Hello. I think i solved it. First show that A1,D1,E2 are on a same line ; then prove that D1D3 ll E2E3 ll A2A3 ; and then show that D1D3 / E2E3 = 11/14 .

So, here is my solution for this problem: First we will show that $A_{n}A_{n+1}\parallel E_{n}E_{n+1}\parallel D_{n+1}D_{n+2}$. Let $G$ be the center of gravity of $A_{1}A_{2}A_{3}$, thus the intersection of the lines $A_{n}B_{n+1}$. By Menelao's Theorem on $A_{1}GB_{1}$ we have: $ \frac{A_{1}E_{1}}{E_{1}G}\cdot \frac{GA_{3}}{A_{3}B_{1}}\cdot\frac{B_{1}C_{1}}{C_{1}A_{1}}=-1$, by midpoints and properties of the gravicenter $ \frac{A_{1}E_{1}}{E_{1}G}=\frac{3}{2}$. So every $E_{n}$ divides each $A_{n}G$ in such proportion, and by Thales each pair $A_{n}A_{n+1}\parallel E_{n}E_{n+1}$. Similar resoning gives us each pair $A_{n}A_{n+1}\parallel D_{n+1}A_{n+2}$, but in now each $D_{n}$ cuts $B_{n}G$ in proportion $ \frac{B_{n}D_{n}}{D_{n}G}=\frac{3}{4}$. So the three triangles are similar, and furthermore, homothetic. Therefore $G$ is also the center of gravity of $D_{1}D_{2}D_{3}$ and $E_{1}E_{2}E_{3}$. Finally, we know that the proportion of two similar triangles is the proportion of one of their corresponding lines, and we have, because of our previous observations on the medians : $ \frac{D_{1}D_{2}D_{3}}{A_{1}A_{2}A_{3}}=\frac{4}{14}$, $ \frac{E_{1}E_{2}E_{3}}{A_{1}A_{2}A_{3}}=\frac{2}{5}$, which finally gives: $ \frac{D_{1}D_{2}D_{3}}{E_{1}E_{2}E_{3}}=\frac{5}{7}$, and since the proportion of the areas is the square of the proportion of the sides we have: $ \frac{|D_{1}D_{2}D_{3}|}{|E_{1}E_{2}E_{3}|}=\frac{25}{49}$ Yay!! My first LaTeX proof!!

one can use barycentric coordinates to trivialize this problem

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO1989Problem3 Vo Duc Dien

By ratio lemma, $\frac{B_nD_n}{A_{n+2}D_n} = \frac{1}{6}$, so $\triangle D_1D_{2}D_{3}$ consists of $\triangle A_1A_2A_3$ scaled by $-\frac{2}{7}$ at the centroid. Similarly, $\frac{A_nE_n}{B_{n+1}E_n} = \frac{2}{3}$, so $\triangle E_1E_2E_3$ consists of $\triangle ABC$ scaled by $\frac{2}{5}$ at the centroid. Hence $\frac{[D_1D_2D_3]}{[E_1E_2E_3]} = \left(\frac{\frac{2}{7}}{\frac{2}{5}}\right)^2 = \frac{25}{49}$.

Let $G$ be the centroid of $\triangle ABC$. By the Ratio Lemma, we have \begin{align*} \frac{B_nD_n}{A_{n+2}D_n} &= \frac{A_nB_n \sin \angle A_{n+1}A_nC_{n+1}}{A_nA_{n+2} \sin \angle A_{n+2}A_nC_{n+1}} \\ &= \frac{1}{2} \cdot \frac{2A_nB_n \sin \angle A_{n+1}A_nC_{n+1}}{A_nA_{n+2} \sin \angle A_{n+2}A_nC_{n+1}} \\ &= \frac{1}{2} \cdot \frac{A_nA_{n+1} \sin \angle A_{n+1}A_nC_{n+1}}{A_nA_{n+2} \sin \angle A_{n+2}A_nC_{n+1}} \\ &= \frac{A_{n+1}C_{n+1}}{2A_{n+2}C_{n+1}} = \frac{1}{6} \end{align*} for $n=1,2,3$. Hence, \[\frac{GD_n}{GB_n} = \frac{4/3}{7/3} = \frac{4}{7},\] which means that \[\frac{[D_1D_2D_3]}{[A_1A_2A_3]} = \frac{[D_1D_2D_3]}{4[B_1B_2B_3]} = \frac{1}{4} \left(\frac{4}{7} \right)^2 = \frac{4}{49}.\] The Ratio Lemma also gives \begin{align*} \frac{A_nE_n}{B_{n+1}E_n} &= \frac{A_nA_{n+2} \sin \angle A_nA_{n+2}C_n}{A_{n+2}B_{n+1} \sin \angle A_{n+1}A_{n+2}C_n} \\ &= 2 \frac{A_nA_{n+2} \sin \angle A_nA_{n+2}C_n}{2A_{n+2}B_{n+1} \sin \angle A_{n+1}A_{n+2}C_n} \\ &= 2 \frac{A_nA_{n+2} \sin \angle A_nA_{n+2}C_n}{A_{n+1}A_{n+2} \sin \angle A_{n+1}A_{n+2}C_n} \\ &= 2 \frac{A_nC_n}{A_{n+1}C_n} = \frac{2}{3} \end{align*} for $n=1,2,3$. This means that \[\frac{GE_n}{GA_n} = \frac{4/3}{10/3} = \frac{2}{5},\]\[\implies \frac{[E_1E_2E_3]}{[A_1A_2A_3]} = \left(\frac{2}{5} \right)^2 = \frac{4}{25},\] upon which dividing gives $\boxed{\tfrac{25}{49}}$.