Let $X$ be the reflection of $A$ on $BC.$ Thus $ABXC$ is a rhombus with incircle $\omega \equiv \odot(M,MH)$ and let the tangent from $E$ to $\omega,$ other than $CX,$ cut $AC,AB$ at $D',F.$ Tangents $D'F,$ $XB,XC$ of $\omega$ induce a proyectivity between $AB,AC,$ thus if $B_{\infty},$ $C_{\infty}$ denote the points at infinity of $AB,AC,$ we get $(H,B,F,B_{\infty})=(A,C_{\infty},D',C)$ $\Longrightarrow$ $\tfrac{HB}{HF}=\tfrac{D'C}{D'A}=\tfrac{D'E}{D'F}$ $\Longrightarrow$ $HD' \parallel BE$ $\Longrightarrow$ $D' \equiv D$ $\Longrightarrow$ $DM$ bisects $\angle CDF \equiv \angle ADE.$

Is there a proof without projective geometry?

Any other solutions?

D' reflect D wrt AM, (DD'M) cuts AC at J, JD cuts C//AB at E so ED/EJ=CD/CA by Thales we need BH/BJ=CD/CA or BH.BA=CD.BJ or BM^2=BJ.BD' true by (DD'M) tangent BC so BE//HD so q.e.d

Cartesian coordinates $$B=(-x,0),C=(x,0),A=(0,y)$$where $x,y\in R_+$. Then $$M=(0,0), H=\left(\frac{-xy^2}{x^2+y^2},\frac{x^2y}{x^2+y^2}\right), D=\left(d,y\cdot\frac{x-d}{x}\right),E=\left(\frac{x^2y^2}{d(x^2+y^2)},y\cdot\frac{xy^2-d(x^2+y^2)}{d(x^2+y^2)}\right)$$where $d\in (0;x)$ is a fixed number. THEOREM Given two lines $l_1:\ y=a_1x+b_1,\ l_2:\ y=a_2x+b_2$ where $a_1,a_2,b_1,b_2\in R$. Then $\tan\angle \left(l_1,l_2\right)=\frac{a_1-a_2}{1+a_1a_2}$. We are talking of course about oriented angle. Assume all non-parallel to $OY$ axis lines $PQ$ are given by equation $y=a(PQ)\cdot x+b(PQ)$. Then$$a(AD)=-\frac{y}{x},\ a(DM)=\frac{y(x-d)}{dx},\ a(DE)=\frac{x^2y^3-2dxy(x^2+y^2)+d^2y(x^2+y^2)}{x^2y^2-d^2(x^2+y^2)}$$So $$\tan\angle \left(AD,DM\right)=\frac{-\frac{y}{x}-\frac{y(x-d)}{dx}}{1-\frac{y}{x}\cdot\frac{y(x-d)}{dx}}=\frac{-x^2y}{d(x^2+y^2)-xy^2}$$and $$\tan\angle \left(DM,DE\right)=\frac{\frac{y(x-d)}{dx}-\frac{x^2y^3-2dxy(x^2+y^2)+d^2y(x^2+y^2)}{x^2y^2-d^2(x^2+y^2)}}{1+\frac{y(x-d)}{dx}\cdot\frac{x^2y^3-2dxy(x^2+y^2)+d^2y(x^2+y^2)}{x^2y^2-d^2(x^2+y^2)}}$$common denominator $dx^2(x^2y^2-d^2(x^2+y^2))$ clearing $$\tan\angle \left(DM,DE\right)=\frac{x^2y(d^2(x^2+y^2)-2xy^2d+x^2y^2)}{(d(x^2+y^2)-xy^2)(-d^2(x^2+y^2)+2dxy^2-x^2y^2)}$$We are actually finished but it would be very nice to check that $-d^2(x^2+y^2)+2dxy^2-x^2y^2$ has no real roots (we don't want to divide by $0$) but that's true since $\Delta_d=-4x^4y^2<0$ since $x,y>0$

$\textbf{Solution by Sir JAWAD(TanR314) orz orz orz IMO Gold 2025}$ $E' \equiv ED \cap AB$ $D'$ be the reflection of $D$ over $AM$ $M' \equiv AM \cap \odot MD'E'$ $$\frac{BH}{BD'}=\frac{ED}{EE'}=\frac{CD}{CA}=\frac{BD'}{AB} $$$$\implies BD'.BE'=BA.BH=BM^{2}$$$\implies BM$ is tangent to $\odot MD'E'$ $$\implies \angle MD'M'=90 \implies \angle MDM'=90$$So, $M,D,M',D',E'$ cyclic Now, $$\angle MDE'=\angle MD'E'=\angle MD'B=\angle MDC \blacksquare$$