@above don't spam, if you think it's easy -- post a solution dude.

https://www.geogebra.org/geometry/myhj3ytc according to the naming in the link for $\triangle{C_0GB}$and $\triangle{B_0CH}$ we have $C_0G\parallel C_B0 $and $BC_0 \parallel B_0H$ so its enough to show that $BG \parallel CH$ which is obvious since $BG\perp AG$ and $CH \perp AG$ and we are done

parmenides51 wrote: Let $ABC$ be a scalene triangle, and $A_o$, $B_o,$ $C_o$ be the midpoints of $BC$, $CA$, $AB$ respectively. The bisector of angle $C$ meets $A_oCo$ and $B_oC_o$ at points $B_1$ and $A_1$ respectively. Prove that the lines $AB_1$, $BA_1$ and $A_oB_o$ concur. Its also vary Bary Bash favourable, just compute $$A_{1} = (a : b : a-b) $$and $$B_{1} = (a : b : b- a)$$, from this $$BA_{1} \cap AB_{1} = (a : -b : a-b) $$, now it's trivial to check that it lies on $A_{0}B_{0}$

Straight-forward harmonics. Note that $B_1,A_1$ does not have to lie on the angle bisector of $\angle ACB$. Let $D=\overline{BC}\cap\overline{AB_1}$ Note that by Menelaus', $\overline{DB_0},\overline{AA_0}$ and $\overline{CB_1}$ are concurrent. Similarly, $\overline{EA_0},\overline{BB_0}$ and $\overline{CA_1}$ are concurrent, where $E=\overline{AC}\cap\overline{A_1B}$. Thus, if $P=\overline{A_1B}\cap\overline{A_0B_0}$, we have $-1=(B,E;A_1,P)=(D,A;B_1,\overline{CP}\cap\overline{AB_1})$. This means that $\overline{CP}\cap\overline{AB_1}$ lies on $\overline{A_0B_0}$, but as $P$ lies on $\overline{A_0B_0}$, we get that $P$ lies also on $\overline{AB_1}$, we are done.