I'm assuming that you mean $A_1 \neq O$. Idea is that the concurrency point lies on the circle centered at $H$ with radius $R$,as well as $A_1,B_1,C_1$.It can be proven by angle chasing(none trivial) that any two of the lines $AA_1,BB_1,CC_1$ intersect on $(H,R)$.if $X=BB_1 \cap CC_1$ then $BXC$ is either $2A$ or $180-2A$,depending on the figure(this follows from angle chasing,I'm too lazy to write the details).Also it can be proven the same way that $HB_1=HC_1=R$ and $B_1HC_1$ is either $4A$ or $360-4A$(depending on $B_1XC_1$),which means that $H$ is the circumcenter of $B_1XC_1$,so any two of $AA_1,BB_1,CC_1$ intersect on $(H,R)$,meaning that they are concurrent.

Here's a shorter proof. Note that $AA_1$ is parallel to the reflection of the Euler line in $BC$, due to the isosceles trapezoid. Note that the reflections of the Euler line in the sidelines are concurrent at the Anti-Steiner point of the Euler line (Euler reflection point). By a reflection at the nine point centre of $ABC$ we're done.

other solutions