Convex hexagon $ABCDEF$ following equalities $AD=BC+EF, BE=AF+CD, CF=DE+AB$. Prove that $\frac{AB}{DE}=\frac{CD}{AF}=\frac{EF}{BC}$ 
A convex hexagon $ABCDEF$ is such that the following equalities $AD=BC+EF, BE=AF+CD, CF=DE+AB$ hold. Prove that
  $$\frac{AB}{DE}=\frac{CD}{AF}=\frac{EF}{BC}$$

I do not know how to begin to solve this problem.
 A: Let $X = BE \cap CF$, $Y=CF \cap AD$, and $Z=AD \cap BE$. Let $T$ be a point such that $BCTE$ is parallelogram. 

Then $ET=BC$ and $CT=BE$. Looking at the quadrilateral $CTEF$ we get
$$CF < CT + TE + EF = BE + BC + EF = BE + AD.$$
Analogously we can show that $$AD < BE + CF \mathrm{\ \ and \ \ } BE < CF + AD,$$
therefore there exists a triangle with sides of lengths $AD, BE, CF$. 
So let $KLM$ be a triangle with $LM=AD, MK=BE, CF=KL$. 
Since $CF=KL$, $CT = MK$, and $FT \le EF + ET = AD = KM$, we find that $\angle TCF \le \angle LKM$. Since $BE \parallel CT$, we have $\angle EXF = \angle TCF$. Therefore 
$$\angle EXF \le \angle LKM.$$
Analogously we get 
$$\angle CYD \le \angle MLK \mathrm{\ \ and \ \ } \angle AZB \le KML.$$
Adding up yields $\pi \le \pi$, therefore we have equalities in all inequalities above.
In particular $EF+ET=FT$ so $E$ lies on the segment $FT$. Therefore 
$$\angle CFT = \angle MLK = \angle CYD$$ 
which means that $AD \parallel EF$. Analogously $AD \parallel BC$, $AB \parallel CF \parallel DE$, and $CD \parallel BE \parallel AF$.
Therefore $ABCY$, $DEFY$ are parallelograms. In particular $AB=CY$, $DE=YF$, $DY=EF$, and $YA=BC$. We get
$$\frac{AB}{DE} = \frac{CY}{YF} = \frac{CD}{AF} = \frac{DY}{YA} = \frac{EF}{BC},$$
where the second and third equality follows from similarity of $CYD$ and $FYA$.
