In the figure $\overleftrightarrow{BE} \parallel \overleftrightarrow{AC}$, $\overleftrightarrow{FG} \parallel \overleftrightarrow{AB}$ and $\frac{AB}{DB}=\frac{BC}{BE}=\frac{CB}{CG}=\frac{CA}{CF}=4$ Prove that $CA=2 \cdot EH$
I know that I need to show that $EH$ is twice as big as $ED$ which I can show is congruent to $CF$. Yet there were so many similar triangles in the figure that I ended up making more work then I feel is necessary. I also have a parallelogram in the figure with $ADHF$ but wasn't sure if that was going to help or not.
Since $FGC\sim HGE$ we have $EH/CF = GE/GC = 2:1$ so $$EH = 2CF =2\cdot {1\over 4}AC$$