Prove that $G,H$ are the poles of the intersections of the opposite sides of the cyclic quad. wrt $(O)$, and then we get a well known orthocenter configuration.

First, let $ E = AB \cap CD $ and $ F = BC \cap DA $ and denote the circumcircle of $ ABCD $ by $ \omega $. Let $ O_1, O_2 $ be the centers of $ \omega_1, \omega_2 $ respectively. Let $ M $ be the midpoint of $ GH $. Consider the inversion about $ \omega $. It is clear that line $ AB $ goes to $ \omega_1 $ and that line $ CD $ goes to $ \omega_2 $ so $ E $ goes to $ G $. Similarly $ F $ goes to $ H $. Moreover, $ \omega_1 $ and $ \omega_2 $ are the circles with diameters $ EG $ and $ FH $ respectively. Now since $ M $ is the midpoint of $ GH $ and since $ O_1 $ is the midpoint of $ GE $ we have that $ O_{1}M \parallel HE $ and similarly $ O_{2}M \parallel GF $. But since $ GF $ is the polar of $ E $ with respect to $ \omega $ we have that $ GF \perp OE $ so $ O_{2}M \perp OE $ which implies $ O_{2}M \perp OO_{1} $. Similarly $ O_{1}M \perp OO_{2} $ and so $ M $ is the orthocenter of triangle $ OO_{1}O_{2} $. This means that $ OM \perp O_{1}O_{2} $ so to show that $ M $ is on the radical axis of $ \omega_1 $ and $ \omega_2 $ it suffices to show that $ O $ is on this radical axis. However, this is clear since both $ \omega_1 $ and $ \omega_2 $ are orthogonal to $ \omega $, so we are done. Interestingly after doing the inversion step, a complex number solution is somewhat doable as well.

This problem is true for all point $P$ lies on line $OQ$ when $Q$ is intersection of $AC$ and $BD$.

Here is a proof of the generalization mentioned by buratinogigle: Let $X \equiv AD \cap CB$ and $Y \equiv AB \cap CD.$ $\omega_1$ and $\omega_2$ are the circles with diameters $\overline{GY}$ and $\overline{HX}.$ $AB,CD,PG$ are pairwise radical axes of $(O),$ $\odot(PAB)$ and $\odot(PCD),$ concurring at their radical center $Y$ $\Longrightarrow$ $YA \cdot YB=YP \cdot YQ$ $\Longrightarrow$ $G$ is on the inverse of $OQ$ under the inversion with center $Y$ and power equal to the power of $Y$ WRT $(O);$ the circle with diameter $\overline{XY}$ $\Longrightarrow$ $\angle XGY=90^{\circ}$ and similarly we have $\angle XHY=90^{\circ}.$ Hence if $R \equiv XG \cap YH,$ the radical axis of $\omega_1,\omega_2$ is just the line passing through $P$ and the orthocenter $Z$ of of $\triangle RGH,$ which bisects $\overline{GH}$ since $PHZG$ is clearly parallelogram.

Luis González wrote: Here is a proof of the generalization mentioned by buratinogigle: Let $X \equiv AD \cap CB$ and $Y \equiv AB \cap CD.$ $\omega_1$ and $\omega_2$ are the circles with diameters $\overline{GY}$ and $\overline{HX}.$ $AB,CD,PG$ are pairwise radical axes of $(O),$ $\odot(PAB)$ and $\odot(PCD),$ concurring at their radical center $Y$ $\Longrightarrow$ $YA \cdot YB=YP \cdot YQ$ $\Longrightarrow$ $G$ is on the inverse of $OQ$ under the inversion with center $Y$ and power equal to the power of $Y$ WRT $(O);$ the circle with diameter $\overline{XY}$ $\Longrightarrow$ $\angle XGY=90^{\circ}$ and similarly we have $\angle XHY=90^{\circ}.$ Hence if $R \equiv XG \cap YH,$ the radical axis of $\omega_1,\omega_2$ is just the line passing through $P$ and the orthocenter $Z$ of of $\triangle RGH,$ which bisects $\overline{GH}$ since $PHZG$ is clearly parallelogram. Can someone please explain in detail how we obtain that $\angle XGY = 90^\circ$. Thanks in Advance

Let $R_G=\overline{AB}\cap\overline{CD}$ and $R_H=\overline{BC}\cap\overline{AD}$. Claim. $O$ lies on the radical axis of $\omega_1,\omega_2$. Proof. By radical axis theorem, it is sufficient to show that $GR_GHR_H$ is cyclic. For this, one could use Miquel point properties to show that $\measuredangle R_HGO=\measuredangle R_GHO=90^\circ$, which yields desired. $\square$ Lemma. Consider triangle $ABC$ and let $D,E$ be points on $\overline{AB},\overline{AC}$, respectively, such that $\measuredangle BDC=\measuredangle BEC=90^\circ$. Let $X,Y,Z,W$ be the midpoints of $\overline{BC},\overline{CE},\overline{ED},\overline{DB}$, respectively. Then, $\overline{AZ}\perp\overline{YW}$. Proof. Observe that $X$ is the antipode of $A$ wrt $(AYW)$ as $\measuredangle AYX=\measuredangle AWX=90^\circ$. Since $\triangle ADE\sim\triangle ACB$, $\overline{AX},\overline{AZ}$ are isogonal wrt $\angle BAC$. Thus, $\overline{AZ}\perp\overline{YW}$, as desired. $\square$ By the claim, we get that perpendicular from $O$ to line connecting the centers of $\omega_1,\omega_2$ is the radical axis of $\omega_1,\omega_2$, which by the lemma also passes through the midpoint of $\overline{GH}$, i.e. the midpoint of $\overline{GH}$ lies on the radical axis of $\omega_1$ and $\omega_2$. $\blacksquare$

Let $E=AB\cap CD, F=AD\cap BC$. From the Brokard's theroem we have $G,H$ are the feet from $F$ to $OE$ and $E$ to $OF$ respectively. Now forget all the things and consider only $\triangle OEF$. We define a function $f:R^2\to R$ such that \[ f(\bullet)=P(\bullet, \omega_1)-P(\bullet, \omega_2) \]We know $f$ is linear. Let $M$ be the midpoint of $GH$. We want $f(M)=0$. By linearity since $f(M)=\frac{1}{2}(f(G)+f(H))$, it is sufficient to prove $f(G)+f(H)=0$. We know \begin{align*} \vec{G}=\frac{GE}{OE}\vec{O}+\frac{OG}{OE}\vec{E},\qquad \vec{H}=\frac{HF}{OF}\vec{O}+\frac{OH}{OF}\vec{F} \end{align*}Therefore by linearity we have \begin{align*} f(G)&=\frac{GE}{OE}f(O)+\frac{OG}{OE}f(E)\\ &=\frac{GE}{OE}\cdot 0+\frac{OG}{OE}(0-EH^2)\\ &=-\frac{OG}{OE}\cdot EH^2 \end{align*}Similarly \begin{align*} f(H)&=\frac{HF}{OF}f(O)+\frac{OH}{OF}f(F)\\ &=\frac{HF}{OF}\cdot 0+\frac{OH}{OF}(GF^2-0)\\ &=\frac{OH}{OF}\cdot GF^2 \end{align*} To conclude we have to show \[ \frac{GF^2}{EH^2}=\frac{OG}{OE}\cdot\frac{OF}{OH} \]which is true (both side are equal to $OF^2/OE^2$) and we are done.

Let $X=AB\cap CD$ and $Y=AD\cap BC.$ By $PoP$ on $X$ and $Y$ one obtains that $O-H-Y$ and $O-G-X,$ and clearly we also have that $X\in\omega_1$ and $Y\in\omega_2.$ I now claim that $\angle XHY=\angle XGY=90^\circ.$ This is in fact well-known by Brocard, as the inversion around $(ABCD)$ sends $X\to G$ and $Y\to H,$ implying that $H$ is in the polar of $X$ and $G$ in the polar of $Y,$ which shows the perpendicularity. Now, by looking at $\triangle OXY,$ the problem tells us to show the following: Claim. Let $ABC$ be a triangle and $B_1\in AC, C_1\in AB$ sastisfy that $BB_1\perp AC, CC_1\perp AB.$ Then if we let $\omega_1$ be the circumference of diameter $BC_1$ and $\omega_2$ be the circumference of diameter $CB_1,$ the radical axis of $\omega_1$ and $\omega_2$ is the $A-symeddian.$ Proof. Let $P$ be the intersection of the tangents through $B$ and $C$ to $(ABC),$ and let $PB\cap \omega_1=B_2$ and $PC\cap \omega_2=C_2.$ Now $Pow_{\omega_1}P=PB\cdot PB_2=PB^2+PB\cdot BB_2$ and $Pow_{\omega_2}P=PC\cdot PC_2=PC^2+PC\cdot CC_2,$ and as $PB=PC,$ if we manage to show that $BB_2=CC_2$ we will be done. For this notice that $\angle B_2B_1B=90^\circ-\gamma$ and $\angle C_2C_1C=90^\circ-\beta,$ so by the Sine Law we get that the problem is equivalent to showing: $$CC_1\cdot \sin{\angle (90^\circ-\beta)}=BB_1\cdot \sin{\angle(90^\circ-\gamma)},$$which is true as $\sin{\angle(90^\circ-\gamma)}=\frac{BB_1}{BC}$ and $\sin{\angle(90^\circ-\beta)}=\frac{CC_1}{BC}.$