Lines AD, BE, and CF are concurrent Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Show that there exist points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$ respectively such that  OD + DH = OE + EH = OF + FH and the lines $AD$, $BE$, and $CF$ are concurrent.
I am trying to solve the problem using this method, but I am not seeing anything.

I can see nothing. I have already suggested the collinear points. Can you solve from a picture?

Hint:  Let $H_a$, $H_b$, and $H_c$ denote the reflections of $H$ about $BC$, $CA$, and $AB$, respectively.  Show that $H_a$, $H_b$, and $H_c$ are on the circumcircle of the triangle $ABC$.  Now, $OD+DH=OD+DH_a$ for example.  Pick $D$, $E$, and $F$ cleverly.  The last hint is: with these good choices of $D$, $E$, and $F$, we have
$$\frac{BD}{CD}=\frac{\sin(2B)}{\sin(2C)}\,,$$
and two more similar equalities.
