I think that something is wrong. I got $AK=EK$ iff $AB=AC$.

oneplusone wrote: I think that something is wrong. I got $AK=EK$ iff $AB=AC$. I think, your claim is not true. Here is the official solution. From $AD=AF, \: BD=BE$ and $DE=DF$ we obtain $AD\sin \left(\frac{A}{2}\right)=BD\sin \left(\frac{B}{2}\right).$ Using this and applying the law of sines to the triangle $AIB,$ we get $\frac{AI}{BI}=\frac{AD}{BD}=\frac{AK}{BE}.$ 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 $AK=EK.$

Oops sorry, I thought I saw $CE=CF$, not $DE=DF$.

there is something strange down there let $ BK $ and $ DE $ meet at $ P $ $ DP=PE $ and $ \angle DPB=90 $ if $ BD=BE $ and $ \angle DBK=\angle EBK $ $ KD=KE $ if $ DP=PE $ and $ \angle DPK=90 $ so $ EK=DK=AK $ i didn't use $ AD=AF $ , $ DF=DE $ or $ AK $ is tangent to the circumcircle of $ AIB $ Where is the problem?

Can i know where can we find the official solution for all national math olympiad and tst for Turkey country????