Vtyjkyuk

Let $ABC$ be a triangle and $S$ its circumcircle. The points $D$, $E$, and $F$ are the feet of altitudes drawn from $A$, $B$, and $C$, respectively. The line $AD$ meets $S$ again at $K$. Let $M$ and $N$ be the midpoints of $DE$ and $DF$, respectively. If $M$ and $N$ respectively intersect $S$ at $I$ and $J$, prove that $D$ is the incentre of the triangle $IJK$.

I tried angle chasing, but failed to preceed something good (though I use scale and compass I failed to construct a proper diagram because there are so many intersections. Actually we don't have many thinks about the line joining the foots of altitudes.
Please help me by giving hints or full answer

Remark: The problem as stated is not entirely correct. Only when both $angle ABC$ and $angle ACB$ are acute angles do we have that $D$ is the incenter of the triangle $IJK$. When $angle ABC$ or $angle ACB$ is a right angle, we obtain a degenerate case, where $M=E=D=B$ so that $IJ=AB$, or $N=F=D=C$ so that $IJ=AC$, respectively. If $angle ABC>dfracpi2$ or $angle ACB>dfracpi2$, then $D$ is an excenter of the triangle $IJK$. See an illustration below. (If we define $I$ and $J$ so that $MI$ and $NJ$ are disjoint, then in the case $angle ABC>dfracpi2$, $D$ is the excenter of the triangle $IJK$ opposite to $I$. On the other hand, if $angle ACB>dfracpi2$, then $D$ is the excenter of the triangle $IJK$ opposite to $J$.)

Let $R$ denote the circumradius of the triangle $ABC$. Write $alpha$, $beta$, and $gamma$ for the angles $angle BAC$, $angle CBA$, and $angle ACB$, respectively. In what follows, we assume that $beta$ and $gamma$ are in the interval $left(0,dfracpi2right)$. First, we assume that $alphainleft(0,dfracpi2right)$. Then, $$angle DFE=pi-2gamma,.$$ Thus, the distance from $D$ to $EF$ is given by $$DF,sin(pi-2gamma)=DF,sin(2gamma),.$$
Since the triangle $ABC$ and the triangle $DBF$ are similar, we get
$$DF=AC,left(fracBDABright)=AC,cos(beta)=2,R,sin(beta),cos(beta)=R,sin(2beta),.$$
Thus, the distance from $D$ to $EF$ is $R,sin(2beta),sin(2gamma)$, whence the distance from $D$ to $IJ$ is
$$rho:=frac12,R,sin(2beta),sin(2gamma),.$$

It is not difficult to see that $OAperp EF$, whence $OA perp IJ$. Therefore, $A$ is the midpoint of the arc $IAJ$. This shows that $$angle DKI=angle AKI=angle AKJ=angle DKJ=:theta,.$$ Ergo, the distance between $O$ and $EF$ is
$$beginalignd:=OA-AE,sin(beta)&=R-AB,left(fracAEABright),sin(beta)\&=R-big(2,R,sin(gamma)big),cos(alpha),sin(beta)\&=Rbig(1-2,cos(alpha),sin(beta),sin(gamma)big),.endalign$$
Thus, the distance $delta$ between $O$ and $IJ$ is
$$beginaligndelta=d-rho&=R,Big(1-2,cos(alpha),sin(beta),sin(gamma)-2,cos(beta),cos(gamma),sin(beta),sin(gamma)Big)
\
&=R,Big(1+2,big(cos(beta+gamma)-cos(beta),cos(gamma)big),sin(beta),sin(gamma)Big)
\
&=R,Big(1-2,sin^2(beta),sin^2(gamma)Big),.endalign$$

Finally, note that $$beginalignDK=DH&=BD,cot(gamma)=big(AB,cos(beta)big),cot(gamma)\&=big(2,R,sin(gamma)big),cos(beta),cot(gamma)\&=2,R,cos(beta),cos(gamma),.endalign$$
Since $delta=R,cos(2,theta)=R,big(1-2,sin^2(theta)big)$, we conclude that
$$sin(theta)=sin(beta),sin(gamma),,$$
whence the distance from $D$ to $IK$ or to $JK$ is
$$DK,sin(theta)=DK,sin(beta),sin(gamma)=2,R,cos(beta),cos(gamma),sin(beta),sin(gamma)=rho,.$$ Hence, $D$ is the incenter of the triangle $IJK$, with inradius $rho=dfrac12,R,sin(2beta),sin(2gamma)$.

The case $alpha inleft(dfracpi2,piright)$ can be dealt with in a similar manner. One of the differences is that $angle DFE=2gamma$.

Consider the special case where $triangle ABC$ is equilateral, with circumradius $OC=1$.

Since $CB=sqrt3$ and $DE=fracCB2$, then $MD=fracsqrt34$, and in $triangle MDR$, $MR=frac12MD=fracsqrt38$, and $$RD=sqrtMD^2-MR^2=sqrtfrac1264-frac364=frac38=.375$$

Now from $D$ make $DGperp IK$. Since chords $AK$ and $IJ$ intersect at $R$, then$$ARcdot RK=IRcdot RJ$$And since $AR=frac98$, and $RK=frac78$, and $IR=RJ$, then$$frac6364=IR^2$$and$$IR=sqrtfrac6364=frac38sqrt7$$Hence$$tanangle KIR=fracRKIR=fracsqrt73=.8819$$and$$angle KIR=arctan .8819=41.41^o$$Therefore$$angle GKD=angle IKR=48.59^o$$And since$$cos48.59^o=.6614=fracGKDK=fracGK.5$$then$$GK=.5times.6614=.3307$$and$$DG=sqrt.5^2-.3307^2=.375$$Therefore, the circle with radius $DR=.375$ is tangent to $IK$ at $G$. And because of the symmetry of the figure about $AK$, the same argument will show that the perpendicular from $D$ to $JK=.375$.

Hence $D$ is the incenter of $triangle IJK$ when $triangle ABC$ is equilateral.

So far so good, but what might be done to generalize this?