Point D is the orthocenter of triangle APQ, AX||YE||PQ=>m(XAQ)=m(AQP)=m(B). By angle chasssing,is easy to show that PY||AB ; remain to prove that AC||BX. How can show that this?

It is trivial that $B,C,P,Q$ are concyclic and the circumcircle of triangle $PDQ$ is tangent to $BC$. Now it remains to prove that $KD^2 = KX \times KY$ where $K$ is the intersection of $PQ$ and $BC$. By projecting this on line $AE$ and using the fact that $AO$ and the perpendicular line from$A$ to $BC$ are isogonal, this can be proved

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Seungjun_Lee wrote: It is trivial that $B,C,P,Q$ are concyclic and the circumcircle of triangle $PDQ$ is tangent to $BC$. Now it remains to prove that $KD^2 = KX \times KY$ where $K$ is the intersection of $PQ$ and $BC$. By projecting this on line $AE$ and using the fact that $AO$ and the perpendicular line from$A$ to $BC$ are isogonal, this can be proved Can you please explain $KD^2=KX\times KY$ in more detail? I can understand the projection part, but I don't get the next part.

let the foot of $D$ to $AC$ be $M$ and the foot of $D$ to $AB$ be $N$, let $PQ$ meet $BC$ at $J$, by Homothety $MN$ is parallel to $BC$ by DDIT on $M,N,P,Q$ and line $BC$ we get $(J, \infty_{BC}), (B,C), (D,D)$ are pairs of involution, so it's an inversion with center $J$ and we know $JD^2=JB\times JC$ use DDIT again on $A,A,E,E, (ABC)$ and line $BC$ to get $(X,Y), (D,D), (B,C)$ are pairs of involution, again it's an inversion, so by calculating the power we get $X,Y,P,Q$ are concyclic and we are done

Let $S \equiv PQ \cap BC,$ $AO$ intersects $(OBC)$ again at $F,$ $G \in BC$ satisfies $FG \perp AO$. We have $\angle{CPD} = 90^{\circ} - \angle{BAC} = \angle{OBC} = \angle{OFC}$. Then $C, P, F, D$ lie on a circle. Similarly, $B, Q, F, D$ lie on a circle. So $\angle{PFQ} = \angle{PFD} + \angle{DFQ} = \angle{ACB} + \angle{ABC} = 180^{\circ} - \angle{BAC}$ or $F \in (APQ)$. But it's easy to see that $D$ is orthocenter of $\triangle APQ$ then $AO \perp PQ,$ so $F$ is reflection of $D$ in $PQ$. Hence $S$ is midpoint of $DG$. We also have $\angle{BDQ} = \angle{ABC} - \angle{BQD} = \angle{ABC} - \angle{BFD} = \angle{ABC} - \angle{OBC} = \angle{OBA} = \angle{OAB} = \angle{DPQ}$. Hence $BC$ tangents $(DPQ)$ at $D$. Since $(AE, DF) = - 1$ and $AX \parallel EY \parallel FG,$ we have $(XY, DG) = - 1$. Combine with $S$ is midpoint of $DG,$ we have $\overline{SX} \cdot \overline{SY} = SD^2 = \overline{SP} \cdot \overline{SQ}$ or $X, P, Q, Y$ lie on a circle

Let $Z = BC \cap PQ$, and $K$ be the reflection of $D$ in $Z$. Let $L$ be the foot of the perpendicular from $K$ to $AO$. [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ pair O = (-3.4548,1.0084); pair A = (-5.319,4.554); pair B = (-7.08994,-0.67453); pair C = (0.19941,-0.63271); pair D = (-2.58354,-0.64868); pair E = (-1.5906,-2.5372); pair P = (1.79388,-2.13134); pair Q = (-9.61514,-8.12996); pair Y = (2.05183,-0.62208); pair X = (-15.35354,-0.72194); pair Z = (4.69323,-0.60693); pair K = (11.97000,-0.56518); pair L = (0.53396,-6.57800); import graph; size(14cm); pen ffttww = rgb(1,0.2,0.4); draw(circle(O, 4.00581), linewidth(1)); draw(B--C, linewidth(1)); draw(A--X, linewidth(1)); draw(E--Y, linewidth(1)); draw(A--Q, linewidth(1)); draw(A--P, linewidth(1)); draw(A--L, linewidth(1)); draw((-1.27024,0.74860)--Q, linewidth(1)); draw((-6.61830,0.71792)--P, linewidth(1)); draw(X--K, linewidth(1)); draw(K--L, linewidth(1)); draw(Z--Q, linewidth(1)); draw(circle((-6.65522,0.08939), 8.73608), linewidth(1) + linetype("4 4") + ffttww); dot("$O$", O, dir((-10, -10))); dot("$A$", A, dir((-10, 10))); dot("$B$", B, dir((-10, 10))); dot("$C$", C, dir((10, 10))); dot("$D$", D, dir((-10, -10))); dot("$E$", E, dir((-3, -10))); dot("$P$", P, dir((10, -10))); dot("$Q$", Q, dir((-10, -10))); dot("$Y$", Y, dir((10, 10))); dot("$X$", X, dir((-10, -10))); dot("$Z$", Z, dir((10, 10))); dot("$K$", K, dir((10, 10))); dot("$L$", L, dir((0, -10))); [/asy][/asy] By easy angle chasing, we see $B, C, P, Q$ are concylic and $(PDQ)$ is tangent to $BC$. This implies $ZD^2 = ZP \cdot ZQ = ZB \cdot ZC$, and $(B, C; D, K) = -1$. From this, we have $\angle BLD = \angle CLD$ so $B, O, C, L$ are concyclic and $OB^2 = OD \cdot OL$. Observe $(X, Y; D, K) = (A, E; D, L) = -1$, so $ZD^2 = ZX \cdot ZY$. From above, we get $ZX \cdot ZY = ZP \cdot ZQ$, so $P, Q, X, Y$ are concyclic. $\blacksquare$

AliAlavi wrote: Solution for this problem: How did you get $RD^2=RX\times RY$ ???

Solved with swynca. Perform $\sqrt{bc}$ inversion and reflect over the angle bisector $\measuredangle CAB$. New Problem Statement: $ABC$ is a triangle with orthocenter $H$. $AH,BH,CH$ intersect $(ABC)$ at $S,P',Q'$ respectively. $SP'$ and $SQ'$ intersect $AC,AB$ at $P,Q$. $X$ lies on $(ABC)$ such that $AX\parallel BC$ and $AS$ meets $BC$ at $D$. $Y$ is chosen on $(ABC)$ with $\measuredangle AYD=90$. Prove that $X,Y,P,Q$ are concyclic. Let $E$ be the antipode of $A$ on $(ABC)$. $XY\cap BC=R$. Pascal at $SSAXYE$ implies $SS\cap XY,D,BC_{\infty}$ are collinear hence $SS\cap XY$ lies on $BC$ which must be $R$. Thus, $RS$ is tangent to $(ABC)$. Since $\measuredangle HRD=\measuredangle DRS=\measuredangle CBS-\measuredangle SAB=\measuredangle B-\measuredangle C$, we observe that $RH$ is the antiparallel to $BC$. Also pascal at $P'SQ'CAB$ gives $P,Q,H$ are collinear. Also if $H_B,H_C$ are the feet of the altitudes from $B,C$ to $AC,AB$, then $PQS$ and $H_BH_CD$ are homothetic so $PQ\parallel H_BH_C$ which yields $PQ$ is antiparallel to $BC$. These imply that $P,Q,H,R$ are collinear. $RQ.RP=RB.RC=RY.RX$ which completes the proof as desired.$\blacksquare$