Let $ \| \begin{array}{c} C_1 \in CC' \cap AB \\ B_1 \in BB' \cap AC \\ E \in AB \cap \omega_{ACC'} \\ F \in AC \cap \omega_{ABB'} \end{array}$. Then quadrilaterals $ AC'EC$ and $ ABFB'$ are cyclic, so: $ \angle{C'AE} = \angle{EAF} = \angle{FAB'}$ and $ \| \begin{array}{c} \angle{C'EC} \\ \angle{B'FB} \end{array} = 180^\circ - 2\angle{A} \Rightarrow \| \begin{array}{cc} \angle{AEC} \\ \angle{AFB} \end{array} = 90^\circ - \angle{A}\Rightarrow \| \begin{array}{cc} \angle{ACE} \\ \angle{ABE} \end{array} = 90^\circ$. Because $ AE$ and $ CF$ are diameters, we have: $ \|\begin{array}{cc} \angle{APE} \\ \angle{APF} \end{array} = 90^\circ \Rightarrow \angle{EPF} = 180^\circ \Rightarrow E,P,F$ are collinear. In triangle $ AEF$, $ EC$, $ FB$ and $ AP$ are altitudes, so they are concurrent. Let $ H$ be ortocenter. Quadrilateral $ ABHC$ is cyclic, so $ H$ lies on the circumcircle of $ ABC$, where $ PA$ is diameter, which conctains $ O$, q.e.d.

$ AB$ is the perpendicular bisector of $ CC^'$ $ \Rightarrow$ the circumcenter of $ \triangle{ACC^'}$ $ \in$ $ AB$. Let it be $ O_1$. Define $ O_2$ analogoulsy for $ \triangle{ABB^'}$. $ \frac{AC^'}{\sin{(90^o - A)}} = 2R_1$ $ \Rightarrow$ $ R_1 = \frac {AC}{2\cos{A}}$ analogously: $ R_2 = \frac {AB}{2\cos{A}}$. $ \Rightarrow$ $ R_1AB = R_2AC$ $ \Rightarrow$ $ BO_1O_2C$ is cyclic .We know $ O_1O_2 \perp AP$, using $ BO_1O_2C$ is cyclic we get $ AP$ passes trough $ O$.

An equivalent enunciation I. Let $ w_1 = C(O_1)$ , $ w_2 = C(O_2)$ be two secant circles in $ A$ , $ D$ so that $ m(\widehat {O_1AO_2}) < 90^{\circ}$ . Denote $ \begin{array}{c} \{A,B\} = AO_1\cap w_2 \\ \ \{A,C\} = AO_2\cap w_1\end{array}$ . Prove that the circumcircle of $ \triangle ABC$ belongs to $ AD$ . Proof. Denote the diameters $ [AX]$ , $ [AY]$ of the circles $ w_1$ , $ w_2$ respectively. Observe that $ D\in XY$ , $ XY\perp AD$ and $ \|\begin{array}{c} XC\perp AY \\ \ YB\perp AX\end{array}$ . Thus, $ BDC$ is the orthic triangle of $ \triangle XAY$ and $ \triangle ABC\sim\triangle AYX$ . Denote the orthocenter $ H$ of the triangle $ ABC$ . Therefore, the rays $ [AD$ , $ [AH$ are isogonals in the angle $ \widehat {BAC}$ . In conclusion, the circumcircle of the triangle $ ABC$ belongs to $ AD$ . Remark. Sorry, in my opinion, the proposed problem is "painted" a bit ! An equivalent enunciation II (evident !). Let $ XAY$ be an acute triangle. Denote the orthic triangle $ DCB$ of it, where $ D\in XY$ , $ C\in YA$ , $ B\in AX$ . Prove that the circumcenter of the triangle $ ABC$ belongs to the altitude $ AD$ .

Equivalent to existence of the orthocenter upon $\sqrt{bc}$ inversion.