5 mins solve. First clearly $H$ lies on $(BDC)$ becuase $\angle BHD=90-\angle BAC=\angle HCD$, now $\angle BDE=\angle BCE=\angle BCD=\angle FHD$ hence $BD$ is tangent to $(FHD)$ as desired, thus we are done

claim 1 : $D$ lies on $(BHC)$ : proof: Let $D'=(BHC) \cap AC$ so we have $\angle ADB=180-\angle BDC=180-\angle BHC = \angle BAD \implies BA=BD' \implies D=D'$ claim 2 : $A,H$ and $F$ are collinear : proof: we have $BA=BF$ and $F \in (BHS)$ so $F$ is the reflexion of $A$ about line $BC$, therfore $F$ lies on $AH$ now we have $\angle HFD = 90- \angle ADF = 90 - \angle EBC = 90-\angle ABC =\angle HCB = \angle HBD$ so $BH$ is a tangent to $(DFH)$

inspired problem : Let $ABC$ be a triangle with $\angle ABC \neq 90^\circ$, and $AB$ its shortest side. Let $H$ be the orthocenter of $ABC$. Let $\Gamma$ be the circle with center $B$ and radius $BA$. Let $D$ be the second point where the line $CA$ meets $\Gamma$. Let $E$ be the second point where $\Gamma$ meets the circumcircle of the triangle $ABC$. Let $F$ be the intersection point of the lines $DE$ and $BH$. Prove that the circumcenter of $(HDF)$ lies on line $AC$.

Here's a link to a video solution https://youtu.be/1p6NAbU8cxw