Team Selection Test for IMO 2010

From $AD=AF$, $BD=BE$ and $DE=DF$ we obtain $AD\sin (\angle A/2)=BD\sin (\angle B/2)$. Using this and applying the law of sines to the triangle $AIB$, we get $AI/BI=AD/BD=AK/BE$. Since $AK$ is tangent to the circumcircle of $AIB$, we also have $\angle KAI=\angle ABI=\angle EBI$. Hence the triangles $KAI$ and $EBI$ are similar. It follows that the points $A,I,E$ are collinear and the points $A,K,E,B$ are concyclic. In particular, $\angle KEA=\angle KBA=\angle EBI=\angle KAE$ and $AK=EK$.