Let $\mathcal{H}$ be a rectanglular hyperbola passes from points $A,B,C,H,P$ with center $T$. Since parallelograms (by symmetry onto point $T$) we have that $Q, S \in \mathcal{H}$. Let $X = AB \cap SH$. By Pascal for $BAQSHP$ points $X,R,T$ are collinear and we are done! Remark. This solution works for all points $P$ on $(ABC)$.

My solution is the same as above, but is there any solution without using conics to the generalization?

Eeightqx wrote: My solution is the same as above, but is there any solution without using conics to the generalization? Of course there is a solution without conics! First observe from the parallelograms that $T$ is the midpoint of $HP$ and $CS$, so lines $AC$ and $QS$ are symmetric wrt to $T$ and $HSPC$ is a parallelogram. Now if $D$ is the reflection of $R$ wrt $T$ then $D\in AC$ (because $R\in QS$) and $PRHD$ is a parallelogram, so $HD\parallel PR\parallel PB$. If $E=SH\cap AB$ then $HE\parallel HS\parallel PC$ and since $R\in TD$, it suffices to prove that $E\in TD$. Define $F=HD\cap PC$ and $G=HE\cap PB$, then $HFPG$ is a parallelogram and therefore $T$ is the midpoint of $FG$. Using Menelaus in $FGH$, it suffices to prove that $\frac{HD}{DF}=\frac{HE}{GE}$. Now let $I\in (BGE)$ be such that $BE$ bisects $\angle GBI$ and let $J=IE\cap BH$. Then$$\measuredangle BIJ=\measuredangle BIE=\measuredangle BGH=\measuredangle FHG=\measuredangle HFC.$$We also have$$\measuredangle EBI=\measuredangle GBE=\measuredangle PBA=\measuredangle PCA=\measuredangle FCD$$and$$\measuredangle HBE=\measuredangle HBA=\measuredangle ACH=\measuredangle DCH.$$Hence$$\measuredangle JBI=\measuredangle HBE+\measuredangle EBI=\measuredangle FCD+\measuredangle DCH=\measuredangle FCH.$$Therefore we conclude $JBI\simeq HCF$. From $\measuredangle GBE=\measuredangle EBI$ and $I\in (BGE)$ we get $IE=EG$. From$$\measuredangle CHJ=\measuredangle BAC=\measuredangle BPC=\measuredangle GPF=\measuredangle FHG$$we get$$\measuredangle JHE=\measuredangle JHF+\measuredangle FHG=\measuredangle JHF+\measuredangle CHJ=\measuredangle CHF\underset{(1)}{=}\measuredangle IJB=\measuredangle EJH,$$where $(1)$ follows from $JBI\simeq HCF$. Hence $HE=EJ$, so$$\frac{HD}{DF}=\frac{JE}{EI}=\frac{HE}{GE}.$$Done! Remark: The only facts about $H$ used here were that $\measuredangle HBA=\measuredangle ACH$ and that $\measuredangle CHB=\measuredangle BAC$. Hence this proof also works if we replace $H$ by the point $A'$ such that $ABA'C$ is a parallelogram.

Very similar to #2. We will prove a stronger result with the case $P$ is any point on $(ABC)$. Claim: If $H,Q$ are the orthocenters of $ABC,PBC$ on cyclic quadrilateral $PABC,$ then $A,B,C,P,H,Q$ are on the same conic.. Proof: Work on the complex plane. By Pascal on $QBHPCA,$ we will show that $BQ\cap PC=F,BH\cap CA=E,PH\cap AQ=M$ are collinear. \[e=\frac{1}{2}(b+a+c+\frac{ac}{b}), \ f=\frac{1}{2}(b+p+c+\frac{pc}{b})\]Also $m=\frac{p+a+b+c}{2}$. \[\frac{e-m}{e-f}=\frac{2(e-m)}{2(e-f)}=\frac{b+a+c+\frac{ac}{b}-a-b-c-p}{b+a+c+\frac{ac}{b}-b-c--\frac{pc}{b}}=\frac{ac-pb}{(a-p)(b+c)}\]Which is real since $\overline{\frac{e-m}{e-f}}=\frac{\frac{1}{ac}-\frac{1}{pb}}{(\frac{1}{a}-\frac{1}{p})(\frac{1}{b}+\frac{1}{c})}=\frac{pb-ac}{(p-a)(b+c)}=\frac{ac-pb}{(a-p)(b+c)}$ which completes the proof.$\square$ Note that $Q$ is the orthocenter of $PBC$ since $q=p+h-a=p+b+c$ on complex plane. Claim: $S\in C(ABCPQH)$. Proof:By Pascal at $HQSPAC$, since $PA_{\infty},AC_{\infty},CH_{\infty}$ are on the line at infinity so $S\in C(HQPAC)\equiv C(ABCPQH)$.$\square$ Pascal at $QSHPBA$ implies $R,SH\cap BA,T$ are collinear as desired.$\blacksquare$

If we take the intersection of \( BHC \cap \) the perpendicular bisector of \( AB = Q \), then it is not a complicated account of angles that it is clear that \( APQH \) is a parallelogram. According to Desargues' theorem for triangles \( HTC \) and \( ASP \), it is clear that \( SP \parallel HC \). Furthermore, according to Desargues' theorem for triangles \( ASH \) and \( QPC \), we understand that \( SH \parallel PC \). Hence, the quadrilateral \( SHCP \) is also a parallelogram. Note that \( \angle PC = 90^\circ - \angle BANK = \angle CAR \) implies that \( \angle KQ = \angle PC = \angle PSA \). Also note that: \[ \angle PC = \angle PHASE + \angle AS = \angle AS + \angle PC = \angle PBQ \implies (PQBS). \] Additionally, we have: \[ \angle ASH = \angle QCP = \angle HBA \implies (AHBS). \] Now note that: \[ RP \cdot RB = RS \cdot RQ \quad \text{and} \quad SN \cdot NH = AN \cdot NB, \] hence points \( N, R \) lie on the radical axis of circles \( (SHQ) \) and \( (ABC) \). By the equality of triangles, these circles are symmetrical with respect to point \( T \), so it also lies on the radical axis.