This was a simple length bash. (I'm pretty certain that there exists a far superior synthetic solution)

By similar triangles $2H_1H=2HH_2$

Also $CO=H_2O$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(7.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.627393585880165, xmax = 8.787132749733011, ymin = -2.2776401813352414, ymax = 5.188313410370774; /* image dimensions */ pair A = (0.,3.), B = (6.,0.), C = (0.,0.), O = (1.1458980337503155,1.1458980337503155), H = (1.2,2.4), O_1 = (0.5124611797498108,2.170820393249937), O_2 = (1.6583592135001262,1.0249223594996215), H_0 = (1.6583592135001264,2.1708203932499366), H_2 = (2.1167184270002526,1.9416407864998737), H_1 = (0.741640786499874,2.6291796067500632); draw(A--B--C--cycle, linewidth(1.)); /* draw figures */ draw(A--B, linewidth(1.)); draw(B--C, linewidth(1.)); draw(C--A, linewidth(1.)); draw(C--H, linewidth(1.)); draw(O--H_0, linewidth(1.)); draw(O_2--H_2, linewidth(1.)); draw(O_1--H_1, linewidth(1.)); /* dots and labels */ dot(A,dotstyle); label("$A$", A,dir(90)); dot(B,dotstyle); label("$B$", B,dir(0)); dot(C,dotstyle); label("$C$", C,dir(-90)); dot(O,linewidth(4.pt) + dotstyle); label("$O$", O,dir(-90)); dot(H,linewidth(4.pt) + dotstyle); label("$H$", (1.2442105642980614,2.4834171164684475), NE * labelscalefactor); dot(O_1,linewidth(4.pt) + dotstyle); label("$O_1$", O_1,dir(-90)); dot(O_2,linewidth(4.pt) + dotstyle); label("$O_2$", O_2,dir(-90)); dot(H_0,linewidth(4.pt) + dotstyle); label("$H_0$", (1.7060221266716298,2.2635068486715104), NE * labelscalefactor); dot(H_2,linewidth(4.pt) + dotstyle); label("$H_2$", (2.156838175655351,2.0326010674847264), NE * labelscalefactor); dot(H_1,linewidth(4.pt) + dotstyle); label("$H_1$", (0.782399001924493,2.7143228976552316), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] As $O_1,O,O_2$ are incenters of the respective triangles and $H_1,H_0,H_2$ are feet of them on one of the side of triangle, $AB$. So we have, \[H_1H=\frac{AH+CH-AC}{2},H_0B=\frac{AB+BC-AC}{2},H_2B=\frac{BH+BC-CH}{2}.\]So, \[HH_0 = BH-\frac{AB+BC-AC}{2},H_0H_2=\frac{AB+BC-AC-BH-BC+CH}{2}=\frac{AB+CH-AC-BH}{2}.\] So we need, \[AH+CH-AC=2BH-AB-BC+AC=AB-AC-BH+CH.\] Replace $AH=AB-BH$, we need $AB-BH+CH-AC=AB-AC-BH+CH$ which is true. Now we need $AH+CH-AC=2BH-AB-BC+AC$. So, we need $AH+CH=2BH-AB-BC+2AC=2BH-AB=BH-AH$ or $2AH+CH=BH$. So we need $2\cdot \frac{AH}{CH}+1=\frac{BH}{CH}$. Now from similarity, $\frac{AH}{CH}=\frac{AC}{BC}=\frac{1}{2}$ and $\frac{BH}{CH}=\frac{BC}{AC}=2$. $2\cdot \frac{AH}{CH}+1=2=\frac{BH}{CH}$. Proved.