Let $ABC$ be a triangle and $D, E$ and $F$ the foots of heights from $A, B$ and $C$ respectively. Let $D_1$ be such a point on $EF$, that $DF = D_1 E$ where $E$ is between $D_1$ and $F$. Similarly, let $D_2$ be such a point on $EF$, that $DE = D_2 F$ where $F$ is between $E$ and $D_2$. Let the bisector of $DD_1$ intersect $AB$ at $P$ and let the bisector of $DD_2$ intersect $AC$ at $Q$.
Prove that, $PQ$ bisects $BC$.
Notice that for any point N on EF, the circumcircle of EDN intersects the line AC a second time at exactly the point at which the perpendicular bisector of ND intersects AC.
Let $K$ be $(NDE)$ intersect $AC$. Then, <KDN=<KEN=<ABC, while <KND=<DEC=<ABC, so NKD isoceles. This proves that K is on the perpendicular bisector of ND.
Thus, since $D_1=N$ and $K=P$ in the above scenario, we find that $D_1F=ED$, $D_1P=DP$, $<FD_1P=<PDE$ This is sufficient to prove that $D_1PF$ is congruent to $DEP$, or $PE=PF$. So P is on the perpendicular bisector of EF. Similar reasoning for Q. Now it suffices to prove that the perpendicular bisector of EF passes through M, midpoint of BC.
This can be shown since M is the center of circle $BFEC$ so $MF=ME$.
This one is trivial once you notice that PQ is actually the bisector of EF, which you get from drawing a good diagram or using geogebra like me.