The segment DE drawn inside triangle ABC is parallel to BC, with D on AB and E on AC. If DE = 8 cm, BC = 20 cm and AE = 6 cm, find EC (in cm).

Since $DE$ is parallel to $BC,\frac{AE}{AC}=\frac{DE}{BC}=\frac{8}{20}=\frac{2}{5}$. With $AE=6$, that gives $AC=6\times \frac{5}{2}=15$ cm, so $EC=15-6=9$ cm. The value 15 answers with $AC$, the whole side, forgetting to subtract $AE$; 2.4 applies $\frac{2}{5}$ to $AE$ itself instead of using it to reach $AC$; 14 comes from subtracting the two given lengths 20 − 6, which mixes a piece of $BC$ with a piece of $AC$.