If we can show that $\triangle A_2HB_2\sim\triangle BHA$, then we know that $A_2B_2\parallel AB$. [asy][asy] import olympiad; import geometry; size(250); currentpen=fontsize(9); pair a,b,c,h,a1,b1,a2,b2; a=dir(105); b=dir(-90-50); c=dir(-90+50); h=a+b+c; a1=foot(a,b,c); b1=foot(b,a,c); a2=foot(a1,b,b1); b2=foot(b1,a,a1); draw(a--b--c--a--a1--a2^^b--b1--b2--a2); label("$A$",a,dir(a)); label("$B$",b,dir(b)); label("$C$",c,dir(c)); label("$H$",h,dir(-30)); label("$A_1$",a1,dir(a1)); label("$A_2$",a2,dir(180-40)); label("$B_1$",b1,dir(b1)); label("$B_2$",b2,dir(b2)); [/asy][/asy] In triangle $BA_1H$, $A_1A_2$ is an altitude, so $\triangle BA_1H\sim \triangle A_1A_2H$. This implies that $\frac{A_2H}{A_1H}=\frac{A_1H}{BH}$, or that $A_2H=\frac{A_1H^2}{BH}$. Similarly, $B_2H=\frac{B_1H^2}{AH}$. In order to show that $\triangle A_2HB_2\sim\triangle BHA$, we will show that $\frac{A_2H}{B_2H}=\frac{BH}{AH}$. By virtue of our previous calculations, we know that $$\frac{A_2H}{B_2H}=\frac{A_1H^2/BH}{B_1H^2/AH}=\left(\frac{A_1H}{B_1H}\right)^2\cdot\frac{AH}{BH}.$$On the other hand, since $AB_1A_1B$ is cyclic, by Power of a Point we know that $(AH)(A_1H)=(BH)(B_1H)\implies \frac{A_1H}{B_1H}=\frac{BH}{AH}$. Thus, $$\frac{A_2H}{B_2H}=\left(\frac{A_1H}{B_1H}\right)^2\cdot\frac{AH}{BH}=\left(\frac{BH}{AH}\right)^2\cdot\frac{AH}{BH}=\frac{BH}{AH},$$which implies that $\triangle A_2HB_2\sim\triangle BHA$ as desired.

Since $AB$ and $A_1B_1$ are antiparallel in $\angle AHB,$ it suffices to show that $A_2B_2$ and $A_1B_1$ are also antiparallel in the same angle. This just follows from the fact that $\angle A_1A_2B_1=\angle A_1B_2B_1=90^\circ,$ so $A_1A_2B_2B_1$ is cyclic. $\blacksquare$

Note that $A_1B_1A_2B_2$ is cyclic since $\angle A_1B_2B_1= 90^\circ=\angle A_1A_2B_1$. It is well known that so is $AA_1BB_1$. So by angle chasing we have, $$\angle B_2A_2H=\angle B_2A_2B_1=\angle B_2A_1B_1=\angle AA_1B_1=\angle ABB_1=\angle ABH$$

Dear Mathlinkers, why not with the Reim's theorem? Sincerely Jean-Louis

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%209.pdf p. 58-59. Sincerely Jean-Louis

solution with formal sum. $A_2B_2+A_1B_1=A_1H+B_1H=AH+BH=AB+A_1B_1$ So $AB=A_2B_2$. We're done.