A question about 4 concyclic points In a triangle $ABC$, let $I$ denote its incenter.  Points $D, E, F$ are chosen on the segments $BC, CA, AB$, respectively, such that $BD + BF = AC$ and $CD + CE = AB$.  The circumcircles of triangles $AEF, BFD, CDE$ intersect lines $AI, BI, CI$, respectively, at points $K, L, M$ (different from $A, B, C$), respectively.  Prove that $K, L, M, I$ are concyclic.
Can someone help me? This was part of the Indian IMO TST in 2013. No one solved it as far as I know. A solution would be nice. Thanks in advance.


Proof.
Three circumcircles of $\triangle AFE$,$\triangle BDF$,$\triangle CED$ concur at $M$ (Miquel point)
$\angle BB'M+\angle AA'M=\angle BFM+\angle AFM=180^{\circ}$ and then $\square IA'MB'$ is concyclic.
Likewise, $\square IB'MC'$ is concyclic and then $I$,$A'$,$B'$,$C'$,$M$ are concyclic.    
Remark.
(1) $I$ need not be incenter.
(2) $BF+BD=AC$, $CD+CE=AB$ are useless.
