Geometry Question - length ratio in a triangle 
In the figure, CD=2AB=2BC and FE = ED
Find AG: HE
This is an Olympic question in China, I tried, still can't figure it out. Please.
 A: Hint: Note that EC is half FA and parallel to it. So HE:HA and FH:FC ratios should be clear. Then apply Menelaus' theorem for triangle AHC slashed by line FGB in order to infer ratio AG:GH.
A: since the point $F$ is free to move, then we may choose the triangle $ADF$ to be right-angled and isosceles. in which case $AG=HE$. however this does not prove the general result unless we can show this ratio invariant under translations of $F$
A: Since $2|AB| = 2|BC| = |CD|$ it follows that $C$ is the midpoint of $AD$. By definition, we also know that $E$ is the midpoint of $FD$. It follows that $FC$ and $AE$ are medians of triangle $\triangle FAD$ and hence $H$ is the centroid of the triangle. The centroid divides any median into a ratio of $1:2$, hence $|AH| = 2|HE|$. 
Now notice that the points $A,G,H,E$ and $A,B,C,D$ are related by a perspectivity centered at $F$. It follows that the cross-ratio is invariant between the two point sets.
Therefore we have:
$$3=\frac{|AC||DB|}{|DC||AB|} = (A,D; C,B) = (A,E; H,G) = \frac{|AH||EG|}{|EH||AG|}=2\frac{|EG|}{|AG|}$$
$$\frac{1}{2}=\frac{|AB||CD|}{|CB||AD|}=(A,C;B,D)=(A,H; G,E)=\frac{|AG||HE|}{|HG||AE|} = \frac{1}{3}\frac{|AG|}{|HG|}$$
Hence we have 
$$\frac{|EG|}{|AG|} = \frac{|AG|}{|HG|}= \frac{3}{2}$$
Finally, we have
$$\frac{3}{2}=\frac{|EG|}{|AG|} = \frac{|GH|}{|AG|}+\frac{|HE|}{|AG|} = \frac{2}{3} + \frac{|HE|}{|AG|}$$
Therefore
$$\frac{|AG|}{|HE|} = \frac{6}{5}$$
A: It is possible to approach the problem through Ceva's and Van Obel's theorems.
Call $G',H'$ the points in which $DG,DH$ cut $AF$. Due to Ceva's theorem we have:
$$\frac{AG'}{G'F}=\frac{1}{3},\qquad \frac{AH'}{H'F}=1,$$
hence Van Obel's theorem gives:
$$\frac{AG}{GE}=\frac{2}{3},\qquad \frac{AH}{HE}=2,$$
and we have:
$$ AG:GH:HE = 6:4:5,$$
hence $\frac{AG}{HE}=\frac{6}{5}$.
