$\angle AXY=\angle AFE=\angle ACB=\angle ADB$ $\Longrightarrow$ $B,X,T,D$ are concyclic. Further, $\angle TEH=\angle THE=\angle BHD=\angle BDH$ $\Longrightarrow$ $D,B,T,E$ are concyclic, thus $D,B,X,T,E$ are concyclic $\Longrightarrow$ $\angle EDX=\angle ETY.$ By similar reasoning, we have $\angle FDY=\angle FTX.$ But $\triangle TEF$ is obviously T-isosceles and $XY \parallel EF$ bisects $\angle ETF$ $\Longrightarrow$ $\angle ETY=\angle FTX$ $\Longrightarrow$ $\angle EDX=\angle FDY$ $\Longrightarrow$ $\angle XDF=\angle YDE.$

@Luis Gonzalez, nice solution! But, i have another solution. Let $l$ be perpendicular line to $AD$ at the point $D$, $CH\cap l=P$, $BH\cap l=Q$, $AB\cap l=M$, $AC\cap l=N$ and $P$ is projection of the point $X$ on the lind $AD$. We have that \[ \frac{MX}{MA}=\frac{DP}{AD}=1-\frac{\cos A\sin B}{2\sin C} .\] $ \triangle DMA$ similar to $ \triangle DPH$ and let a point $X'$ lie on $PH$ and $ \angle XDX'=90^\circ $ and $X'DFX$ is cyclic. So \[ X'F=\frac{2R\sin A \cos (B-C)}{\sin C} \] and similar that we have $Y'$ lie on the $QH$, $Y'DEY$ is cyclic and \[ Y'E=\frac{2R\sin A \cos (B-C)}{\sin B} .\] Hence $ \frac{FX'}{EY'}=\frac{AC}{AB}=\frac{AX}{AY}=\frac{FX}{EY} $ $ \Rightarrow $ $ \triangle X'FX$ is similar to $ \triangle Y'EY$ and $ \angle FX'X=\angle EY'Y $ $ \Rightarrow $ $ \angle FDX=\angle EDY $.

Nice , it's 5'th solution for this nice problem

Let $EF$ intersect $AD$ at $M$. We have $BDMF$ and $BDTX$ cyclic so we easily get $\angle FDX=\angle TBM$. By analogy we get $\angle EDY=\angle MCT$ so we need to prove $\angle MBT=\angle MCT$ or in other word that $M$ is the orthocenter of $\triangle BTC$ or $RM \cdot RT=RB \cdot RC=RH \cdot RA$ where $R$ is the intersection of $AD$ and $BC$. This is well know to be true because $R,H,M,A$ are harmonic so we are finished.

Nice problem ! Assume that $CH$ intersects $\odot(ABC)$ at point $C'$ and $BH$ intersects $\odot(ABC)$ at point $B'$ Claim: Quadrilaterals $TBDE$ and $TFDC$ are cyclic Proof: It is well -known that $HC'=2HF$ and $HB'=2HE$. From Power of Point we have: \begin{align*} AH \cdot HD = BH \cdot HB' \\ 2HT \cdot HD = BH \cdot 2HE \\ TH \cdot HD = BH \cdot HE \end{align*}This proves that quadrilateral $TBDE$ is cyclic. Similarly we can obtain that quadrilateral $TFDC$ is cyclic. Claim: Quadrilaterals $TXBD$ and $TYCD$ are cyclic Proof: Note that: $$ \angle AXY = \angle AFE = \angle AHE = \angle ACB = \angle ADB $$This proves that quadrilateral $TXBD$ is cyclic. Similarly we can obtain that quadrilateral $TYCD$ is cyclic. Now to finish the problem note that: \begin{align*} \angle TDE = \angle TBE =\alpha \\ \angle TDX =\angle XBT =\beta \\ \angle TDF =\angle TCF = \alpha_1 \\ \angle TDY =\angle TCY =\beta_1 \end{align*}Obviously $\angle FBE = \alpha + \beta = \angle ECF =\alpha_1 + \beta_1$. This implies: \begin{align*} \alpha -\beta_1 = \alpha_1 -\beta \\ \angle TDE - \angle TDY =\angle TDF - TDX \\ \angle YDE = \angle XDF \end{align*}as desired.