle ABC \sim \triangle BXY$). We need to calculate the enlargement factor.
From the diagram, we can identify corresponding sides. The triangles are oriented such that B is the center of enlargement. Side AC corresponds to side XY, and side AB corresponds to side XB.
Given lengths:
AC = 8 cm
XY = 20 cm

The enlargement factor ($k$) is the ratio of a side length in the enlarged triangle to the corresponding side length in the original triangle. Since XY (20 cm) is longer than AC (8 cm), $\triangle BXY$ is the enlarged triangle and $\triangle ABC$ is the original.

$$k = \frac{\text{Length of side in enlarged triangle}}{\text{Length of corresponding side in original triangle}}$$ $$k = \frac{XY}{AC}$$ $$k = \frac{20 \text{ cm}}{8 \text{ cm}}$$

Simplify the fraction:

$$k = \frac{20 \div 4}{8 \div 4}$$ $$k = \frac{5}{2}$$

The enlargement factor is $\boxed{\frac{5}{2}}$.

The problem states that triangles ABC and XYZ are similar ($\triangle ABC \sim \triangle XYZ$). We need to find the length of side BC.
Given lengths:
In $\triangle ABC$: AB = 2 cm, BC = $x$
In $\triangle XYZ$: XY = 5.6 cm, YZ = 14 cm, XZ = 11.2 cm

Since the triangles are similar, the ratio of their corresponding sides is equal.
The corresponding sides are:
AB corresponds to XY
BC corresponds to YZ
AC corresponds to XZ

We can set up a proportion using the known side lengths and the unknown side $x$:

$$\frac{AB}{XY} = \frac{BC}{YZ}$$

Substitute the given values:

$$\frac{2 \text{ cm}}{5.6 \text{ cm}} = \frac{x}{14 \text{ cm}}$$

To solve for $x$, multiply both sides by 14:

$$x = \frac{2}{5.6} \times 14$$ $$x = \frac{28}{5.6}$$ $$x = 5$$

The length of BC is $\boxed{5 \text{ cm}}$.

We need to show that $\angle ABD = \angle ACD$ in the given figure.
Consider $\triangle ABD$ and $\triangle ACD$.

1. Side AB is equal to side AC (given by the single tick marks).
2. Side BD is equal to side CD (given by the double tick marks).
3. Side AD is common to both triangles.

Since all three corresponding sides are equal ($AB=AC$, $BD=CD$, $AD=AD$), the triangles $\triangle ABD$ and $\triangle ACD$ are congruent by the Side-Side-Side (SSS) congruence criterion.

$$\triangle ABD \cong \triangle ACD \quad (\text{SSS})$$

Because the triangles are congruent, their corresponding angles are equal. Therefore, $\angle ABD$ corresponds to $\angle ACD$, and thus:

$$\angle ABD = \angle ACD$$

This shows that the angle ABD is equal to the angle ACD.

The problem states that triangles DEF and PQF are similar ($\triangle DEF \sim \triangle PQF$). We are given the area of $\triangle PQF$ and need to calculate the area of $\triangle DEF$.
Given lengths:
EF = 15 cm
FQ = 10 cm
Area of $\triangle PQF = 25 \text{ cm}^2$

When two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
From the similarity statement $\triangle DEF \sim \triangle PQF$, side EF corresponds to side QF.
The ratio of corresponding sides ($k$) is:

$$k = \frac{EF}{QF}$$ $$k = \frac{15 \text{ cm}}{10 \text{ cm}}$$ $$k = \frac{3}{2}$$

The ratio of the areas is $k^2$:

$$\frac{\text{Area}(\triangle DEF)}{\text{Area}(\triangle PQF)} = \left(\frac{EF}{QF}\right)^2$$ $$\frac{\text{Area}(\triangle DEF)}{25 \text{ cm}^2} = \left(\frac{3}{2}\right)^2$$ $$\frac{\text{Area}(\triangle DEF)}{25 \text{ cm}^2} = \frac{9}{4}$$

Now, solve for Area($\triangle DEF$):

$$\text{Area}(\triangle DEF) = \frac{9}{4} \times 25 \text{ cm}^2$$ $$\text{Area}(\triangle DEF) = \frac{225}{4} \text{ cm}^2$$ $$\text{Area}(\triangle DEF) = 56.25 \text{ cm}^2$$

The area of triangle DEF is $\boxed{56.25 \text{ cm}^2}$.