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.
 A: 
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.
A: 
Lemma. Given $\overline{XP} \cong \overline{XQ}$ and $\overline{XY}\cong\overline{ZQ}$ (and $\overline{XZ} \cong\overline{YP}$) as in the diagram, let $K$ be isosceles $\triangle XPQ$'s circumcenter (which necessarily lies on the bisector of $\angle X$). Then $X$, $Y$, $Z$, $K$ are concyclic.


Proof of Lemma. Evidently, $\triangle XYK \cong \triangle QZK$ (and $\triangle XZK \cong \triangle PYK$). Thus, $\angle XYK$ and $\angle XZK$ are supplementary, so that $\square XYKZ$ is cyclic. $\square$
Note that the Lemma implies that perpendiculars from $K$ meet $\overline{XP}$ and $\overline{XQ}$ at midpoints $M$ and $N$.

Thus,

$$|\overline{XM}| = |\overline{XN}| = \frac{1}{2}\left(\;|\overline{XY}| + |\overline{XZ}|\;\right)\quad\quad(\star)$$


In the problem at hand, we observe that the relations
$|\overline{BD}|+|\overline{BE}| = b$ and $|\overline{CE}| + |\overline{CD}| = c$ imply the counterpart $|\overline{AE}| + |\overline{AF}| = a$, which, by $(\star)$, tells us (for instance) that a perpendicular from $K$ meets $\overline{AC}$ at $K^\prime$ such that $|\overline{AK^\prime}| = a/2$. The location of $K^\prime$ (and $K$) is independent of the positions of $D$, $E$, $F$.

Writing $H$ for the circumcenter of $\triangle ABC$, and $H^\prime$ for the foot of the perpendicular from $H$ to $\overline{AB}$, we have $|\overline{BH^\prime}| = c/2$. From here, one can show that 
$$|\overline{HK}|^2 = R(R-2r) = |\overline{HI}|^2 \qquad (\star\star)$$
where $R$ and $r$ are, respectively, the circumradius and inradius of $\triangle ABC$, and where the second equality is Euler's Triangle Formula. Since $|\overline{HI}|$ is symmetric in the triangle's elements, we can replace $K$ with $L$ and $M$ in $(\star)$, so that:
$$|\overline{HI}| = |\overline{HK}| = |\overline{HL}| = |\overline{HM}|$$
showing that $I$, $K$, $L$, $M$ lie on a circle with center $H$.

My derivation of the first equality in $(\star\star)$ involves tedious and complicated trig. I'm still seeking an elegant solution.
