Since $AB \parallel DE$, triangles $ABC$ and $EDC$ are similar (AA similarity).

In right triangle $EDC$: $\angle D = 90°$, $CD = 3$, $DE = 4$, so $CE = \sqrt{9 + 16} = 5$.

The similarity ratio: $\frac{AB}{ED} = \frac{BC}{DC} = \frac{AC}{EC}$.

From the figure, $AB = 5$, $ED = 4$:

$\frac{AC}{EC} = \frac{AB}{ED} = \frac{5}{4}$

$AC = 5 \times \frac{5}{4} = \frac{25}{4}$