\[Рисунок\ в\ учебнике.\]
\[Дано:\]
\[DE \parallel AC;\]
\[BE = 10\ см;\]
\[BD = 2AD.\]
\[Решение.\]
\[1)\ Согласно\ теореме\ Фалеса:\]
\[\frac{\text{BD}}{\text{AD}} = \frac{\text{BE}}{\text{CE}}\]
\[CE = \frac{BE \bullet AD}{\text{BD}}\]
\[CE = \frac{10 \bullet AD}{2AD} = 5\ см.\]
\[2)\ BC = BE + CE = 15\ см.\]
\[Ответ:\ \ 15\ см.\]