Let $K$ be on line $DE$ such that $BK$ is the internal angle bisector of $\angle ABC$. Note that $DA=DB=DK$ by angle-chasing, so thus $H,K$ lie on the circle with diameter $AB.$ $F$ also lies on this circle since it's the reflection of $H$ over a line through the center, $D$. Hence, $BHKF$ is cyclic, while $KF=KH$ implies that $BK$ also bisects $\angle FBH$. Hence, $BF$ is isogonal to $BH$ and passes through $O$.

Here is my solution for this problem Solution Let $S$ be midpoint of $BC$; $K$ $\equiv$ $BH$ $\cap$ $DE$ We have: ($DB$; $DF$) $\equiv$ ($DB$; $DE$) + ($DE$; $DF$) $\equiv$ ($BA$; $BC$) + ($DH$; $DE$) $\equiv$ ($BA$; $BC$) + ($DH$; $DS$) + ($DS$; $DE$) $\equiv$ ($BA$; $BC$) + ($SD$; $SE$) + ($CA$; $CB$) $\equiv$ ($BA$; $BC$) + ($AC$; $AB$) + ($CA$; $CB$) $\equiv$ 2 ($CA$; $CB$) $\equiv$ 2 ($EA$; $ED$) $\equiv$ ($EA$; $EF$) $\equiv$ ($EA$; $EF$) $\equiv$ ($KB$; $KF$) (mod $\pi$) So: $B$, $D$, $K$, $F$ lie on a circle Then: ($BF$; $BD$) $\equiv$ ($KF$; $KE$) $\equiv$ ($KE$; $KH$) $\equiv$ ($BC$; $BH$) (mod $\pi$) or $BF$, $BH$ are isogonal conjugate with respect to $\triangle$ $ABC$ Hence: $BF$ passes through circumcenter of $\triangle$ $ABC$

It can be done using Complex bash also, Set $\odot(ABC)$ as the unit circle, so, $h=\frac{1}{2}(a+b+c-\frac{ac}{b})$, also $d=\frac{a+b}{2}$ and $e=\frac{a+c}{2}$. Now by the reflection lemma you get $f$. Now we see that $\frac{b-f}{b-o}\in\mathbb R$, where $o=0$. So, $B,F,O$ are collinear.

We only need to prove ∠FBA = 90 - ∠C. ∠AHB = 90 ---> DA = DH = DB F is refection of H about DE ---> DH = DF so DA = DH = DB = DF and it means AHFB is cyclic so ∠FBA = ∠FHE which is 90 - ∠c. we're Done.

Interesting problem Here is my solution: We extend $BF$ which intersects the circumcircle of $\bigtriangleup ABC$ at $I$.Then join $IC,EF,DF,DH$ Claim 1 : $BAHF$ is cyclic Proof : In $\bigtriangleup ABH$, We have $DA=DB=DH$.Since $DE$ is the perpendicular bisector of $HF$, $DHEF$ is a kite. Thus, $DH=DF=DA=DB$ implies that indeed quadrilateral $BAHF$ is cyclic with center $D$. Claim 2 : $\angle BCI=90^{\circ}$ Proof : Extend $HF$ which intersects $BC$ at $G$.Then we have $\angle HBG=\angle GHC\equiv \angle FHC=\angle ABF $. Then $\angle ABH=\angle ABF-\angle HBF=\angle HBG-\angle HBF=\angle FBG\equiv \angle IBC$ Now $\angle ABH=\angle IBC$ and $\angle BAC\equiv \angle BAH=\angle BIC$.Thus $\bigtriangleup ABH\sim\bigtriangleup IBC$ implies that $\angle BCI=\angle BHA=90^{\circ}$ So, from claim 2 we conclude that $BI$ is the diameter of $(ABC)$ and $BF$ passes through the circumcenter of $\bigtriangleup ABC$ $\square$ Quote: This is my first ever post with LATEX . So feel free to spot any error .Thank you

Let $\alpha, \beta, \gamma$ be the angles $\angle A, \angle B, \angle C,$ respectively. Since $D$ is the circumcenter of $AHB,$ we must have that $DA = DF = DH = DB.$ In particular, $DF = DB.$ Now, we have that $$\angle ADF = \angle ADE + \angle EDF = 2 \angle ADE - \angle ADH = 2 \beta - (180 - 2 \alpha) = 180 - 2 \gamma.$$This means that $\angle BDF = 2 \gamma.$ Since $DF = DB,$ we conclude that $\angle ABF = \angle DBF = 90^\circ - \gamma.$ But $\angle CBH = 90^\circ - \gamma,$ so the lines $BH$ and $BF$ are isogonal. It is well-known that the $B$-altitude is isogonal to the $B$-diameter line, so $BF$ passes through the circumcenter of $\triangle ABC.$ Hence done.