i use two lemma to proof problem and proof it soon first lemma: $BK$is perpendicular$AI$ and $CL$is perpendicular to $AI$ second lemma: $LM$is parallel to $AB$ proof of problem: by lemma 1 $A,K,H,B$are concyclic and $A,H,C,L$are concyclic so $\angle KHM = \angle BAI$ and by lemma 2 we know that $\angle MLA = \angle BAL $ so $\angle MLK = \angle MHK$ i write the lemma proof soon

Dear Mathlinkers, always nice... based 1. on a result of Altshiller-Court that you can see in my article : An unlikely concurrence... vol. 4 p. 5 2. on the Lieutenant Calabre...: Symétrique de (OI)... vol. 4 p. 6 in http://perso.orange.fr/jl.ayme Sincerely Jean-Louis

Of course, $\angle LFE=\angle DFE=90^{\circ}-C$ so $AI\perp EF$ we have $\angle FLI=\angle\frac{C}{2}$. So $\angle FLI=\angle ICD$. This implies $ICDL$ is cyclic, and clearly $E$ also lies on this circle, which has $IC$ as diameter. Thus $\angle ALC=90^{\circ}$. Let $CL$ meet $AB$ at $C'$. Then $AI\perp CC'$ and since $\angle C'AL=\angle CAL$, it is clear that $L$ is the midpoint of $CC'$. Then since $M$ is the midpoint of $BC$, it follows that $BC'||LM$ i.e. $AB||LM$. By a symmetric argument, using the point $B'$ etc, we can prove that $MK||AC$. Now $ALHC$ is cyclic since $\angle ALC=\angle AHC=90^{\circ}$. Thus $\angle LHM=\angle LAC=\frac{1}{2}A$. But since $MK||AC$, we also have $\angle MKL=\angle LAC=\frac{A}{2}$. So $\angle MHL=\angle MKL$ and the conclusion follows.

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Another solution: Let $AI$ cut the circumcircle of $ABC$ at $G$, the middle point of the lower arc $BC$. It is obvious that $O, M, G$ are collinear. $\angle AEK = 90^{\circ} + \angle \frac{ACB}{2}$, thus $\angle AKE = \angle \frac{ABC}{2}$, hence $BIDK$ is cyclic, from which we obtain a very well known fact that $BK\perp AK$. Similarly $CL\perp AL$. $ALHC$ is cyclic, hence $\angle KLH = \angle HCA = \angle BGA$. But $BMKG$ is also cyclic, hence $\angle BGA = \angle KMH$, thus $\angle KLH = \angle KMH, MKHL$ is cyclic.

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Before giving a solution I shall present two prerequisite lemmas. Lemma 1: $CK \perp AK$. Solution: $\angle{CIA}=90+\frac{B}{2}$ and $\angle{KDB}=\angle{EDB}=90-\frac{B}{2}$ so ${CIKD}$ concyclic.So $\angle{CKA}=\angle{CKI}=\angle{IDC}=90^{\circ}$. Lemma 2: $BL \perp AL$ Solution: $\angle{BIL}=90-\frac{C}{2}$ and $\angle{BDL}=\angle{FDC}=90-\frac{C}{2}$.Thus $\angle{ALB}=\angle{ILB}=\angle{IDB}=90^{\circ}$. Now I come back to the main problem.Let $N$ be the midpoint of $AB$.Then clearly $N$ is the circumcenter of $ALB$ so $\angle{LNB}=2\angle{LAB}=A$.In other words $NL \parallel AC$ or $N,M,L$ are collinear.Thus $\angle{HML}=\angle{CML}=C$.Also note that $\angle{AHC}=\angle{AKC}=90^{\circ} \implies ACHK$ cyclic.So $\angle{HKL}=\angle{ACH}=C$.Combining these we have $\angle{HKL}=\angle{HML}=C$ or $MKHL$ is cyclic,as desired.

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We have that: $$\angle GLD = \angle FLA = 180-\left( \angle LAF + \angle LFA\right) = 180-\left( \frac{1}{2} \angle A + 90 + \frac{1}{2}\angle B\right) = \frac{1}{2} \angle C$$this implies that $CDLIE$ is a cyclic quad. Similarly we have that $\angle DKI = \frac{1}{2} \angle B$, which implies that $BKDIF$ is cyclic as well. The statements above imply that $BK \perp AI$ and $CK \perp AI$. This implies that $BKHA$ and $CHLA$ are both cyclic. Now we have the following lemma, which ends the problem. $\color{red}\rule{25cm}{0.5pt}$ Lemma: $LM \parallel AB$. Proof (Sketch): Throw this onto the complex plane with the circumcircle being the unit circle and let $c=\frac{1}{b}$. By this way we have that $t=-1$ Since we have that $CL \perp AI$ and $L \in AI$ we easily get that: $$l=\frac{2b^4-2ab^4+(1+ab^3)(b^2+1)}{2b(b^3+1}$$but this implies that: $$\frac{m-l}{\overline{m-l}}=\frac{m-n}{\overline{m-n}}$$thus we have that $M,L$ and $N$ are colinear. $\color{red}\rule{25cm}{0.5pt}$ Now back to our problem. We have that: $$\angle LMH = \angle LMG = \angle GBA = \angle HBA = \angle HKA = \angle HKL$$thus we have that $KMLH$ is a cyclic quadrilateral.