There is a $\Delta ABC$ with $AB < BC$ , with orthocenter $H$ and circumcenter $O$. Point $D$ is on line $BO$ such that $O$ is between $B$ and $D$ and $\angle ADC = \angle ABC$. The semi-line starting at $H$ and parallel to $BO$ meets the circumscribed circle at points $E$ and $F$ respectively. Prove that $BH = DE$.
What I Tried: Here's a picture in Geogebra to get an idea of the problem :-
Really Tough. Spent around $1.5$ hours on this problem but no progress, as I am a little weak at Geometry I tried elementary methods like angle-chasing , similarity and so on, but no success. Another problem is that $DABC$ is not a parallelogram as it may seem, so angle-chasing and similarity do not work to any triangles at all till now, so that I could have found something useful.
We basically need to prove that $DEHB$ is an Isosceles Trapezium, and I have no idea. I did not even any reasonable cyclic quadrilaterals to work on and gain some information. I am quite stuck on this problem.
Can anyone help me?
Let $O', H'$ be the circumcenter and the orthocenter of $\triangle ADC$ respectively. Denote, furthermore, by $G$ the orthogonal projection of $O$ onto $CA$.
First of all, observe that the Law of Sines implies that the circumcircle of $\triangle ADC$ has the same radius as the circumcircle of $\triangle ABC$, since $R'=\frac{CA}{\sin\angle CDA}=\frac{CA}{\sin\angle ABC}=R$. It follows that $G$ is the midpoint of $OO'$.
Besides, it is well-known (at least considering the ratios of the Euler line), that $$2OG=BH\implies OO'=BH$$ and since both $OO'$ and $BH$ are perpendicular to $CA$, we have that $OO'HB$ is a parallelogram. Yet this implies that $OB\parallel O'H$, and, hence, $O',E,H$ are collinear. Notice now that, since $OE'\parallel OD$ and $DO'=R'=R=EO$, the quadrilateral $DO'EO$ is an isosceles trapezium; we infer that the diagonals have the same length, i.e. $DE=OO'$. Thus $$DE=OO'=BH$$
Some additional observations. As I noticed while playing with Geogebra, this problem could lead to other nice relationships such as