From Desargue's theorem, it follows that we need to show that the triangles formed by the tangents to the circumcircle in $A\",B\",C\"$ on the one hand, and by the perpendiculars from $A'$ to $AO$ and the like on the other hand are perspective. Let the perpendiculars through $B',C'$ to $BO,CO$ respectively intersect at $T$, and let the tangents to the circumcircle in $B\",C\"$ intersect at $S$. I'll show that $ST$ passes through $G$ (the centroid). Let $P$ be $B\"C\"\cap BC$. $SG$ is the polar of $P$ wrt $(O)$, so we need to show that $T$ lies on this polar. A quick angle chase shows that $T$ is, in fact, the intersection of the tangents through $B',C'$ to the nine-point circle. We thus need to show that if $T'$ is the intersection pt of the tangents in $B,C$ to $(O)$, then $T'$ lies on the polar of $P$ wrt $(O)$, but this is just a degenerate case of Pascal's theorem.

The solution of the problem is a strightforward consequence of the following lemma RL110904 and an opportune iterate application of Desargues theoreme. Lemma RL110904 Given a ciclic quadrialteral ABCD, let K be the intersection of AC and BD and L be the intersection between the tangent to c trhough C and D. If R and S are two point on BD and AC respectively s.t. RS//AB, then the intersection of the perpendiculars from R to the diameter through B and from S to the diameter through A stay on line KL. Proof. this lemma derive from the well-known fact that (which is in turn consequence of Pascal theorem) the diagonals of two dual quadrilaterals are concurrents.

Let me restate the problem so that it can fit my solution: "Let $ ABC$ be triangle inscribed in circumcircle $ (O)$. Denote $ A_1,B_1,C_1$ the intersections by the medians from $ A,B,C$ and $ (O)$, respectively. Let $ A_3,B_3,C_3$ be the midpoints of $ BC,CA,AB$, respectively. The perpendicular from $ A_3$ to $ AO$ intersects the tangent from $ A_1$ wrt $ (O)$ at $ X_a$. Define the same for $ X_b,X_c$. Prove that $ X_a,X_b,X_c$ are collinear." Proof: Lemma $ 1$: Let $ AB$ be the chord of $ (O)$. Denote $ C$ is an arbitry point on the greater arc $ AB$. Let $ M$ be the midpoint of $ AB$. Denote $ E$ the intersection by $ CM$ and $ (O)$, $ D$ by the intersection of $ CO$ and $ (O)$. The perpendicular from $ M$ to $ CD$ intersects the tangent from $ E$ wrt $ (O)$ at $ K$. Let $ KN$ be another tangent form $ K$ to $ (O)$ ($ N\in (O)$). Then $ N,M,D$ are collinear. Indeed, let $ S$ be the projection of $ M$ onto $ CD$. Denote $ X$ by the pole of $ MS$ wrt $ (O)$. Since $ K\in$ the polar of $ X$ wrt $ (O)$ hence the polar of $ K$ passes $ X$, which implies that $ X\in NE$, which also implies $ X\equiv CD\cap NE$. Let $ Y\equiv NE\cap MS\Longrightarrow (XYEN) = (XSDC) = (XSCD) = - 1\Longrightarrow YS,EC,DN$ are concurrent. But $ M$ already is the intersection of $ CE,YS$, thus $ M\in ND$. Lemma $ 2$: Let $ ABC$ be a triangle inscribed in circumcircle $ (O)$, $ H$ is its orthocenter. Denote $ A_3,B_3,C_3$ the midpoints of $ BC,CA,AB$, respectively. $ A_1,B_1,C_1$ be the intersections of $ AA_3,BB_3,CC_3$ with $ (O)$, respectively. Denote $ A_0,B_0,C_0$ be the second intersections of circumcircles of diameters $ AH,BH,CH$ with $ (O)$, respectively. Prove that: $ A_1A_0,B_1B_0,C_1C_0$ are concurrent. Indeed, let $ D,E,F$ be the projections of $ H$ onto $ BC,CA,AB$, respectively. Consider the inversion through pole $ H$, power $ \overline {HD}.\overline {HA} = \overline {HE}.\overline {HB} = \overline {HF}.\overline {HC} = k^2$. It is very well- known that $ \mathcal {I}(H,k^2): A_3\mapsto A_0$, $ B_3\mapsto B_0$, $ C_3\mapsto C_0$. Therefore, it is so obvious that $ \{B_3,C_3,B_0,C_0\}$ is the set of concyclic points. Also, $ (C_3C_1,C_3B_3)\equiv (C_3C,C_3B_3)\equiv (CC_1,CB)\equiv (B_1C_1,B_1B_3) \pmod \pi$, which implies $ \{C_1,C_3,B_3,B_1\}$ is a set of concyclic points. Now, we have $ B_1C_1,B_3C_3,B_0C_0$, respectively are the radical axes of $ \{(B_1C_1B_3C_3), (O)\}$, $ \{(B_1C_1B_3C_3),(B_3C_3B_0C_0)\}$, $ \{(B_0C_0B_3C_3),(O)\}$. Therefore $ B_1C_1,B_3C_3,B_0C_0$ are concurrent, let $ Y_1$ be the intersection of these lines. We have $ \overline {Y_1B_3}.\overline {Y_1C_3} = \overline {Y_1B_0}.\overline {Y_1C_0}\Longrightarrow \mathcal {P}_{(Y_1/(A_3B_3C_3))} = \mathcal {P}_{(Y_1/(O))}$. With the same define and argument for $ Y_2,Y_3$. We conclude that $ Y_1,Y_2,Y_3$ are collinear and further the line contains them is the radical axes of the Euler circle and $ (O)$. We have $ Y_1\equiv B_0C_0\cap B_1C_1$, $ Y_2\equiv A_0B_3\cap A_1B_1$, $ Y_3\equiv A_0C_0\cap A_1C_1$, $ Y_1,Y_2,Y_3$ are collinear therefore $ \triangle A_0B_0C_0$ and $ \triangle A_1B_1C_1$ are $ 2$ perspective triangles. Then by Desgaurte theorem, $ A_1A_0,B_1B_0,C_1C_0$ are concurrent. Back to our problem. Let $ X_aA_0,X_bB_0,X_cC_0$ let other tangents form $ X_a,X_b,X_c$, respectively to $ (O)$. Let $ A',B',C'$ be the intersections of $ AO,BO,CO$, respectively with $ (O)$. Then by lemma $ 1$, we conclude that $ A'A_0,B'B_0,C'C_0$ passes through $ A_3,B_3,C_3$, respectively, moreover, they also concurrent at $ H$- the orthocenter of $ \triangle ABC$. Thus, $ A_0,B_0,C_0$ now also play roles that they are respectively the second intersections of circumcircles of diameters $ AH,BH,CH$ with $ (O)$. Apply lemma $ 2$, we conclude that $ C_0C_1,A_0A_1,B_0B_1$ concurrent. But, $ C_0C_1,A_0A_1,B_0B_1$, respectively are the pole of $ X_c,X_a,X_b$ wrt $ (O)$. We conclude $ X_a,X_b,X_c$ are collinear. Our proof is completed then.

Yep, very nice problem. Take a look at a very nice solution by Nguyen Minh Ha here.