A line drawn parallel to the side BC of triangle ABC cuts AB at D and AC at E, splitting AB so that AD : DB = 4 : 5. Given BC = 27 cm, how long is DE (in cm)?

From $AD:DB=4$ : 5, the whole side $AB$ is 4 + 5 = 9 parts, so $\frac{AD}{AB}=\frac{4}{9}$. Because $DE$ is parallel to $BC$, triangle $ADE$ is similar to triangle $ABC$ and $\frac{DE}{BC}=\frac{AD}{AB}=\frac{4}{9}$. Hence $DE=\frac{4}{9}\times 27=12$ cm. The trap answer 21.6 is $\frac{4}{5}\times 27$, using $AD:DB$ where $AD:AB$ is needed — the classic BPT slip; 15 is the leftover 27 − 12; 13.5 halves $BC$.