It's easy to see that $I$ is the midpoint of $AB$. Furthermore, the equal angles in an isosceles triangle imply that $(ADEB)$ is cyclic. Keeping this and mind (as well as the cyclicity of $DHKF$ and the fact that $KH$ is a midline in triangle $DBE$) we get $\angle DFH = \angle DKH = \angle DBE = \angle DAE$, from where we get that $F$ is the midpoint of $AD$. Similarly, $G$ is the midpoint of $BE$. Furthermore $IJ || BE || KH$ (midlines), and similarly $IK || AD || JH$, therefore $IKHJ$ is a parallelogram. Similarly we obtain that $IHGF$ is a parallelogram whose diagonals intersect at $M$, therefore $M$ is the midpoint of $IH$, meaning it is also the intersection point of the diagonals in $IKHJ$, therefore $J, M,$ and $K$ are collinear.

Lemma: In a cyclic quadrilateral $ABCD$, let the midpoints of $AB,BC,CD,DA,AC,BD$ be $E,F,G,H,I,J,K$ respectively. Then $FH,IJ,EG$ are concurrent. Proof: Let $IJ\cap EG = K$. Because $EIGJ$ is a parallelogram, we have $K$ is the midpoint of $IJ$. Similarly $FIHJ$ is a parallelogram implying that $FH$ pass through the midpoint of $IJ$ which is $K$. $\blacksquare$ Let $M'$ be the midpoint of $KJ$. Since $IK=\frac{AD}{2} = JH$ and $IK\parallel AD\parallel JH$, we have $IKHJ$ is a parallelogram implying $I,M',H$ are collinear. Because $CD=DE$, we have $\angle BAC = \angle ACB = \angle DEC \implies BADE$ is cyclic. Let $G'$ and $F'$ be the midpoints of $BE$ and $AD$ respectively. We have $\angle EJH = \angle EAD = \angle EBD = \angle EGH \implies GEHJ$ is cyclic implying $G'=G$. Similarly we have $\angle HFD = \angle EAD = \angle EBD = \angle HKD \implies KHDF$ is cyclic implying $F'=F$. Finally applying the lemma to the cyclic quadrilateral $BADE$, we get $JK,IH,GF$ pass through a common point which is $M'$ so $M'=M$ as desired. $\blacksquare$

$H$ is the midpoint of the $DE$ $<=>$ $DH$=$EH$ $J$ is the midpoint of the $AE$ $<=>$ $AJ$=$EJ$ $K$ is the midpoint of the $BD$ $<=>$ $BK$=$DK$ $KI$ $\parallel$ $AC$ $=>$ $KI$ $\parallel$ $AD$ Since $BK$=$DK$ and $KI$ $\parallel$ $AD$ we get that $I$ is the midpoint of $AB$ $<=>$ $AI$=$BI$ $AJ$=$EJ$ , $DH$=$EH$ $<=>$ $JH$ is the midsegment of the $\triangle$ $ADE$ $=>$ $HJ$ $\parallel$ $AD$ $KI$ $\parallel$ $AD$ $\parallel$ $HJ$ $=>$ $KI$ $\parallel$ $HJ$ $...(1)$ $AI$=$BI$ , $AJ$=$EJ$ $<=>$ $IJ$ is the midsegment of the $\triangle$ $BAE$ $=>$ $IJ$ $\parallel$ $BE$ $BK$=$DK$ , $DH$=$EH$ $<=>$ $KH$ is the midsegment of the $\triangle$ $BDE$ $=>$ $KH$ $\parallel$ $BE$ $IJ$ $\parallel$ $BE$ $\parallel$ $KH$ $=>$ $IJ$ $\parallel$ $KH$ ...(2) Combining $(1)$ with $(2)$ we get that $\square$ IJKH is a parallelogram $=>$ Points $I-M-H$ and points $J-M-K$ are collinear $=>$ $J-M-K$ are collinear

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Proposed by Petar Filipovski.