There are lot of properties I think I found about this figure: For example, the properties of $E$, implies the ones of $F$. Also it implies $AC=BD$, and also the lines trought midpoints of $AB$ and $CD$, and $AD$ and $BC$ are perpendicular! This properties are not hard to prove. If someone is interested I can post them and show how they easily imply the result.

Ok can anyone show that property... Davron

standard using complex nos. take one pt. of intersection of perpendicular bisectors as origin,right given conditions using mod a\modb=a\b8eto the power ic,c-angle bet a and b.

I think there's a more basical solution, but I can't find it .

Problem: $E$ and $F$ are interior points of convex quadrilateral $ABCD$ such that $AE=BE$, $CE=DE$, $\angle AEB = \angle CED$, $AF=DF$, $BF=CF$, $\angle AFD = \angle BFC$. Prove that $\angle AFD+\angle AEB = \pi$. Solution: Notice that, because $E$ lies in the interior of convex quadrilateral $ABCD$, \[\angle AEC = \angle AED+\angle CED;\] \[\angle AEC = \angle AED+\angle AEB;\] \[\angle AEC = \angle BED,\] and that $AE = BE$ and $CE = DE$, so $\triangle AEC \cong \triangle BED$. Similarly, we show that $\triangle AFC \cong \triangle DFB$. Let $S$ be the intersection of $AC$ and $BD$. Then $\angle EAS = \angle EBS$. As $ABCD$ is convex, and thus that $S$ (as well as $E$) lies in the interior of $ABCD$, $A$, $B$, $E$ and $S$ are concyclic. Thus $\angle AEB = \angle ASB$. Similarly, we show that $\angle AFD = \angle ASD$. Notice that $B$, $S$ and $D$ are collinear, so $\angle ASD+\angle ASB = \pi$. Thus $\angle AFD+\angle AEB = \pi$. $\Box$

Attachments:
Solution CTST 01 1.tex (3kb)

Let $\angle BEA=x$ and $\angle AFC=y$. WLOG the vertices are labelled in anticlockwise order. Consider the following transformation: Anticlockwise rotation centered at $E$ with angle $x$, followed by anticlockwise rotation centered at $F$ with angle $y$. This transformation is a rotation with angle $x+y$, mapping point $A$ to point $C$, and mapping point $C$ to point $A$. Thus we see that $x+y$ is of the form $(180 + 360k)^\circ$, and a simple bound yields $x+y=180^\circ$.

I think that it is easy for China TST.

My solution: Let $ X $ be the intersection of $ AC $ and $ BD $ . Easy to see $ \triangle ACE \cong \triangle BDE $ . Since $ E $ is the center of spiral similar which map $ \triangle ACE \mapsto \triangle BDE $ , so we get $ X \in \odot (ABE) \Longrightarrow \angle CXB=2\angle CXE=2\angle EAB $ . ... $ (1) $ Similarly, we can prove $ X \in \odot (DAF) $ and $ \angle DXC=2\angle DXF=2\angle DAF $ . ... $ (2) $ Since $ \angle DXC+\angle CXB=\angle DXB=180^{\circ} $ , so from $ (1), (2) $ we get $ \angle EAB+\angle DAF =90^{\circ} \Longrightarrow \angle AFD+\angle AEB=180^{\circ} $ . Q.E.D