解:由题图及题设,显然 | A F | = | F E | = 1 2 | A E | {\displaystyle \left\vert AF\right\vert =\left\vert FE\right\vert ={\frac {1}{2}}\left\vert AE\right\vert } . 又 ∵ D E ∥ B C {\displaystyle \because DE\parallel BC} , ∴ △ A D E ∼ △ A B C {\displaystyle \therefore \triangle ADE\sim \triangle ABC} , 显然相似比为 2 3 {\displaystyle {\frac {\sqrt {2}}{\sqrt {3}}}} , ∴ | A E | | A C | = 2 3 {\displaystyle \therefore {\frac {\left\vert AE\right\vert }{\left\vert AC\right\vert }}={\frac {\sqrt {2}}{\sqrt {3}}}} , 即 | A E | | E C | = 2 3 − 2 {\displaystyle {\frac {\left\vert AE\right\vert }{\left\vert EC\right\vert }}={\frac {\sqrt {2}}{{\sqrt {3}}-{\sqrt {2}}}}} , ∴ | A F | : | F E | : | E C | = 1 : 1 : ( 6 − 2 ) {\displaystyle \therefore \left\vert AF\right\vert :\left\vert FE\right\vert :\left\vert EC\right\vert =1:1:({\sqrt {6}}-2)} .