We are going to show that these three points lie on the radical axis of $ \omega_1$ and $ \omega_2$ . First we show this for $ H $ . Let $ D' = AH \cap BC ~,~ E' = BH \cap CA $ . Then power of $ H $ to $ \omega_1 = - AH\cdot D'H = - BH\cdot E'H = $ power of $ H $ to $ \omega_2 $ so $ H $ is on the radical axis . Now we can actually forget about the original figure ( at least the triangle ) and owing to brevity we assume that these two circles are disjoint . To verify my claim for $ X $ , it suffices to show that $ P_1 , Q_1 , P_2 , Q _2 $ are concyclic , which can be done by angle chasing : $ \angle Q_1P_1D = \angle Q_1FD = \angle GEP_2 = \angle GQ_2P_2 $ ( $ F,D,E,G$ are concyclic . ) For point $ Y $ , we consider the inversion about $ H $ that sends $P_i $ to $R_i$ , $Q_i$ to $S_i$ . ( In this case $ H $ is inside the circles , we will consider inversion about $ H $ + reflection in $ H$ .) Since $ P_1 , Q_1 , P_2 , Q _2 $ lie on a circle , so do $ R_1 , S_1 , R_2 , S_2 $ . Therefore , $ Y$ also lies on the radical axis , the proof is complete .

Wow! I got scared after drawing the configuration, but it turned out to be much simpler than I thought. . Anyways Ig my solution is a bit different from the above one. math154 wrote: In $\triangle ABC$, $H$ is the orthocenter, and $AD,BE$ are arbitrary cevians. Let $\omega_1, \omega_2$ denote the circles with diameters $AD$ and $BE$, respectively. $HD,HE$ meet $\omega_1,\omega_2$ again at $F,G$. $DE$ meets $\omega_1,\omega_2$ again at $P_1,P_2$ respectively. $FG$ meets $\omega_1,\omega_2$ again $Q_1,Q_2$ respectively. $P_1H,Q_1H$ meet $\omega_1$ at $R_1,S_1$ respectively. $P_2H,Q_2H$ meet $\omega_2$ at $R_2,S_2$ respectively. Let $P_1Q_1\cap P_2Q_2 = X$, and $R_1S_1\cap R_2S_2=Y$. Prove that $X,Y,H$ are collinear. Ray Li. First we begin with a Well Known Lemma Lemma:- $ABC$ be a triangle and $AD,BE$ are arbiatry cevians and $\omega_1$ and $\omega_2$ are the circles with diameters $AD,BE$ respectively, then the Orthocenter $H$ of $\triangle ABC$ lies on the Radical Axis of $\{\omega_1,\omega_2\}$ Proof of Lemma:-Let the foot of altitudes from $A,B$ to $CB,CA$ respectively be $M,N$ respectively. So, $M\in\omega_1$ and $N\in\omega_2$ and $Pow_{(\omega_1)}H=HM\cdot HA=HB\cdot HN=Pow_{(\omega_2)}H\implies H\in\text{Radical Axis of } \omega_1\text{ and }\omega_2$. Let $\omega_1\cap\omega_2=\{U,V\}$ respectively. So from Lemma we get that $\overline{U-H-V}$. So, $HD\cdot HF=HU\cdot HV=HE\cdot HG\implies DEFG$ is a cylic quadrilateral. So, $\angle P_2Q_2Q_1=\angle DEG=\angle DFQ_1=\angle DP_1Q_1\implies P_2P_1Q_2Q_1$ is a cyclic quadrilateral $\implies XQ_1\cdot XP_1=XQ_2\cdot XP_2\implies X$ has equal Power WRT $\{\omega_1,\omega_2\}$. Similary we get that $S_1S_2R_1R_2$ is a cyclic quadrilateral $\implies YS_1\cdot YR_1=YS_2\cdot YR_2\implies Y$ has equal Power WRT $\{\omega_1,\omega_2\}$. So, $\{X,Y\}$ lies on the Radical Axis of $\{\omega_1,\omega_2\}$. But from our Lemma we get that $H$ lies on the Radical Axis of $\omega_1,\omega_2$ too. So, $\{X,Y,H\}$ lies on Radical Axis of $\{\omega_1,\omega_2\}$. So, $\overline{X-Y-H}$. $\blacksquare$