We proceed using Cartesian coordinates. Let $\mathcal H$ be the rectangular circumhyperbola of $ABCD$, which is nondegenerate as lines $AP,BQ,CR$ are distinct. Let $\mathcal H$ be given by $xy=1$ and $A=(a,a^{-1}),B=(b,b^{-1}),C=(c,c^{-1}),D=(d,d^{-1})$ with $t=abcd$. Points $(e,e^{-1}),(f,f^{-1}),(g,g^{-1}),(h,h^{-1})$ on $\mathcal H$ form an orthocentric system iff $efgh=-1$ and are concyclic iff $efgh=1$. This fact is easy to check, so we leave the details to the reader. It remains to show that $t=1$. We have that $P,Q,R\in\mathcal H$, so if $P=(p,p^{-1}),Q=(q,q^{-1}),R=(r,r^{-1})$ then $p=-at^{-1},q=-bt^{-1},r=-ct^{-1}$. The equations of lines $AP,BQ,CR$ are then given by $x+apy=a+p,x+bqy=b+q,x+cry=c+r$, so since they are concurrent we have that \[\begin{vmatrix}1&-a^2t^{-1}&a(1-t^{-1})\\1&-b^2t^{-1}&b(1-t^{-1})\\1&-c^2t^{-1}&c(1-t^{-1})\end{vmatrix}=\begin{vmatrix}1&ap&a+p\\1&bq&b+q\\1&cr&c+r\end{vmatrix}=0.\]Then, $a,b,c$ are distinct roots of $ut^{-1}x^2+v(1-t^{-1})x+w$ for some $u,v,w$ which are not all $0$, which means that $ut^{-1}=v(1-t^{-1})=w=0\implies t=1$, as desired.

I'm sort of conflicted. I'm happy that I was able to use this sort of stuff on an actual test, but also, what?

My problem. I was very surprised to see it chosen as a problem 6. Here's my solution: Let $T$ be the concurrency point. We have $AQ\parallel BP$ because both lines are perpendicular to $CD$. Since these lines are distinct there is a homothety $\Psi$ centered at $T$ with $\Psi(A)=P, \Psi(Q)=B$. By repeating this argument, we also have $\Psi(C)=R$, and then $\Psi(P)=A$. Hence $\Psi$ must be reflection about $T$ because $A\neq P$. Next, we have $PD\perp BC$ and $BC\parallel QR$, so $D$ is the orthocenter of $PQR$ by symmetry. Hence if $H$ is the orthocenter of $ABC$, then $T$ is the midpoint of $DH$. Now our setup is symmetric: we have four points $A, B, C, D$, forming four orthocenters $H, P, Q, R$ which are reflections of the previous four points about $T$. Let $S$ be the center of mass of the four points $A, B, C, D$, and $O$ be the reflection of $T$ through $S$. We claim that $A, B, C, D$ are equidistant from $O$. Let $A', O', S', T', D'$ be the projections of the corresponding point onto line $BC$. Then $T'$ is the midpoint of $A'D'$, so $O'$ is the midpoint of $BC$ (because $S'=\frac{A'+D'+B+C}{4}$). Hence $OB=OC$. By the aforementioned symmetry, we easily show that $O$ is equidistant from $A, B, C, D$.

We have that if $T_A$ is the foot of the altitude from $D$ to $BC$, we have $$\frac{\sin(BAP)}{\sin(CAP)}=\frac{BP}{CP}\cdot \frac{\sin(ABP)}{\sin(ACP)}=\frac{BT_A}{CT_A}\cdot\frac{CD}{BD}\cdot \frac{\sin(ABP)}{\sin(ACP)}=\frac{BT_A}{CT_A}\cdot\frac{CD}{BD}\cdot \frac{\cos(\angle(AB,CD))}{\cos(\angle(AC,BD))},$$so multiplying the similar terms, we get that $$\frac{BT_A}{CT_A}\cdot \frac{CT_B}{AT_B}\cdot \frac{AT_C}{BT_C}=1,$$where $T_B,T_C$ are defined similarly to $T_A$. Now, we get that $T_A,T_B,T_C$ are collinear by menalauas, and we are done by simson lines.

I almost was going to use partagura's theorem, but then i realized it didnt work. Thus, I used lines that never met and fractions. Then I used fake numbers. This was a tough problem. I also tried using Seward’s Theory of Triangular Motion, as well as Potomac Quadrangle Conjecture. The Circular Quadra-angle was pretty hard to find ngl.

Fun fact: Seward's Theory of Triangular Motion is named after William H. Seward, the Secretary of State during the Civil War who narrowly escaped assassination!

Another similar triangles/trigonometry solution(not my solution during the test): Suppose the three lines meet at a point \(O\). Then since \(AQ || BP\) as both are perpendicular to \(CD\), \(\triangle AOQ \sim \triangle POB\). Thus, \(\frac{AO}{OP}=\frac{QO}{OB}\). We can set up a similar fomula on the other pairs \((B,C)\) and \((C,A)\) to get \[\frac{AO}{OP}=\frac{QO}{OB}=\frac{CO}{OQ}=\frac{OP}{AO},\] so \(AO=OP\). Note that \(\frac{AO}{OP}=\frac{AQ}{BP}\), so \(AQ=BP\). \(\dfrac{AQ}{2|\cos \angle CAD|}\) and \(\dfrac{CD}{2\sin \angle CAD}\) are both the diameter of \((ACD)\), so \(CD=2 AQ |\tan \angle CAD| =2BP |\tan \angle CBD|\). Thus \(\angle CAD\) and \(\angle CBD\) are either supplementary or equal. A quick case check on whether \(A\) and \(B\) are on the same side of \(CD\) finishes.

This is probably similar to others; but I decided to post it anyway. Really nice problem ! Let $S$ be the desired concurrency point. Note that $BR \perp AD$ and $CQ \perp AD$, so $BR\parallel CQ$. This means $\triangle BSR \sim \triangle SQC$. So we have : $$ \frac {BS}{QS} = \frac {RS}{CS} \stackrel {\triangle RSA \sim \triangle CSP}{=} \frac {AS}{PS} \stackrel {\triangle BSP \sim \triangle QSA}{=} \frac {QS}{BS} \implies QS=SB $$ Similarly, $AS=SP$ and $CS=SR$. Note that we now have : $$ BR \stackrel {\parallel}{=} CQ \quad BP \stackrel {\parallel}{=} AQ \quad CP \stackrel {\parallel}{=} AR$$$$ BC \stackrel {\parallel}{=} RQ \quad AB \stackrel {\parallel}{=} PQ \quad CA \stackrel {\parallel}{=} PR$$ So $PD\perp QC$ , and so on, so $D$ is the orthocenter of $PQR$. (Note that this provides a shorter finish to the sol in #2) Hence since $PQR \cong ABC$, we must have $H,S,D$ collinear with $HS=SD$. Finally note that the problem is now symmetric in $A,B,C,D$, so we can WLOG assume that $A,D$ lie on opposite sides of $BC$. Let $R_A$ and $R_D$ denote the circumradii of $\triangle ABC$ and $\triangle DBC$ respectively. Let $\measuredangle XYZ$ denote the acute angle $XYZ$. We have : $$ BC = 2R_A \times \sin \measuredangle BAC = AH \times \tan \measuredangle BAC$$ Similarly $BC=PD \times \tan \measuredangle BDC$ Since $BC=PD$, we must have $\tan \measuredangle BDC = \tan \measuredangle BAC$. If $\angle BAC + \angle BDC =180$, we are done. The other possibility is $\angle BAC = \angle BDC$. The second claim implies $P\in \odot (ABC)$, similarly for $Q,R$. But then $D$ lies inside $ABC$ so the angle condition can't hold. $\square$

Solved this in under two minutes—disappointing . Let $D_1$ be the isogonal conjugate of $D$ w.r.t. $\triangle ABC$. Assume for the contradiction that $D_1$ is not an infinity point. Let $X=QR\cap BC$, $Y=PR\cap AC$, and $Z=PQ\cap AB$; they are colinear due to Desargues. However, from Property 1 here, we get that $\triangle XYZ$ is the pedal triangle of $D_1$ w.r.t. $\triangle ABC$. But this means by Simson lines that $D_1\in\odot(ABC)$ or $D$ is an infinity point, contradiction.

We have $AP,BQ,CR$ concurrent and $AR\parallel CP$, $AQ\parallel BP$ and $BR\parallel CQ$ so they concur at the midpoint of $AP,BQ,CR$(same idea as Evan's proof from Twitch stream till now). Now, we also get $D$ is the orthocenter of $PQR$ and thus, $PQRD$ is $ABCH$ reflected across some point. But, as they all lie on a rectangular hyperbola, and $PQAB$ is now a parallelogram, the intersection point is the center of the hyperbola and thus, is the Poncelet point of $D$ and thus lies on the NPC. Thus, $H$ reflected around it, must lie on $(ABC)$ so $D$ is on $(ABC)$.

USA TSTST 2020 P6 wrote: Let $A$, $B$, $C$, $D$ be four points such that no three are collinear and $D$ is not the orthocenter of $ABC$. Let $P$, $Q$, $R$ be the orthocenters of $\triangle BCD$, $\triangle CAD$, $\triangle ABD$, respectively. Suppose that the lines $AP$, $BQ$, $CR$ are pairwise distinct and are concurrent. Show that the four points $A$, $B$, $C$, $D$ lie on a circle. W H A T Let $O',O'',O'''$ be the circumcenters of triangles $ABD,ACD,BCD$ respectively. If two of them coincide, we are done. Suppose for the sake of contradiction that $O' \neq O'' \neq O''' \neq O'$. Also let $O$ be the concurrency point of $AP,BQ,CR$. Since $CQ \parallel BR$ (both perpendicular to $AD$) and similar relations, we have $$\frac{AR}{CP}=\frac{AO}{PO}=\frac{AQ}{BP}=\frac{QO}{BO}=\frac{CQ}{BR}$$hence $$\frac{CQ}{BR}=\frac{AQ}{BP}=\frac{AR}{CP}$$ Let $X,Y,Z$ be the midpoints of $AD,BD,CD$. Using the $AH=2OM$ Lemma, we obtain $\frac{CQ}{BR}=\frac{O''X}{O'X}$ and similar relations. Cyclically we have $$\frac{O''X}{O'X}=\frac{O''Z}{O'''Z}=\frac{O'Y}{O'''Y}$$The first equality implies $XZ \parallel O'O'''$ and since $XZ \parallel AC$ and $O'O''' \perp BD$ we obtain $AC \perp BD$. Similarly the second implies $O'O'' \parallel YZ$ hence $AD \perp BC $. Therefore $D$ is the orthocenter of $ABC$, contradiction to the problem hypothesis. Hence, done.

spartacle wrote: Let $A$, $B$, $C$, $D$ be four points such that no three are collinear and $D$ is not the orthocenter of $ABC$. Let $P$, $Q$, $R$ be the orthocenters of $\triangle BCD$, $\triangle CAD$, $\triangle ABD$, respectively. Suppose that the lines $AP$, $BQ$, $CR$ are pairwise distinct and are concurrent. Show that the four points $A$, $B$, $C$, $D$ lie on a circle. Let $X$ be the point of concurrence, and let $H$ be the orthocenter of $\triangle ABC$. By Desargues on $\triangle AQR$ and $\triangle PBC$, we get $QR \parallel BC$. But we also know $BR \parallel CQ$ $\implies$ $\square BCQR$ is a parellelogram. Therefore $X$ is the common midpoint of $AP, BQ, CR$ $\implies$ $\triangle PQR$ is the reflection of $\triangle ABC$ in $X$. We have $PD \perp BC$ and so on $\implies$ $D$ is the orthocenter of $\triangle PQR$ $\implies$ $D$ is reflection of $H$ in $X$. Let $\mathcal{H}$ be the unique rectangular hyperbola passing through $A,B,C,D$. Then $P,Q,R,H$ also lie on $\mathcal{H}$ $\implies$ $X$ is the center of $\mathcal{H}$ $\implies$ $X$ lies on nine-point circle of $\triangle ABC$ $\implies$ $D$ lies on $\odot (ABC)$. $\blacksquare$

Is no one going to comment on how the title might potentially spoil the problem? To me, the title might do a wonderful job in skewing the thought process for other users trying this problem. Also, the censorship on this community is ridiculous; someone also posted on the TSTST P4 thread about a potentially spoiler title, but the post was deleted without any sort of address to it.

mofumofu wrote: Is no one going to comment on how the title might potentially spoil the problem? To me, the title might do a wonderful job in skewing the thought process for other users trying this problem. Also, the censorship on this community is ridiculous; someone also posted on the TSTST P4 thread about a potentially spoiler title, but the post was deleted without any sort of address to it. This is a good point. I thought about this, but I thought that nearly everyone either would not know how to use that method or find it easy regardless of whether or not it was in the title. I suppose this was probably bad judgement on my part. I've changed the title to "Converse of a classic orthocenter problem." (Re TSTST P4 thread: I think that was a simple mistake, that thread has had some issues with posts being incorrectly deleted.)

Here is the current draft of the official solution; the solution here is the author's submitted approach. Let $T$ be the concurrency point, and let $H$ be the orthocenter of $\triangle ABC$. [asy][asy] pair A = dir(130); pair B = dir(210); pair C = dir(330); pair D = dir(97); pair P = B+C+D; pair Q = C+A+D; pair R = A+B+D; filldraw(R--B--D--cycle, invisible, palecyan); filldraw(Q--A--C--cycle, invisible, palered); filldraw(C--B--D--cycle, invisible, palegreen); draw(R--foot(R,B,D), lightgrey+dashed); draw(B--foot(B,D,R), lightgrey+dashed); draw(D--foot(D,R,B), lightgrey+dashed); draw(A--foot(A,C,Q), lightgrey+dashed); draw(C--foot(C,Q,A), lightgrey+dashed); draw(Q--foot(Q,A,C), lightgrey+dashed); draw(C--foot(C,B,D), lightgrey+dashed); draw(B--foot(B,D,C), lightgrey+dashed); draw(D--foot(D,C,B), lightgrey+dashed); draw(A--foot(A,B,C), lightgrey+dashed); draw(B--foot(B,C,A), lightgrey+dashed); draw(C--foot(C,A,B), lightgrey+dashed); pair T = midpoint(A--P); pair S = (A+B+C+D)/4; pair O = 2*S-T; /* A' = foot A B C R270 T' = foot T B C R270 D' = foot D B C R300 O' = foot O B C R300 S' = foot S B C R270 T--Tp lightgrey dashed S--Sp lightgrey dashed O--Op lightgrey dashed */ draw(T--O, lightblue); filldraw(A--B--C--cycle, invisible, lightgrey); draw(A--C, palered); draw(A--Q, red+1); draw(B--P, red+1); draw(A--P, blue); draw(B--Q, blue); draw(C--R, blue); pair H = A+B+C; draw(D--H, lightblue); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(80)); dot("$P$", P, dir(285)); dot("$Q$", Q, dir(Q)); dot("$R$", R, dir(R)); dot("$T$", T, 1.8*dir(90)); dot("$S$", S, 1.8*dir(75)); dot("$O$", O, dir(315)); dot("$H$", H, dir(45)); /* TSQ Source: A = dir 130 B = dir 210 C = dir 330 D = dir 97 R80 P = B+C+D R285 Q = C+A+D R = A+B+D R--B--D--cycle 0.1 lightcyan / palecyan Q--A--C--cycle 0.1 lightred / palered C--B--D--cycle 0.1 lightgreen / palegreen R--foot(R,B,D) lightgrey dashed B--foot(B,D,R) lightgrey dashed D--foot(D,R,B) lightgrey dashed A--foot(A,C,Q) lightgrey dashed C--foot(C,Q,A) lightgrey dashed Q--foot(Q,A,C) lightgrey dashed C--foot(C,B,D) lightgrey dashed B--foot(B,D,C) lightgrey dashed D--foot(D,C,B) lightgrey dashed A--foot(A,B,C) lightgrey dashed B--foot(B,C,A) lightgrey dashed C--foot(C,A,B) lightgrey dashed T = midpoint A--P 1.8R90 S = (A+B+C+D)/4 1.8R75 O = 2*S-T R315 T--O lightblue A--B--C--cycle 0.1 yellow / lightgrey A--C palered A--Q red+1 B--P red+1 A--P blue B--Q blue C--R blue H = A+B+C R45 D--H lightblue */ [/asy][/asy] Claim: [Key claim] $T$ is the midpoint of $\overline{AP}$, $\overline{BQ}$, $\overline{CR}$, $\overline{DH}$, and $D$ is the orthocenter of $\triangle PQR$. Proof. Note that $\overline{AQ} \parallel \overline{BP}$, as both are perpendicular to $\overline{CD}$. Since lines $AP$ and $BQ$ are distinct, lines $AQ$ and $BP$ are distinct. By symmetric reasoning, we get that $AQCPBR$ is a hexagon with opposite sides parallel and concurrent diagonals as $\overline{AP}$, $\overline{BQ}$, $\overline{CR}$ meet at $T$. This implies that the hexagon is centrally symmetric about $T$; indeed \[ \frac{AT}{TP} = \frac{TQ}{BT} = \frac{CT}{TR} = \frac{TP}{AT} \]so all the ratios are equal to $+1$. Next, $\overline{PD} \perp \overline{BC} \parallel \overline{QR}$, so by symmetry we get $D$ is the orthocenter of $\triangle PQR$. This means that $T$ is the midpoint of $\overline{DH}$ as well. $\blacksquare$ Corollary: The configuration is now symmetric: we have four points $A$, $B$, $C$, $D$, and their reflections in $T$ are four orthocenters $P$, $Q$, $R$, $H$. Let $S$ be the centroid of $\{A, B, C, D\}$, and let $O$ be the reflection of $T$ in $S$. We are ready to conclude: Claim: $A$, $B$, $C$, $D$ are equidistant from $O$. Proof. Let $A'$, $O'$, $S'$, $T'$, $D'$ be the projections of $A$, $O$, $S$, $T$, $D$ onto line $BC$. Then $T'$ is the midpoint of $\overline{A'D'}$, so $S' = \tfrac14(A'+D'+B+C)$ gives that $O'$ is the midpoint of $\overline{BC}$. Thus $OB = OC$ and we're done. $\blacksquare$

Let $\mathcal{H}$ the rectangular circumhyperbola passing through $A,B,C,D,P,Q,R$ and let $X$ the concurrence point. By Pascal’s Theorem on $APDRCB$, we have that $X,DP \cap BC, DR \cap AC$ are colinear. Similarly, by Pascal’s Theorem on $APDRCB$, $X, DP \cap BC, DR \cap AB$ are collinear. Thus, $DP \cap BC, DQ \cap AC, DR \cap AB$ are collinear. Hence, since these points are the orthogonal projections of $D$ WRT $BC,CA,AB$ (because $P,Q,R$ are the ortocenter of $BCD, CDA, DAB$, respectively), we are done by the converse of the Simson Line. $\blacksquare$

Consider the rectangular circumhyperbola \(\mathcal H\) through the points \(A\), \(B\), \(C\), \(D\). Then \(\mathcal H\) also contains \(P\), \(Q\), \(R\). Let \(\overline{AP}\), \(\overline{BQ}\), \(\overline{CR}\) concur at \(O\). Claim: \(O\) is the midpoint of segments \(AP\), \(BQ\), \(CR\). Proof. First I contend \(BCQR\) is a parallelogram: indeed, \(\overline{BR}\) and \(\overline{CQ}\) are both perpendicular to \(\overline{AD}\), so \(\overline{BR}\parallel\overline{CQ}\). Similarly, \(\overline{BP}\parallel\overline{AQ}\) and \(\overline{CP}\parallel\overline{AR}\), so \(\overline{BC}\parallel\overline{QR}\) by Desargue's theorem on \(\triangle PBC\) and \(\triangle AQR\). Then \(\overline{BQ}\cap\overline{CR}=O\) is the midpoint of \(\overline{BQ}\) and \(\overline{CR}\), so the claim follows symmetrically. \(\blacksquare\) Claim: \(O\) is the center of \(\mathcal H\). Proof. Let \(\mathcal H'\) be the image of \(\mathcal H\) under reflection in \(O\). Then \(\mathcal H\) and \(\mathcal H'\) both contain \(A\), \(B\), \(C\), \(D\), \(P\), \(Q\), \(R\), so they are the same hyperbola. \(\blacksquare\) Finally \(\overline{PD}\perp\overline{BC}\parallel\overline{QR}\), so \(D\) is the orthocenter of \(\triangle PQR\). Therefore the reflection of \(D\) in \(O\) is the orthocenter \(H\) of \(\triangle ABC\). But it is well-known that the Poncelet point \(O\) lies on the nine-point circle of \(\triangle ABC\), so \(D\) lies on the circumcircle.

Basically like MarkBcc168's solution, but eh. Solution: We will use the same notations as MarkBcc168 as Simson lines will come in handy, i.e. $X=QR\cap BC$, $Y=PR\cap AC$, $Z=PQ\cap AB$. Now, by Desargues on $\triangle{AQR}$ and $\triangle{PBC}$, we get that $X, Y, Z$ are collinear. Furthermore, from Delta 7.1, we have that points which have isogonal conjugates at infinity are the points on the circumcircle, but the Six Point Circle Theorem notes that this happens iff the pedal triangle is degenerate, which is the case here! That is, $\triangle{XYZ}$ is the degenerate pedal triangle of $D*$ with respect to $\triangle{ABC}$ where $D*$ is the isogonal conjugate of $D$ with respect to $\triangle{ABC}$, so we are done. $\square$

Let $\mathcal{H}$ be the rectangular circumhyperbola passing through $A,B,C,D,P,Q,R$ and let $T$ be the concurrency point. Let $X=BR\cap CQ,Y=CP\cap AR,Z=AQ\cap BP$ (they all lie on the line at infinity). By Brocard's Theorem we know that the polar of $T$ WRT $\mathcal{H}$ passes through $X,Y,Z$, so $T$ is the center of $\mathcal{H}$. Then we get $BR=CQ$. By Carnot's Theorem we get the (directed) distance from the circumcenter of $\triangle ABC$ and the circumcenter of $\triangle DBC$ to AD is equal. Therefore the circumcenter of $\triangle ABC$ and the circumcenter of $\triangle DBC$ coincide, so we are done. $\square$

Idk if this sol has been posted before Consider the two triangles $\triangle ABR$ and $\triangle PQC$. Note that $AR \parallel CP$ as both are perpendicular to $BD$. Similarly, $BR \parallel CQ$ are both are perpendicular to $AD$. Next, we perform Desargues. Note that by the problem, $AP, BQ, CR$ concur, hence $AB \cap PQ, AR \cap CP, BR \cap CQ$ are collinear. But by the previously derived parallellisms, the latter two are points at infinity, hence $AB \parallel PQ$. This tells us that $ABPQ$ is a parallelogram, since $AQ \parallel BP$ is clear as they are both perpendicular to $CD$. Therefore, by parallelogram diagonal bisection, it is clear that $T = AP \cap BQ \cap CR$ is actually the midpoint of each of the three segments. Now, by orthocentric lengths,\[CD\cot{\angle DBC} = BP = AQ = CD\cot{\angle DAC}\]hence $\angle DBC = \angle DAC$ so $A, B, C, D$ indeed cyclic. $\blacksquare$ edit: oh, nvm, more or less, it has :/ I'm not liking this circumrectangular hyperbola stuff :/

Am I doing this right, idk. There's a lot of "by symmetry"s in this solution that I didn't add because I was lazy. $\overline{AQ},\overline{PB}$ are both perpendicular to $\overline{CD}$, so they are both parallel. Desargues on $\triangle AQR$ and $\triangle PBC$ means that $\overline{QR},\overline{BC}$ meet at the line at infinity, so they are also parallel. Thus $\overline{PQ}||\overline{AB}$, but it was already said that $\overline{AQ}||\overline{PB}$, so $AQPB$ is a parallelogram. Let $S$ be the intersection of $\overline{AP},\overline{BQ},\overline{CD}$. Then $S$ is the midpoint of $AP,BQ,CR$. $\overline{PD}\perp\overline{BC}$, so $\overline{PD}\perp\overline{QR}$. By symmetry $D$ is the orthocenter of $\triangle PQR$, so by a reflection around $S$, if $H$ is the orthocenter of $\triangle ABC$, then $HD$ has midpoint $S$. Plot this on the complex plane. Let $A,B,C,D$ be represented by $a,b,c,d$, where $a,b,c$ lie on the unit circle. $H$ is mapped to $a+b+c$, and $P$ is mapped to $b+c+d$ because $AHPD$ is a parallelogram. The centroid of $\triangle BCD$ is $\frac{b+c+d}{3}$, so the circumcenter of $\triangle BCD$ is $0$, as desired.

Let $Y$ be the concurrence point. We begin with the following key claim: $\textbf{Claim:}$ $\triangle ABC$ and $\triangle PQR$ are symmetric about $Y$. This is ratios. Observe that we have \begin{align*} \frac{YA}{YP} = \frac{YQ}{YB} = \frac{AQ}{BP}, \frac{YQ}{YB} = \frac{YC}{YR} = \frac{QC}{BR}, \frac{YA}{YP} = \frac{YR}{YC} = \frac{AR}{CP}. \end{align*}Now focus on \begin{align*} \frac{YQ}{YB} = \frac{YA}{YP} = \frac{YC}{YR} = \frac{YR}{YC}, \end{align*}which yields $YR = YC$ and everything else follows. $\square$ Now we throw everything onto the complex plane. Let $a$, $b$, $c$ be on the unit circle with origin $0$, and let $d$ be the coordinate for $D$. Let $D'$ denote the reflection of $D$ over $Y$. Observe that by our parallel lines that $D$ is the orthocenter of $\triangle PQR$, and so $D'$ is the orthocenter of $\triangle ABC$. Thus we can calculate $d' = a + b + c$. However, since $CDRD'$ is a parallelogram, we see that $r + c = d' + d$. Thus we have $r = a + b + d$. Since $A$ and $B$ lie on the unit circle, we know that this holds if and only if $D$ lies on the unit circle too. Thus we are done.

[asy][asy] size(8cm); pen prim = blue; pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = dir(70) * 3/2; pair E = foot(D, B, C); pair F = foot(C, B, D); pair G = foot(D, A, C); pair H = foot(C, A, D); pair P = extension(D, E, C, F); pair Q = extension(D, G, C, H); draw(A--B--C--cycle); draw(CP( (C+D)/2, C)); draw(D--E); draw(C--F); draw(D--G);draw(C--H); draw(A--D); draw(D--B); draw(G--E, prim); draw(F--H, prim); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, S); dot("$F$", F, dir(20)); dot("$G$", G, dir(220)); dot("$H$", H, dir(H-Q)); dot("$P$", P, dir(290)*2); dot("$Q$", Q, dir(350)); draw(B--Q, prim); draw(A--P, prim); [/asy][/asy] Let $E, G$ be the feet from $D$ to $BC, AC,$ $H$ be the foot from $C$ to $AD,$ and $F$ be the foot from $C$ to $DB.$ Claim: $AP \cap BQ$ lies on $GE.$ Proof. Because of right angles, $DHGFEC$ is cyclic with diameter $DC.$ Now, we apply Pascals on $CEGDFH, CFHDEG$ to get $B, EG \cap FH, Q$ and $P, HF \cap GE, A$ are collinear. Therefore, $AP, GE, HF, BQ$ are concurrent. $\blacksquare$ Applying this claim symmetrically tells us that if the foot from $D$ to $AB$ is $I$ then $EG, GI, EI$ all pass through $AP \cap BQ \cap CR,$ which implies $EGI$ is collinear. By converse of Simson line $ABCD$ is cyclic as desired.

Hmm this was easier than expected ... 7200th post! [asy][asy] size(250); pair A = (-3.6488936772554, 4.7629376368042), B = (4.064712684455, 4.4134012725823), C = (4.709553074972, -3.7175408315205), D = (0.4691154449004, -5.9816327787115); pair P = orthocenter(B, C, D), Q = orthocenter(A, C, D), R = orthocenter(A, B, D); pair X = extension(B, Q, A, P); draw(A--B--C--D--cycle, blue); draw(circumcircle(A, B, C), lightblue); draw(A--C, green); draw(B--D, green); draw(A--P, orange); draw(B--Q, orange); draw(C--R, orange); draw(B--R--Q--C, red+dashed); draw(A--R, magenta+dashed); draw(C--P, magenta+dashed); dot("$A$", A, NW); dot("$B$", B, NE); dot("$C$", C, SE); dot("$D$", D, S); dot("$Q$", Q, SE); dot("$P$", P, SE); dot("$R$", R, N); dot("$X$", X, E); [/asy][/asy] Let the concurrency point be $X$. The crux of the problem lies in the following: Claim. $BRQC$ is a parallelogram. Proof. We will show that $RQ \parallel BC$, because $BR \parallel CQ$ is trivial (because they are both perpendicular to $\overline{AC}$ by definition. First, observe that since $BP \parallel AQ$, we have $\triangle XAQ \sim \triangle XPB$. Similarly, $CP \parallel AR$ implies $\triangle ARX \sim \triangle PCX$. From these two relations, we have $$\frac{CP}{AR} = \frac{PH}{AH} = \frac{BP}{AQ}.$$However, because $AR \parallel PC$ and $AQ \parallel PB$, $\overline{CP} \cap \overline{AQ}, A, \overline{AR} \cap \overline{BP}, P$ are the vertices of a parallelogram. From this, we devise that $\angle QAR = \angle CPB$, so we have $\triangle ARQ \sim \triangle PCB$. It follows that $\angle ARQ = \angle BCR$. Combined with $\angle ARX = \angle XCR$ which is already known, we have $\angle QRX = \angle HCX$, implying $\overline{QR} \parallel \overline{BC}$, as desired. $\blacksquare$ To finish, observe that $$\frac{CQ}{\cos \angle ACD} = \frac{CQ}{\sin \angle QDC} = \frac{CD}{\sin \angle DQC} = \frac{CD}{\sin \angle DAC},$$so $$CQ = \cos \angle ACD \left(\frac{CD}{\sin \angle DAC}\right) = \cos \angle ACD \left(\frac{AD}{\sin \angle ACD}\right) = \frac{AD}{\tan \angle ACD}.$$Similarly, $$\frac{BR}{\cos \angle ABD} = \frac{BR}{\sin \angle RAB} = \frac{AB}{\sin \angle ARB} = \frac{AB}{\sin \angle ADB},$$so $$BR = \cos \angle ABD\left(\frac{AB}{\sin \angle ADB}\right) = \cos \angle ABD\left(\frac{AD}{\sin \angle ABD}\right) = \frac{AD}{\tan \angle ABD}.$$But because $BQ=CR$ due to parallelogram $BQRC$, we must have $\tan \angle ABD = \tan \angle ACD$. This implies that $\angle ABD = \angle ACD$, or $ABCD$ is cyclic, as desired. $\square$ Remark. When $\angle ABD$ or $\angle ACD$ is obtuse, it actually follows that $\sin \angle RAB = -\cos \angle ABD$, in which case we obtain $\tan \angle ABD = -\tan \angle ACD$, implying $\angle ABD + \angle ACD = 180^{\circ}$. This resolves any potential configuration issues.

Let $\mathcal H$ be the rectangular circumhyperbola of $ABC$ including $D$. Clearly $\mathcal H$ includes the orthocenters of $\triangle ABC, \triangle BCD, \triangle CDA, \triangle DAB$. Let the orthocenter of $\triangle ABC$ be $H$. Let the common intersection of $AP, BQ, CR$ be $T$. Note that $BR\perp AD\perp CQ$ so $\triangle TRB\sim \triangle TCQ$. In particular, we have $RT/TB = CT/CQ$, so by analogous results, we have \[\frac{RT}{TC} = \frac{RT}{TB}\cdot \frac{TB}{TP} \cdot \frac{TP}{TC} = \frac{CT}{CQ}\cdot \frac{TQ}{TA}\cdot \frac{TA}{TR} = \frac{CT}{TR}.\]Thus $CT=TR$, meaning $T$ is the midpoint of $CR$. But $T$ is similarly the midpoint of $AP$ and $BQ$, meaning that reflection through $T$ must be an involution on $\mathcal H$. Thus $T$ is the Poncelet point of $\mathcal H$, meaning that, as the reflection of $R$ over $T$, $C$ must lie on $(ABD)$, so we are done.

Define $S$ as the orthocenter of $ABC$, and $M$ as the concurrency point of $AP, BQ, CR$. First, observe that $AQ\perp CD$, $BP\perp CD$, so $AQ || BP$. Similarly, $AR || CP, BR || CQ$, so \[\frac{AM}{MQ} = \frac{MP}{BM}, \frac{AM}{MR} = \frac{MP}{CM}\Rightarrow \frac{MR}{MQ} = \frac{CM}{BM}\]However $\frac{CM}{MQ} = \frac{MR}{BM}$ so $MB = MQ$. Therefore, $M$ is the midpoint of $CR, BQ, AP$. Now, if we denote $S'$ as the reflection of $D$ over $M$, then $AS' || DP, BS' || DQ, CS' || DR$, but we also have $AS || DP, BS || DQ, CS || DR$, so $S = S'$. Therefore, $M$ is also the midpoint of $DS$. However, consider the circum-rectangular hyperbola going through $ABCDPQRS$. Since $M$ is the midpoint of $4$ different line segments on the hyperbola, this means $M$ is the poncelet point. This means $M$ lies on the nine point circle of $ABC$, so the reflection of $S$ over $M$ lies on the circumcircle of $(ABC)$. Therefore, $A,B,C,D$ are cyclic.

Solved with rama1728 Let the lines $AP$, $BQ$, $CR$ concur at $X$ We begin by proving some claims: Claim 1: X is midpoint of $AP$, $BQ$, $CR$ Proof: Note, $AQ||BP$ as both are perpendicular to $CD$. Also, $AP \cap BQ = X$. $ \Rightarrow \frac{PX}{XA} =\frac{BX}{XQ} $ Similarly, $BR||CQ$ as both are perpendicular to $AD$. Also $ BQ \cap CR = X$.$\Rightarrow \frac{BX}{XQ} = \frac{RX}{XC} $ Hence, $\frac{PX}{XA} = \frac{RX}{CR} \Rightarrow AC || PR $ Also, $ AR||CP$ as both are perpendicular to $BD$. Hence, $ \square ARCP $ is a parallelogram! Hence, $X$ which is intersection of it's diagonals is indeed the midpoint of the diagonal. Hence, $X$ is midpoint of $AP$ and $CR$. We can similarly prove that $X$ is midpoint of $BQ$ Hence proved claim 1.$\blacksquare$ Claim 2: $D$ is orthocenter of $PQR$. Proof: We proved that $ \square ARCP $ is a parallelogram. We can similarly prove that $\square AQBP $ and $ \square BRQC $ is parallelogram. $PD$ is perpendicular to $BC$, as $P$ is orthocenter of $BCD$. Hence, $PD$ is perpendicular to $ QR $, as $ BC||QR$ using the parallelogram $\square BRQC $ We can similarly prove that $QD$ is perpendicular to $PR$, and $RD$ is perpendicular to $PQ$. hence, $D$ is orthocenter of $PQR$. hence Proved claim 2.$\blacksquare$ Claim 3: $X$ is midpoint of $DS$ where $S$ is orthocenter of $\triangle ABC$ Proof: From the three parallelograms, $ \square ARCP, \square AQBP, \square BRQC $, we get that,$PQ=AB, QR=BC, PR=AC$ Hence, $ \triangle ABC \sim \triangle PQR $ Further, $AP, BQ, CR$ intersect at $X$ Hence, there is a homothety centered at $X$ mapping $ \triangle ABC $ to $ \triangle PQR $ with scale factor $1$. Hence, under this homothety, the orthocenter of $ \triangle ABC$ gets mapped to orthocenter of $\triangle PQR $ . Hence, using claim 2, $S$ gets mapped to $D$ under this homothety. Hence, $SD$ has midpoint X due to the homothety at $X$ with scale factor $1$. Hence proved claim 3.$\blacksquare$ Claim 4: $\square PQRS $ is cyclic quadrilateral. Proof: We present a proof using complex numbers. We let $a$ be the complex number representing $A$ and similar notation throughout the proof. We can wlog assume that $a,b,c$ lie on the unit circle. It's equivalent to prove that $d$ also lies on the unit circle. As $a,b,c$ lie on the unit circle, orthocenter of $\triangle ABC $ which is $S$ will be $s=a+b+c$. $X$ is midpoint of $SD$ using claim 3. Hence, $x = \frac{s+d}{2} = \frac{a+b+c+d}{2}$ Also, $X$ is midpoint of $AP$, $BQ$, $CR$ using claim 1. Hence, we get that $ p=b+c+d, q=a+c+d , r=a+b+d $ (Note $a,b,c$ lie on the unit circle ) Hence, obviously $d$ lies on the unit circle. Hence, $A,B,C,D$ lie on the unit circle and hence are cyclic. $\blacksquare$ Note that, a Homothety centered at $X$ with scale factor $1$ sends $\square ABCD $ to $\square PQRS $. As $\square PQRS $ is cyclic from claim 3, hence so is $\square ABCD $! Hence proved ! $\blacksquare$

First, let's rename the points to make things more clear. Replace $D$ with $P$ and let $H_A$, $H_B$, and $H_C$ be the orthocenters of $\triangle PBC$, $\triangle PCA$, and $\triangle PAB$. The problem statement tells us that $AH_A$, $BH_B$, and $CH_C$ all concur at a point $X$ and we would like to show $P$ lies on the circumcircle $\Omega$ of $\triangle ABC$. Lemma: If hexagon $ABCDEF$ obeys $AB \parallel DE$, $BC \parallel EF$, and $CD \parallel FA$, and $AD$, $BE$, and $CF$ concur at $X$, then hexagon $ABCDEF$ is rotationally symmetric around $X$. Indeed, by similar triangles, we have \[ \frac{AX}{XD} = \frac{BX}{XE} = \frac{CX}{XF} = \frac{DX} {XA} \implies AX = DX\]So, $X$ is the midpoint of $AD$, $BE$, and $CF$ which clearly implies the result. Notice that $AH_B \parallel BH_A$, $BH_C \parallel CH_B$, and $CH_A \parallel AH_C$ and $AH_A$, $BH_B$, and $CH_C$ concur. Thus, hexagon $AH_BCH_ABH_C$ is rotationally symmetric around $X$. Thus, $\triangle ABC$ and $\triangle H_AH_BH_C$ are reflections of each other over $X$. Note that $BC \parallel H_BH_C$. Now, assume for the asked of contradiction that $P$ does not lie on $\Omega$. Let $K_B$ and $K_C$ be reflections of $H_B$ and $H_C$ over line $AP$. We now have that $BAPK_B$, $CAPK_P$, and $BCK_BK_C$ are all cyclic. Since $P$ does not lie on $\Omega$, it must follow that $BAPK_B$ and $CAPK_C$ have distinct circumcircles. Since $BCK_BK_C$ is cyclic, it follows that $BK_B$ and $CK_C$ intersect on the radical axis of $BAPK_B$ and $CAPK_C$ which is $AP$. But, this is clearly a contradiction since $BK_B \parallel CK_C$.

Step 1: $AR\parallel PC, QA \parallel BP, CQ\parallel RB$. Proof: We have $AR\perp BD$ and $PC\perp BD$ due to orthocentres, so $AR\parallel BD$ and the rest follow by symmetry. Step 2: $ARPC$ is a parallelogram Proof: Let $AP, BQ, CR$ concur at $X$. Due to the parallel lines: $$\frac{AX}{PX}=\frac{QX}{BX}=\frac{CX}{RX}=\frac{PX}{AX}$$so $AX=PX$ giving $RX=CX$ by symmetry so $ARPC$ is a parallelogram. Step 3: $ABCD$ cyclic. Proof: We have $$\cot \angle BAD=\frac{AR}{BD}=\frac{CP}{BD}=\cot \angle BCD$$so $\angle BAD=\angle BCD$ so $ABCD$ is cyclic.

Notice how distance $AQ=DC\cot \angle DAC,$ and similar for all the other $Vertice-Orthocenter$ kinda distances. We'll show that $AQ=BP$ and $AR=PC,$ which will then solve the problem. Denote the interesection of $AP,BQ,CR$ by $S.$ Notice that $$CP\parallel AR\implies \frac{SC}{SR}=\frac{SP}{SA},$$and $$AQ\parallel BP\implies \frac{SP}{SA}=\frac{SB}{SQ}.$$Thus $\frac{SC}{SR}=\frac{SB}{SQ},$ so we get that $\triangle SRQ\sim \triangle SCB, $ which implies that $RQ\parallel BC \implies RBCQ$ parallelogram, which implies that $S$ is the midpoint of $BQ.$ So now, $AQ\parallel BP\implies BP=AQ,$ and $AR\parallel BC\implies RA=PC,$ which ends the problem.

Let $T$ be the concurrency point. Claim: $ARBPCQ$ has opposite sides parallel. Proof: Note that both $\overline{AR}$ and $\overline{CP}$ are both perpendicular to $\overline{BD}$, so $\overline{AR} \parallel \overline{CP}$. The other parallelisms follow similarly. $\blacksquare$ Claim: $T$ is the midpoint of $\overline{AP},\overline{BQ},\overline{CR}$. Proof: From the parallelism, we have $\triangle TAR \sim \triangle TPC$, so $\frac{TA}{TP}=\frac{TR}{TC}$. Continuing this cyclically, we have $$\frac{TA}{TP}=\frac{TR}{TC}=\frac{TB}{TQ}=\frac{TP}{TA}=\frac{TC}{TR}=\frac{TQ}{TB},$$so the ratios are all $1$. $\blacksquare$ We now use complex numbers, denoting points with their lowercase versions and setting $(ABC)$ as the unit circle. We can calculate the circumcenter of $(BCD)$ as $$\frac{\begin{vmatrix}b&1&1\\c&1&1\\d&d\overline{d}&1\end{vmatrix}}{\begin{vmatrix}b&\frac{1}{b}&1\\c&\frac{1}{c}&1\\d&\overline{d}&1\end{vmatrix}}=\frac{\begin{vmatrix}b&0&1\\c&0&1\\d&d\overline{d}-1&1\end{vmatrix}}{\begin{vmatrix}b&\frac{1}{b}&1\\c&\frac{1}{c}&1\\d&\overline{d}&1\end{vmatrix}}=\frac{(b-c)(d\overline{d}-1)}{\frac{(b-c)(b+c)}{bd}-\frac{d(b-c)}{bc}-\overline{d}(b-c)}=\frac{bc(d\overline{d}-1)}{b+c-d-bc\overline{d}}.$$Since $H=3G-2O$, it follows that $P=b+c+d-\tfrac{2bc(d\overline{d}-1)}{b+c-d-bc\overline{d}}$, and $T$, calculated as the midpoint of $\overline{AP}$, is thus $$\frac{a+b+c+d}{2}-\frac{bc(d\overline{d}-1)}{b+c-d-bc\overline{d}}.$$We obtain cyclic variations for $T$ from $\overline{BQ}$ and $\overline{CQ}$. Therefore, if $d\overline{d}-1 \neq 0$, we require $$\frac{ab}{a+b-d-ab\overline{d}}=\frac{ac}{a+c-d-ac\overline{d}} \iff ab+bc-bd-abc\overline{d}=ac+bc-cd-abc\overline{d} \iff (a-d)(b-c)=0,$$absurd. Thus $d\overline{d}-1=0$, so $ABCD$ cyclic. $\blacksquare$

Let $\mathcal{H}$ denote the circumrectangular hyperbola through $A,B.C,D,P,Q,R$ Firstly, we claim $T$ is midpoint of $AP,BQ,CR$, which follow from length chasing with parallel lines. Next, we claim that $T$ is center of $\mathcal{H}$, which follows from the fact that if we reflect everything about $T$ then we get same hyperbola. So $D$ must be the orthocenter of $PQR$, now since it is well known poncelet points lie on nine point circle, $D \in (ABC)$. Done.

Call $X$ the concurrency point. Since $AQ\perp CD\perp BP,$ we get $\frac{AX}{PX}=\frac{QX}{BX}.$ Similarly, $\frac{QX}{BX}=\frac{CX}{RX}=\frac{PX}{AX},$ thus all ratios must be $-1$ in directed lengths as clearly they cannot be $1.$ Then $ARBPCQ$ is centrally symmetric about $X.$ Now consider the rectangular hyperbola through $A,B,C,D.$ It must then pass through $P,Q,R.$ Now if its center is some point $Y\ne X,$ the reflection of hexagon $ARBPCQ$ over $Y$ is a different hexagon that still has all vertices on the hyperbola. However, this reflected hexagon is also a translation of $ARBPCQ.$ Since there is at most one conic through six points, this implies that the hyperbola is invariant under some translation, which is impossible. Thus $X$ is the center of the hyperbola. Now we know that $X$ lies on the nine-point circle of $BCD,$ so the reflection of $P$ over $X$ lies on $(BCD).$ But this point is $A$ so we are done.

$\triangle ARB$ and $\triangle PCQ$ are perspective and $BR\parallel CQ$ and $AR\parallel CP$, so there is a homothety from $\triangle ARB$ to $\triangle PCQ$. Therefore, $AB\parallel QP$. Thus, $ABPQ$ is a parallelogram. So, using signed lengths, $BP=AQ$. Note that using directed angles, $AQ=CD\cot \measuredangle DAC$ and $BP = CD \cot \measuredangle DBC$. Thus, $\measuredangle DAC = \measuredangle DBC$.

I don't think its been noted here but the reason the center of the circle is the reflection of the concurrence over the centroid follows by complex bash to get that $s = \frac{1}{2}(a + b + c + d)$ which is also called the anticenter. This is also a motivation for considering such a point, by knowing a priori the converse solution. Anyways, ye old hyperbola nuke. Let $\mathcal{H}$ be the the circumrectangular hyperbola through $A, B, C, D$. Then $P, Q, R$ must lie on the hyperbola. Then note that $AQ \parallel BP$, $BR \parallel CQ$, $CP \parallel AR$. Claim: If the diagonals of equiangular hexagon $ABCDEF$ concur at some point $O$, then hexagon is symmetric about $O$. Proof. Let $X = AB \cap CD$ and $Y = AF \cap DE$. Note that $AXDY$ is a paralellogram. We claim that $G = BE \cap CF$ lies on $XY$. Let $G_B = BE \cap XY$ and $G_C = CF \cap XY$. Then it follows that \[ \frac{XG_B}{G_BY} = \frac{XB}{YE} = \frac{XC}{YF} = \frac{XG_C}{YG_C} \]so $G_B = G_C = G$. Anyways, if $G$ lies on $AD$, it follows that $G = XY \cap AD$, which is the midpoint of $AD$. By symmetry, it follows that $G$ is the midpoints of $BE, CF$ which suffices. $\blacksquare$ As such, since $AQCPBR$ is equiangular, it follows that their concurrence point $K$ is their symmetry point. This implies that $(AQCPBR)$ is fixed under reflection about $K$, which implies that $K$ is the Poncelet point. Then $A$ is the reflection of $P$ across $K$, and thus $ABCD$ must be cyclic.

Pure moving points? spartacle wrote: Let $A$, $B$, $C$, $D$ be four points such that no three are collinear and $D$ is not the orthocenter of $ABC$. Let $P$, $Q$, $R$ be the orthocenters of $\triangle BCD$, $\triangle CAD$, $\triangle ABD$, respectively. Suppose that the lines $AP$, $BQ$, $CR$ are pairwise distinct and are concurrent. Show that the four points $A$, $B$, $C$, $D$ lie on a circle. Andrew Gu Fix a line $\ell$ passing through the orthocentre $H$ of $\triangle ABC$. Suppose it meets $(ABC)$ at $X$ and $Y$. Animate point $D$ on line $\ell$. We will show that $D$ must equal one of $X, Y, \ell_{\infty}$ for the concurrence to hold. This combined with varying $\ell$ implies the conclusion. Each of these four cases work is obvious: $X$ and $Y$ work for analogous reasons as the midpoints of the segments $AP, BQ, CR$ in that case coincides with the point with vector $(A+B+C+D)/2$, $\ell_{\infty}$ works trivially as each of $P, Q, R$ coincide with the point at infinity in the direction perpendicular to $\ell$. Finally, observe that as $H \in \ell$, the locii of each of $P, Q, R$ are hyperbolas $\mathcal{H}_a, \mathcal{H}_b, \mathcal{H}_c$ passing through $A, B, C$ respectively with the maps $D \mapsto P, D \mapsto Q, D \mapsto R$ all projective, hence lines $AP, BQ, CR$ are each degree $1$ moving lines. The concurrence is a cubic condition which can have at most $3$ solutions, else it is an identity. However, the concurrence is never a concurrence: this can be seen by letting $D = \ell \cap BC$ --- points $Q, R$ lie on the $A$-altitude and $AP$ becomes the $A$-altitude, so concurrence requires $Q=R=D=AH \cap BC$. Thus, for every choice of $\ell$ other than the $A$ altitude, the concurrence is not always true, similarly taking cases for $B$ and $C$ tells that the concurrence is never identically true for any line $\ell$. This completes the proof.

Let's first prove the converse of the problem. Assume $ABCD$ is cyclic and define $H_A$ orthocenter of $\triangle BCD$ and similar. Then in complex numbers midpoint of $AH_A$ is $\frac{a+h_A}{2}=\frac{a+b+c+d}{2}$ which is the same for other points. Therefore $AH_A,BH_B,CH_C,DH_D$ are concurrent. Let $\mathcal{H}$ be a conic which passes through $A,B,C,D,Q$. This is a rectangular hyperbola so orthocenters $P,R\in \mathcal{H}$. Since $AP,CR,DQ$ are concurrent there is an involution $\Phi:\mathcal{H}\rightarrow \mathcal{H}$ which swaps $(A,P),(C,R),(B,Q)$. Therefore $(DB,DQ;DA,DC)\stackrel{\Phi}{=}(DQ,DB;DP,DR)=(DB,DQ;DR,DP)$. Now let $D'=(ABC)\cap BD$ and suppose $D'\neq D$ and define $P',Q',R'$ accordingly. Since converse of the problem is true we have $BQ',AP',CR'$ concurrent so analogously $(D'B,D'Q';D'A,D'C)=(D'B,D'Q';D'R',D'P')=(DB,DQ;DR,DP)=(DB,DQ;DA,DC)$. This means that $A,B,C,D,D',DQ_{\infty}$ are on a conic. $B,D,D'$ are collinear so this conic must be union of 2 lines which is impossible unless $BD_{\infty}=DQ_{\infty}$ (since certainly $DQ_{\infty}\neq AC_{\infty}$) which would mean $BD\perp AC$. Then we could just do the same thing for $C$ and $A$. If both fail then we have $AB\perp CD,CB\perp AD$ which is a contradiction since $ABCD$ is not an orthocentric system.

I'm sure this solution was found already, but posting for storage. Let $T$ be the concurrency point of $AP$, $BQ$, $CR$. Note that $AR \cap PC$, $BP\cap AQ$, $BR \cap CQ$ are all points at infinity, so they lie on the line at infinity. Hence, by the converse of Pascals theorem, the points $A$,$B$,$C$,$D$,$P$,$Q$,$R$ lie on a conic $\mathcal{H}$, which incidentally is a rectangular hyperbola. Next, note that $$\frac{AT}{TP} = \frac{RT}{TC} = \frac{BT}{TQ} = \frac{TP}{TA}$$This implies that $T$ is the midpoint of $AP$, $BQ$ and $CR$. We now show that $T$ is the center of $\mathcal{H}$. Consider the triangles $\triangle BRQ$ and $\triangle BCQ$. It's easy to see that the nine point circle of these two are tangent to each other at $T$. On the other hand, we know that the center of $\mathcal{H}$ lies on these nine point circles. This means that $T$ is the center of $\mathcal{H}$. Let $S$ be the orthocenter of $\triangle ABC$. Then, the two triangles $\triangle PQR$ and $\triangle ABC$ are reflections of each other at $T$. This means that their orthocenters, which are $D$ and $S$ respectively, are also reflections of each other through $T$. To finish, since $T$ lies on the nine point circle of $\triangle ABC$, the reflection of $S$, which is $D$, must lie on $(ABC)$. We're done.

Let $QR \cap BC=\{X\}$, $PR \cap AC=\{Y\}$, $PQ \cap AB=\{Z\}$. Now we'll introduce the following lemma. Lemma: Let $l_a$ be the line through $X$ and perpendicular to $BC$. Define $l_b$ and $l_c$ similarly. Then the lines $l_a, l_b$, and $l_c$ are concurrent on the isogonal conjugate of $D$. Proof: Let $\mathcal{H}$ be the circumrectangular hyperbola passing through $A, B, C, D$. Then from the property of the rectangular hyperbola, $P, Q, R \in \mathcal{H}$. Now consider Pascal's theorem on the points $(PQRABC)$. From this, we obtain that $XZ, RA$, and $PC$ are concurrent. (possibly on the point at infinity) Since both of the lines $RA$ and $PC$ are perpendicular to $BD$, we obtain that $XZ$ is perpendicular to $BD$. Similarly, $XY$ is perpendicular to $CD$ and $YZ$ is perpendicular to $AD$. Thus $XYZ$ is the pedal triangle of the isogonal conjugate of $D$ wrt. triangle $ABC$. From Desargues' theorem; $AP$, $BQ$ and $CR$ are concurrent if and only if $X$, $Y$ and $Z$ are collinear. Thus this concurrency is equivalent to the six-point circle of $D$ being a line, which is equivalent to $A, B, C, D$ being concyclic as desired.