is there any non-bashing solution i.e. synthetic one?

Consider the reflection $F'$ of $F$ over $BC$. It is sufficient to prove that $K = BF' \cap OH$ lies on $(BHC)$. Then $\angle HBK = \angle HBF' = \angle HBC - \angle F'BC= 90 - \angle C - \angle FBC = 90 - \angle C - (\angle FBO - \angle OBC) = 90 - \angle C + 90 - \angle A - \angle FBO = \angle B - \angle BHO + 90$. Then $\angle BHK = \angle BHO$, so $\angle HKB = 180 - \angle BHK - \angle HBK = 90 - \angle B$, so $\angle HKB = \angle HCB$ and $(BHKC)$ is cyclic as desired.

Note that $EH$ and $ED$ are symmetric with respect to $\overline{EBC}$. Let $B'$ be the antipode of $B$ on $(BOH)$. Let $G'$ be the reflection of $G$ over $BC$. In order to prove the collinearity of $B,B',G$, it sufficies to show $\measuredangle G'BC=\measuredangle CBB'$. Let $(BHO)\cap BC=Q$. \[\measuredangle G'BC=\measuredangle G'HC=\measuredangle OHB-\measuredangle CHB=\measuredangle OHB-180+\measuredangle A=\measuredangle QHB-90=\measuredangle QBB'\]As desired.$\blacksquare$

ezpotd wrote: Consider the reflection $F'$ of $F$ over $BC$. It is sufficient to prove that $K = BF' \cap OH$ lies on $(BHC)$. Then $\angle HBK = \angle HBF' = \angle HBC - \angle F'BC= 90 - \angle C - \angle FBC = 90 - \angle C - (\angle FBO - \angle OBC) = 90 - \angle C + 90 - \angle A - \angle FBO = \angle B - \angle BHO + 90$. Then $\angle BHK = \angle BHO$, so $\angle HKB = 180 - \angle BHK - \angle HBK = 90 - \angle B$, so $\angle HKB = \angle HCB$ and $(BHKC)$ is cyclic as desired. oops! i forgot that there exists a symmetry in this problem. thanks anyway!

iStud wrote: is there any non-bashing solution i.e. synthetic one? No Let $O'$ be the reflection of $O$ over $BC$; now since $OH$ and $O'D$ are also reflections, $E$ must lie on line $O'D$, so it suffices to prove that $O'D$ and $BF$ meet on $\omega$. Complex bash with $(ABC)$ the unit circle so that $o = 0$, $o' = b + c$, $h = a + b + c$, $d = -\frac{bc}{a}$. Now \begin{align*} f &= \begin{vmatrix}b & b \overline{b} & 1 \\ h & h \overline{h} & 1 \\ 0 & 0 & 1\end{vmatrix} \div \begin{vmatrix}b & \overline{b} & 1 \\ h & \overline{h} & 1 \\ 0 & 0 & 1\end{vmatrix} \\ &= \frac{bh \overline{h} - hb \overline{b}}{b \overline{h} - h \overline{b}} \\ &= \frac{b(a + b + c)(\frac 1a + \frac 1b + \frac 1c) - (a + b + c)}{b(\frac 1a + \frac 1b + \frac 1c) - \frac{a + b + c}{b}} \\ &= \frac{(a + b + c)(\frac ba + 1 + \frac bc - 1)}{\frac ba + 1 + \frac bc - \frac ab - 1 - \frac cb} \\ &= \frac{(a + b + c)(\frac ba + \frac bc)}{\frac ba + \frac bc - \frac ab - \frac cb} \\ &= \frac{(a + b + c)(b^2c + ab^2)}{b^2c + ab^2 - a^2c - ac^2} \\ &= \frac{b^2(a + b + c)(a + c)}{(a + c)(b^2 - ac)} \\ &= \frac{b^2(a + b + c)}{b^2 - ac}. \end{align*}Therefore \begin{align*} \frac{d - o'}{b - f} \div \frac{d - a}{b - a} &= \frac{-\frac{bc}{a} - b - c}{b - \frac{b^2(a + b + c)}{b^2 - ac}} \cdot \frac{b - a}{-\frac{bc}{a} - a} \\ &= \frac{\frac{bc}{a} + b + c}{\frac{b^2(a + b + c)}{b^2 - ac} - b} \cdot \frac{a - b}{\frac{bc}{a} + a} \\ &= \frac{b^2 - ac}{a} \cdot \frac{ab + bc + ca}{ab^2 + b^3 + b^2c - b^3 + abc} \cdot \frac{a}{1} \cdot \frac{a - b}{bc + a^2} \\ &= \frac{(b^2 - ac)(ab + bc + ca)(a - b)}{b(ab + bc + ca)(bc + a^2)} \\ &= \frac{(b^2 - ac)(a - b)}{b(bc + a^2)}. \end{align*}If we let $q$ denote this value, then \begin{align*} \overline{q} &= \frac{(\frac{1}{b^2} - \frac{1}{ac})(\frac 1a - \frac 1b)}{\frac 1b(\frac{1}{bc} + \frac{1}{a^2})} \\ &= \frac{\frac{ac - b^2}{ab^2 c} \cdot \frac{b - a}{ab}}{\frac 1b \cdot \frac{a^2 + bc}{a^2 b c}} \\ &= \frac{\frac{(ac - b^2)(b - a)}{a^2 b^3 c}}{\frac{a^2 + bc}{a^2 b^2 c}} \\ &= \frac{(b^2 - ac)(a - b)}{b(bc + a^2)} \\ &= q. \end{align*}Therefore $\frac{d - o'}{b - f} \div \frac{d - a}{b - a} \in \mathbb R$ and so $\measuredangle(O'D, BF) = \measuredangle DAB$, as desired.

Let $F'$ be the reflection of $F$ across $BC$. It is well known that $D$ is the reflection of $H$ across $BC$. So, we wish to prove $G = OH \cap BF'$ lies on $(BHC)$. Working in directed angles mod $180^\circ$, we have \begin{align*} \angle HBG &= \angle HBF' \\&= \angle HBC + \angle CBF' \\ &= 90^\circ - \angle BCA - \angle CBF \\ &= 90^\circ - \angle BCA - (\angle CBO + \angle OBF) \\ &= 90^\circ - \angle BCA - (90^\circ - \angle BAC + \angle OBF) \\ &= \angle BAC - \angle BCA - \angle OBF \\ &= -\angle CAB - \angle BCA - (90^\circ - \angle BHO) \\ &= \angle ABC + 90^\circ + \angle BHO \\ &= \angle ABC + 90^\circ - \angle OHB \\ &= \angle ABC + 90^\circ - \angle GHB, \\ -\angle HBG - \angle GHB &= 90^\circ - \angle ABC\\ \angle BGH &= \angle BCH. \end{align*}So, $BGCH$ is cyclic, and we're done.