I think the statement obscures what's really going on: if you replace the circumcircle with a circumconic, and $O,H$ with any two points $U,V$, you still have a valid problem . So let's restate it (it won't be the exact analog, but something equivalent): Problem: Let $BB_1B''B',CC_1C''C'$ be two quadrilaterals inscirbed in a conic $\mathcal T$ having the same intersection $V$ of the diagonals (i.e. $V=B_1B'\cap BB''=C_1C'\cap CC''$). Put $U=CC_1\cap BB_1,T=C'C''\cap B'B''$. Then $U,V,T$ are collinear. If we define $A_1,A',A''$ in a similar manner and prove the analogous statements for the pairs of configurations corresponding to $(A,B),(A,C)$, we will have shown that each two among the lines $A'A'',B'B'',C'C''$ intersect on $UV$, which, except maybe for some uninteresting limit cases, is different from all of them, meaning that all three of them actually concur on $UV$, as desired. Now for a proof of the problem stated above: Fix the quadrilateral $BB_1B''B'$ and move $C$ on $\mathcal T$. This gives rise to a self-map $C'\mapsto C''$ of $\mathcal T$ which is clearly projective. When $C$ reaches the position that $C_1$ initially had, $C'$ and $C''$ are interchanged, so this map is an involution, meaning that $C'C''$ passes through a fixed point. When $C=B_1$, the line $C'C''$ coincides with $B''B'$, so this fixed point is actually the intersection $T$ between $C'C''$ and $B'B''$. Finally, when $C=B''$, the collinearity of $U,V,T$ follows from a direct application of Pascal's Theorem applied to the hexagon $CC_1C'C''B_1B'$ (in this configuration $C=B'',C''=B$). There's probably a slicker and more direct proof of the problem based on a couple of applications of Pascal's Theorem as well, but I didn't want to go point-chasing .

what is point chasing?

shobber wrote: what is point chasing? I believe it is not mathematically defined. Anyway, grobber's meaning of point chasing is understandable here: Applying Pascal Theorem to one orientation of six of $B,B_1,B'',B',C,C_1,C'',C'$. I've tried it last night(GMT+8), and applying two times is enough. Anyway, I prefer using trigo rather than Pascal to solve it-they are just the same

Just wanted to note: The points H, O, H lie on the Euler line of triangle ABC. The triangle ABC is the circumcevian triangle of the point H with respect to triangle A''B''C''. The triangle $A_1B_1C_1$ is the circumcevian triangle of the point O with respect to triangle ABC. The triangle A'B'C' is the circumcevian triangle of the point H with respect to triangle $A_1B_1C_1$. Thus, the statement of the problem, namely that the lines A''A', B''B', C''C' concur at one point on the Euler line of triangle ABC, follows from the "circumcevian pingpong theorem". Darij

here is my solution: if we use Pascal theorem about inscribed hexagons we will have: in $A_1AA'C_1CC'$ $\rightarrow$ $[(A_1A , C_1C)=O]$& $[(AA',CC')=H]$ & $[(A_1C' , C_1A')=X]$ are on one line.(2) and in $C_1A'A''A_1C'C''$ $\rightarrow$ $[(C_1A' , C'A_1)=X]$ & $[(A'A'' , C'C'')=Q]$ & $[(C_1C'' , A_1A'')=H]$ are on one line.(1) and we have $[(C_1A' , C'A_1)=(A_1C' , C_1A')]$ AND $[(C_1C'' , A_1A'')=(AA' , CC')=H]$ (3) and from (1),(2) and (3) we get that $H,O,Q,X$ are on one line. and it means that $A'A'',C'C'',OH$ are concurrent. Similary we get that $B'B'',C'C'',OH$ are concurrent and we R done

PRETTY EASILY USING COMPLEX

Trivial cause using a composition of three involutions in $A'$ to $A$ to $A_1$ to $A''$ with focus $H,O,H$ resp, makes $A',A"$ an involution with focus at $OH$

Another way to do this is poles and polar: Let $X_a$ be the pole of $A'A"$.Define $X_b,X_c$ similarly.Since $(A',A";B,C)=-1 \implies X_a \in BC$.Note that $$X_aO^2 -X_aH^2=X_aA'^2+R^2-X_aH^2=R^2$$So by Carnot's theorem,$X_a,X_b,X_c$ are collinear and the line passing through them is perpendicular to $OH$,so their polar are concurrent on $OH$ and we are done.

It can also be proven very easily using inversion .

Perform an inversion centered at $H$ with radius $\sqrt{-HA\cdot HA''}$. So this inversion fixes the circumcircle. After this inversion, it is sufficient to show circumferences $(A''H)$, $(B''H)$, $(C'C''H)$ are coaxial for the concurrency of $A'A''$, $B'B''$, $C'C''$. Notice that \[ \angle A'A''H=\angle A'A_1O=\angle A_1A'O \]So $OA'$ is tangent to $(A'A''H)$, similarity for the other two. Hence $O$ has equal power wrt these three circles, so $OH$ is the common radical axis of these three circles. So these circles are coaxial. Done for the first part. Now suppose $A'A''$, $B'B''$, $C'C''$ concur at $X$. By power of point notice that $X$ lies on their common radical axis which is $OH$. Done.