South African Mathematics Olympiad Second Round

If two triangles have the same height, then the ratio of their areas is equal to the ratio of their bases. It follows that $\frac{DE}{EC}=\frac{6}{10}=\frac{3}{5}$, so $\frac{DE}{DC}=\frac{3}{3+5}=\frac{3}{8}$. Similarly, $\frac{DF}{FC}=\frac{17}{7}$, so $\frac{DF}{DC}=\frac{17}{17+7}=\frac{17}{24}$. Therefore $\frac{EF}{DC}=\frac{DF-DE}{DC}=\frac{17}{24}-\frac{3}{8}=\frac{17-9}{24}=\frac{8}{24}=\frac{1}{3}$. It follows that $△BEF=\frac{1}{3}△BDC=\frac{1}{3}(17+7)=8$.