Given that BA is parallel to CE, we can identify two similar triangles: DCE and DBA.

Given that BA is parallel to CE, we can identify two similar triangles: $\triangle DCE$ and $\triangle DBA$.

• $\angle CDE = \angle BDA$ (Common angle)
• $\angle CED = \angle BAD$ (Corresponding angles, since CE || BA and AD is a transversal)
• $\angle DCE = \angle DBA$ (Corresponding angles, since CE || BA and BD is a transversal) Therefore, $\triangle DCE \sim \triangle DBA$ by the Angle-Angle-Angle (AAA) similarity criterion.

From the similarity $\triangle DCE \sim \triangle DBA$, the ratio of corresponding sides is equal:

$$\frac{CE}{BA} = \frac{DE}{DA}$$

We are given:
BA = $9 \text{ cm}$
CE = $4 \text{ cm}$
AE = $3 \text{ cm}$
Let DE = $x$.
Then DA = DE + AE = $x + 3 \text{ cm}$.

Substitute these values into the ratio:

$$\frac{4}{9} = \frac{x}{x+3}$$

Cross-multiply the equation:

$$4(x+3) = 9x$$

Distribute the 4 on the left side:

$$4x + 12 = 9x$$

Subtract $4x$ from both sides:

$$12 = 9x - 4x$$ $$12 = 5x$$

Divide by 5:

$$x = \frac{12}{5}$$ $$x = 2.4$$

So, the length of DE is $2.4 \text{ cm}$.