a1267ab wrote: Let scalene triangle $ABC$ have altitudes $AD, BE, CF$ and circumcenter $O$. The circumcircles of $\triangle ABC$ and $\triangle ADO$ meet at $P \ne A$. The circumcircle of $\triangle ABC$ meets lines $PE$ at $X \ne P$ and $PF$ at $Y \ne P$. Prove that $XY \parallel BC$. Proposed by Daniel Hu Let $ABCA'$ be an isosceles trapezoid with $\overline{AA'} \parallel \overline{BC}$. Claim. $P,H,A'$ are collinear. (Proof) Reflect $O$ in the $A$-midline of $\triangle ABC$ to obtain point $G$. Then $AH=2OM$ where $M$ is the midpoint of $\overline{BC}$; yields $H,G,A'$ are collinear. Redefine $P$ as $P \overset{\text{def}}{:=} \overline{GH} \cap \odot(ABC)$ with $P \ne A'$. Then $\measuredangle APG=\measuredangle OAD=\measuredangle ADG$ hence $ADGP$ are concyclic; proving the claim. $\blacksquare$ Now observe $\overline{PA}, \overline{PH}$ are isogonal in angle $BPC$. Apply isogonality lemma to obtain $\overline{PE}, \overline{PF}$ are isogonal in angle $BPC$ and we're done!

$\textbf{LEMMA:1}$ In a $\Delta ABC$ let $I,I_A$ be the incenter and $A-$excenter and let $X$ be the reflection of $I$ in $BC$ and let $Y$ be the feet of $A-$ altitude then $I_A,X,Y$ are collinear. $\textbf{PROOF:}$ let $T$ be the projection of $I$ in $BC$ ,it is well known that $I_AT$ bisects $AY$ now since $IX\parallel AY$ and $IT=TX$ we have $I_A,X,Y$ are collinear $\blacksquare$ $\textbf{LEMMA:2}$ In a $\Delta ABC$ let $O$ be the circumcenter and let $AO\cap \odot (BOC)=O_A$ and let $X$ be the projection of $O_A$ in $BC$ and let $A_1,A_2$ be the reflection of $A$ in $BC$ and antipode of $A$ in $\odot (ABC)$ ,then $A_1,A_2,X$ are collinear. $\textbf{PROOF:}$ We can easily see that $A_2,A$ are the incenter and $A-$ excenter of $\Delta O_ABC$ now reflecting in $BC$ and by $\textbf{LEMMA:1}$ we are done $\blacksquare$. Now coming back to the problem by isogonality lemma all we need to prove is $PA,PH$ are isogonal in $\angle BPC$ ,let $A'$ be the reflection of $A$ in the perpendicular bisector of $BC$ so we need $P,H,A'$ are collinear. Now invert at $A$ with power $\sqrt{AF.AB}$ , then the problem becomes as $\textbf{LEMMA:2}$ ,so we are done $\blacksquare$

Here is my complex bash from testsolving: Let $ABC$ be the complex unit circle. Then $D=\frac{1}{2}(a+b+c-\frac{bc}{a})$, and we know \[\frac{p-a}{p-o}\cdot\frac{d-o}{d-a}\in\mathbb{R}\]\[ \frac{p-a}{p}\cdot\frac{a+b+c-\frac{bc}{a}}{b+c-a-\frac{bc}{a}}=\frac{\frac{1}{p}-\frac{1}{a}}{\frac{1}{p}}\cdot\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a}{bc}}{-\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a}{bc}}\]\[\frac{1}{p}\cdot\frac{a+b+c-\frac{bc}{a}}{b+c-a-\frac{bc}{a}}=\frac{-1}{a}\cdot\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a}{bc}}{\frac{1}{b}+\frac{1}{c}-\frac{1}{a}-\frac{a}{bc}}\]\[\frac{-a}{p}\cdot\frac{a^2+ab+ac-bc}{ab+ac-a^2-bc}=\frac{bc+ab+ab-a^2}{ab+ac-bc-a^2}\]\[p=a\cdot\frac{a^2+ab+ac-bc}{a^2-ab-ac-bc}\]Now note that $p+x=e+px\overline{e}$, so $x=\frac{p-e}{p\overline{e}-1}$ But we compute that \[p-e=a\cdot\frac{a^2+ab+ac-bc}{a^2-ab-ac-bc}-\frac{1}{2}(a+b+c-\frac{ac}{b})\]\[=\frac{a^3b+a^3+2a^2b^2+a^2bc+ab^3+ab^2c+b^3c+b^2c^2-a^2c^2}{2b(a^2-ab-ac-bc)}\]\[=\frac{(a+b)(b+c)(a^2+ab-ac+bc)}{2b(a^2-ab-ac-bc)}\]And also compute \[p\overline{e}-1=a\cdot\frac{a^2+ab+ac-bc}{a^2-ab-ac-bc}\cdot\frac{1}{2}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{b}{ac})-1\]\[=\frac{a^3b+a^3c+a^2bc+a^2c^2+ab^2c+2abc^2+b^3c+b^2c^2-ab^3}{2bc(a^2-ab-ac-bc)}\]\[=\frac{(a+b)(b+c)(a^2+ac+bc-ab)}{2bc(a^2-ab-ac-bc)}\]So \[x=\frac{\frac{(a+b)(b+c)(a^2+ab-ac+bc)}{2b(a^2-ab-ac-bc)}}{\frac{(a+b)(b+c)(a^2+ac+bc-ab)}{2bc(a^2-ab-ac-bc)}}=c\cdot\frac{a^2+ab+bc-ac}{a^2+ac+bc-ab}\]By symmetry, \[y=b\cdot\frac{a^2+ac+bc-ab}{a^2+ab+bc-ac}\]Now note that $xy=bc$ to finish.

Here is a solution by me and Stanley Wang that generalizes to arbitrary isogonal conjugates. Let $H$ be the orthocenter of $\triangle ABC$ and $A'$ be the point on $\odot(ABC)$ with $\overline{AA'} \parallel {BC}$; we claim that $\overline{PHA'}$ are collinear. Let $O'$ be the antipode of $A$ in $\odot(ABC)$; we want to show that $$(AH;H'\infty) = A'(AP;H'O') = A(ZP;DO)$$where $\{A,Z\} \equiv \overline{AA} \cap \odot(ADO)$. Performing a $\sqrt{bc}$ inversion at $A$ and denoting inverses with a $*$ yields $$(ZP;DO) = (Z^*P^*;O'O^*) = \infty_{BC}(AD;H'O^*) $$so the conclusion is equivalent to showing that \begin{align*} \frac{AH'}{HH'} = \frac{AH' \cdot DO^*}{DH' \cdot AO^*} &\iff \frac{HH'}{DH'} = \frac{AO^*}{DO^*}\\ &\iff \frac{DH}{DH'} = \frac{DA}{DO^*}\\ &\iff DH\cdot DO^* = DH'\cdot DA \end{align*}which follows by Power of a Point (note that the conclusion already follows at the first if and only if since both ratios are equal to $2$, but the computation above is given for the generalization as well). Finally, by Dual of Desargues' Involution theorem on complete quadrilateral $BFEC$, $\{\overline{PE}, \overline{PF}\}, \{\overline{PH}, \overline{PA}\}, \{\overline{PB}, \overline{PC}\}$ form pairs of an involution, so projecting through $P$ onto $\odot(ABC)$ yields the desired.

For convenience, let $\omega$ be $(ABC)$. Gen Gen G5 wrote: In $\triangle ABC$, $P$ and $Q$ are isogonal conjugates, and $\triangle DEF$ is the cevian triangle of $P$. Let $(ADQ)$ intersect $\omega$ again at $S$, and suppose that $SE$ and $SF$ intersect $\omega$ again at $X$ and $Y$. Let $Q'$ be the second intersection of $AQ$ with $\omega$, and let $Z\neq Q'$ be the point on $\omega$ for which $AP \parallel Q'Z$. Then $AZ$, $XY$ and $BC$ concur. Taking $P$ as the orthocenter and $Q$ as the circumcenter gives us the original problem. Let $P'$ be the second intersection of $AP$ and $\omega$. [asy][asy] import olympiad; import cse5; pen skyBlue = rgb(0,0.6,1), neonMagenta = rgb(1,0,1), deepBlue = rgb(0,0.1,0.6), fillMagenta = rgb(0.995, 0.9, 0.96); pointfontsize = 10; defaultpen(fontsize(10pt)); pointpen = black; pathpen = black; size(11cm); pair A = dir(130); pair B = dir(215); pair C = dir(325); real a = abs(B-C); real b = abs(C-A); real c = abs(A-B); real s = (a+b+c)/2; pair bary(real x, real y, real z) { real k = x+y+z; x /= k; y /= k; z /= k; return x*A + y*B + z*C; } path omega = circumcircle(A,B,C); real u = 0.42; real v = 0.39; real w = 1-u-v; pair P = bary(u, v, w), Q = bary(a*a/u, b*b/v, c*c/w), Pp = IP(Line(A,P,10,10),omega,1), Qp = IP(Line(A,Q,10,10),omega,1); pair D = bary(0, v, w), E = bary(u, 0, w), F = bary(u, v, 0); path gamma = circumcircle(A,Q,D); pair S = IP(omega, gamma,1), X = IP(omega, Line(S,E,10,10), 0), Y = IP(omega, Line(S,F,10,10),0); pair K = extension(X, Y, B, C), Z = IP(Line(A, K, 10, 10), omega, 0); filldraw(S--Pp--Qp--cycle, fillMagenta); filldraw(S--D--Q--cycle, fillMagenta); filldraw(S--P--A--cycle, fillMagenta); D(omega,blue); D(gamma,gray+linewidth(0.3)); DPA(A--Pp^^A--Qp, pink+linewidth(0.4)); D(S--Z, dashed+skyBlue); D(A--B--C--cycle); DPA(K--B^^K--Z^^K--X,dotted+skyBlue); D(S--Pp--Qp--cycle,neonMagenta+linewidth(0.8)); D(S--D--Q--cycle,neonMagenta+linewidth(0.8)); D(S--P--A--cycle,neonMagenta+linewidth(0.8)); D("A", A, dir(100)); D("B", B, dir(B)); D("C", C, dir(C)); D("P",P,dir(230)); D("Q",Q,dir(30)); D("P'",Pp,dir(Pp)); D("Q'",Qp,dir(Qp)); D("D",D,dir(305)); D("E",E,dir(60)); D("F",F,dir(F)); D("S",S,dir(260)); D("X",X,dir(X)); D("Y",Y,dir(320)); D(" ",K,dir(K)); D("Z",Z,dir(Z)); [/asy][/asy] It is clear that $S$ is the spiral center taking $DQ \to P'Q'$, so it is also the spiral center taking $DP' \to QQ'$. Now, consider the following lemma: Lemma. $\frac{DP}{DP'} = \frac{QA}{QQ'}$. Proof. Reflection is a projectivity, so reflection about the angle bisector of $\angle ABC$ preserves cross-ratio. Hence, if $\infty_P$, $\infty_Q$ are the infinity points on $AP$, $AQ$, then \[ \frac{DP}{DP'} = (\overline{BD}, \overline{B\infty_P}; \overline{BP}, \overline{BP'}) = (\overline{BA}, \overline{BQ'}; \overline{BQ}, \overline{B\infty_Q}) = \frac{QA}{QQ'}, \]as desired. $\blacksquare$ Returning to our original problem, it is now clear that the spiral similarity at $S$ sending $DP' \to QQ'$ also sends $P \to A$. Hence we have \[\measuredangle ZSA = \measuredangle ZQ'A = \measuredangle P'AQ' = \measuredangle P'SQ' = \measuredangle PSA,\]where the second equality follows by definition and the last equality follows from spiral similarity. Hence $Z$ lies on $SP$. Finally, we finish by DDIT on $S$ with respect to $BECF$, whence \[ (SB, SC), (SE, SF), (SA, SP) \]are pairs of the same involution. Projecting onto $\omega$, we conclude that $XY, BC$ and $AZ$ concur as desired.

anantmudgal09 wrote: a1267ab wrote: Let scalene triangle $ABC$ have altitudes $AD, BE, CF$ and circumcenter $O$. The circumcircles of $\triangle ABC$ and $\triangle ADO$ meet at $P \ne A$. The circumcircle of $\triangle ABC$ meets lines $PE$ at $X \ne P$ and $PF$ at $Y \ne P$. Prove that $XY \parallel BC$. Proposed by Daniel Hu Let $ABCA'$ be an isosceles trapezoid with $\overline{AA'} \parallel \overline{BC}$. Claim. $P,H,A'$ are collinear. (Proof) Reflect $O$ in the $A$-midline of $\triangle ABC$ to obtain point $G$. Then $AH=2OM$ where $M$ is the midpoint of $\overline{BC}$; yields $H,G,A'$ are collinear. Redefine $P$ as $P \overset{\text{def}}{:=} \overline{GH} \cap \odot(ABC)$ with $P \ne A'$. Then $\measuredangle APG=\measuredangle OAD=\measuredangle ADG$ hence $ADGP$ are concyclic; proving the claim. $\blacksquare$ Now observe $\overline{PA}, \overline{PH}$ are isogonal in angle $BPC$. Apply isogonality lemma to obtain $\overline{PE}, \overline{PF}$ are isogonal in angle $BPC$ and we're done! Is it $D$ , $G$ , $A’$ collinear ?

Let $H$ be the orthocenter of $\triangle{ABC}$ and $H'$ be it's reflection in $BC$. Let $OD$ intersect $\odot(OBC)$ at $K$. Clearly $A, P$ and $K$ are colinear. Let $A'$ be the second intersection of $PH$ with $\odot(ABC)$. $$\measuredangle{PKH'}+2\measuredangle{H'PK}=\angle{AOH'}-2\angle{APH'}=0 \implies KP=KH' \implies DP=DH'\implies \angle{HPH'}=90^{\circ}\implies AA'||BC$$It is enough to show that $PE$ and $PF$ are isogonal wrt $\angle{APH}$, but this follows from isogonality lemma applied on $PB$ and $PC$ wrt $\angle{APH}$.

Very nice problem.Here is my solution: We apply inversion with center $A $ and radius $\sqrt {AF\cdot AB} $.Now , our problem becomes: Inverted version of G5/ELMO SL 2018 wrote: Let scalene triangle $ABC$ have altitudes $AD, BE, CF$ and let the reflection of $A $ wrt $EF $ be $Q$. The lines $QH $ and $EF$ meet at $P$. Prove that internal bisector of $\angle BPC $ is perpendicular to $EF $. Note:The above is not exactly what we get after inverting.I did some angle chasing to see that the above is equivalent to the inverted version. Let $EF\cap BC=K $.I claim that $DP $ is perpendicular to $EF $. Proof: $P (B,C;D,K)= (A,Q;EF\cap AQ,DP\cap AQ) =-1$ ,but $EF $ bisects $AQ $ $\implies$ $DP\parallel AQ $ ,so $DP $ is perpendicular $EF $. Now $(B,C;D,K)$ is harmonic , $DP \perp EF $ so by famous lemma we have $DP $ is bisector of $\angle BPC$ , which is perpendicular to $EF $ ,as desired.

Anantmudgal09, Can you give me a detail proof for "isogonality lemma to obtan PE, PF are isogonal in angle BPC." thanh you very much.

Here's a proof that links this problem to a well-known configuration, for example in this paper from here. Let the reflection of $H$ (the orthocenter of $ABC$) in $D$ be $H'$, and the reflection of $H'$ in $OD$ be $P'$. Firstly, I claim that Claim : $P = P'$. To see this, note that $P'$ is on $(ABC)$, due to the reflection in $OD$. Also, we have by angle chasing, that $\angle DPO = \angle DH'O = \angle AH'O = \angle DAO$, so $P=P'$, as claimed. This means that $DH = DH' = DP$, so $P$ is on the circle with diameter $HH'$. Thus, if $A'$ is the point where the line through $A$ and parallel to $BC$ meets $(ABC)$, then $H, P, A'$ are collinear. So it suffices to show that $PE, PF$ are isogonal with respect to $\angle APH$. Now we take a homothety at $H$ with ratio $\frac 12$, motivated by the desire to make the problem similar to the mentioned well-known configuration. The problem then becomes the following Modified ELMO SL 2018 G5 wrote: Let there be a triangle $ABC$, with incenter $I$. $(AI)$ meets $(ABC)$ at $A, K_A$. The intouch triangle of $ABC$ is $DEF$. $AI$ meets $(ABC)$ again at $M_A$. Then prove that the lines joining $K_A$ to the circumcenters of $IDB, IDC$ are isogonal with respect to the lines $K_ADM_A, K_AI$. Note that we used the fact that $K_A, D, M_A$ are collinear, which is proved in the linked paper as Lemma 3.5. But this result is fairly obvious, since the paper gives us the fact that the required circumcenters are inverses in the circumcircle of $K_AID$ because of a homothety at $I$ with ratio $\frac 12$ and, in the notations of the paper, the fact that $IY^2 = YB \cdot YC$. This finishes the proof of the problem.

adamov1 wrote: Here is my complex bash from testsolving: Let $ABC$ be the complex unit circle. Then $D=\frac{1}{2}(a+b+c-\frac{bc}{a})$, and we know \[\frac{p-a}{p-o}\cdot\frac{d-o}{d-a}\in\mathbb{R}\]\[ \frac{p-a}{p}\cdot\frac{a+b+c-\frac{bc}{a}}{b+c-a-\frac{bc}{a}}=\frac{\frac{1}{p}-\frac{1}{a}}{\frac{1}{p}}\cdot\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a}{bc}}{-\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a}{bc}}\]\[\frac{1}{p}\cdot\frac{a+b+c-\frac{bc}{a}}{b+c-a-\frac{bc}{a}}=\frac{-1}{a}\cdot\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a}{bc}}{\frac{1}{b}+\frac{1}{c}-\frac{1}{a}-\frac{a}{bc}}\]\[\frac{-a}{p}\cdot\frac{a^2+ab+ac-bc}{ab+ac-a^2-bc}=\frac{bc+ab+ab-a^2}{ab+ac-bc-a^2}\]\[p=a\cdot\frac{a^2+ab+ac-bc}{a^2-ab-ac-bc}\] Here's a slightly faster way to find the coordinates of $P$. Let $AD$ and $AO$ mean $(ABC)$ at $K$ and $L$. Then, $k = -bc/a$ and $l = -a$. Since $P$ is the center of the spiral similarity taking $DO$ onto $KL$, we have $$p = \frac{dl-ok}{d+l-o-k} = \frac{dl}{d+l-k}.$$Applying $d=\frac{1}{2}\left(a+b+c - \frac{bc}{a}\right)$, gives $$p = \frac{\frac{1}{2}\left(a+b+c - \frac{bc}{a}\right)(-a)}{\frac{1}{2}\left(a+b+c - \frac{bc}{a}\right) -a + \frac{bc}{a}} = \frac{-a(a+b+c)+bc}{b+c-a + \frac{bc}{a}} = \frac{-a^2(a+b+c) + abc}{ab+bc+ca - a^2}$$

anantmudgal09 wrote: a1267ab wrote: Let scalene triangle $ABC$ have altitudes $AD, BE, CF$ and circumcenter $O$. The circumcircles of $\triangle ABC$ and $\triangle ADO$ meet at $P \ne A$. The circumcircle of $\triangle ABC$ meets lines $PE$ at $X \ne P$ and $PF$ at $Y \ne P$. Prove that $XY \parallel BC$. Proposed by Daniel Hu Let $ABCA'$ be an isosceles trapezoid with $\overline{AA'} \parallel \overline{BC}$. Claim. $P,H,A'$ are collinear. (Proof) Reflect $O$ in the $A$-midline of $\triangle ABC$ to obtain point $G$. Then $AH=2OM$ where $M$ is the midpoint of $\overline{BC}$; yields $H,G,A'$ are collinear. Redefine $P$ as $P \overset{\text{def}}{:=} \overline{GH} \cap \odot(ABC)$ with $P \ne A'$. Then $\measuredangle APG=\measuredangle OAD=\measuredangle ADG$ hence $ADGP$ are concyclic; proving the claim. $\blacksquare$ Now observe $\overline{PA}, \overline{PH}$ are isogonal in angle $BPC$. Apply isogonality lemma to obtain $\overline{PE}, \overline{PF}$ are isogonal in angle $BPC$ and we're done! enhanced wrote: $\textbf{LEMMA:1}$ In a $\Delta ABC$ let $I,I_A$ be the incenter and $A-$excenter and let $X$ be the reflection of $I$ in $BC$ and let $Y$ be the feet of $A-$ altitude then $I_A,X,Y$ are collinear. $\textbf{PROOF:}$ let $T$ be the projection of $I$ in $BC$ ,it is well known that $I_AT$ bisects $AY$ now since $IX\parallel AY$ and $IT=TX$ we have $I_A,X,Y$ are collinear $\blacksquare$ $\textbf{LEMMA:2}$ In a $\Delta ABC$ let $O$ be the circumcenter and let $AO\cap \odot (BOC)=O_A$ and let $X$ be the projection of $O_A$ in $BC$ and let $A_1,A_2$ be the reflection of $A$ in $BC$ and antipode of $A$ in $\odot (ABC)$ ,then $A_1,A_2,X$ are collinear. $\textbf{PROOF:}$ We can easily see that $A_2,A$ are the incenter and $A-$ excenter of $\Delta O_ABC$ now reflecting in $BC$ and by $\textbf{LEMMA:1}$ we are done $\blacksquare$. Now coming back to the problem by isogonality lemma all we need to prove is $PA,PH$ are isogonal in $\angle BPC$ ,let $A'$ be the reflection of $A$ in the perpendicular bisector of $BC$ so we need $P,H,A'$ are collinear. Now invert at $A$ with power $\sqrt{AF.AB}$ , then the problem becomes as $\textbf{LEMMA:2}$ ,so we are done $\bl[quote=enhanced]$\textbf{LEMMA:1}$ In a $\Delta ABC$ let $I,I_A$ be the incenter and $A-$excenter and let $X$ be the reflection of $I$ in $BC$ and let $Y$ be the feet of $A-$ altitude then $I_A,X,Y$ are collinear. $\textbf{PROOF:}$ let $T$ be the projection of $I$ in $BC$ ,it is well known that $I_AT$ bisects $AY$ now since $IX\parallel AY$ and $IT=TX$ we have $I_A,X,Y$ are collinear $\blacksquare$ $\textbf{LEMMA:2}$ In a $\Delta ABC$ let $O$ be the circumcenter and let $AO\cap \odot (BOC)=O_A$ and let $X$ be the projection of $O_A$ in $BC$ and let $A_1,A_2$ be the reflection of $A$ in $BC$ and antipode of $A$ in $\odot (ABC)$ ,then $A_1,A_2,X$ are collinear. $\textbf{PROOF:}$ We can easily see that $A_2,A$ are the incenter and $A-$ excenter of $\Delta O_ABC$ now reflecting in $BC$ and by $\textbf{LEMMA:1}$ we are done $\blacksquare$. Now coming back to the problem by isogonality lemma all we need to prove is $PA,PH$ are isogonal in $\angle BPC$ ,let $A'$ be the reflection of $A$ in the perpendicular bisector of $BC$ so we need $P,H,A'$ are collinear. Now invert at $A$ with power $\sqrt{AF.AB}$ , then the problem becomes as $\textbf{LEMMA:2}$ ,so we are done $\blacksquare$ What is the isogonality lemma? Help me please

barbie_girl_pink_is_life wrote: What is the isogonality lemma? Help me please See here

It is also equivalent to the Dual Desargue involution theorem. http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZS9jLzYzYWY3MmU0MmJjNjhiZmYwMDVhMjEzOWQzYmZjMGVmODVlOTZjLnBkZg==&rn=ZGVzYXJndWVzLWludm9sdXRpb24tdGhlb3JlbS5wZGY=

Extend $AH$ to meet $(ABC)$ again at $H'$, and extend $OD$ to meet $(BOC)$ again at $D'$. By inversion about $(ABC)$, we have $A, P, D'$ collinear and $A, O, H', D'$ concyclic. Since $OA = OH'$, we have $\angle OD'H' = \angle OD'P$. In particular, $\triangle OPD' \cong \triangle OH'D'$ so it follows that $D'P = D'H'$ and thus $DP=DH'=DH$. Now we have $\triangle PDH, \triangle POA$ both isosceles and furthermore $\angle PDH = \angle POA$, so $P$ is the spiral center taking $DH \mapsto OA$. This implies $\angle APH = \angle OPD = \angle OAH'$. Extend $AO$ and $PH$ to meet $(ABC)$ again at $A'$ and $P'$; then $O, H$ isogonal implies $A'H' \parallel BC$, and since arcs $A'H'$ and $AP'$ have equal measure we get $A'H' \parallel AP'$. But then this implies $A, H$ isogonal wrt $\angle BPC$. Finally, apply the isogonal line lemma on $\{PA, PH\}$ and $\{PB, PC\}$ wrt $\angle BPC$ to get $E, F$ isogonal, so then $XY \parallel BC$, as desired. [asy][asy]import olympiad; size(12cm); pair A = dir(110); pair B = dir(215); pair C = dir(325); draw(B--A--C--B); pair O = circumcenter(A, B, C); draw(circumcircle(A, B, C)); draw(circumcircle(B, O, C)); pair H = orthocenter(A, B, C); pair D = foot(A, B, C); pair E = foot(B, A, C); pair F = foot(C, A, B); draw(B--E); draw(C--F); pair P = intersectionpoints(circumcircle(A, O, D), circumcircle(B,A,C))[1]; draw(circumcircle(A, O, D)); pair d = extension(A, P, O, D); draw(A--d); draw(O--d); pair h = 2*D - H; path blah = arc(circumcenter(A, O, h), circumradius(A, O, h), -62, 40); draw(blah); pair x = intersectionpoints(circumcircle(O, B, C), circumcircle(A, O, h))[1]; draw(A--h); draw(P--H, dashed); draw(P--D, dashed); draw(P--O, dashed); draw(A--O, dashed); string[] names = {"$A$", "$B$", "$C$", "$D$", "$E$", "$F$", "$H$", "$P$", "$H'$", "$O$", "$D'$"}; pair[] pts = {A, B, C, D, E, F, H, P, h, O, x}; pair[] labels = {A, B, C, dir(315), E, F, dir(135), P, dir(270), dir(330), x}; for(int i=0; i<names.length; ++i){ dot(names[i], pts[i], dir(labels[i])); } [/asy][/asy]

[asy][asy] unitsize(1.5inches); pair A=dir(100); pair B=dir(220); pair C=dir(-40); pair D=foot(A,B,C); pair E=foot(B,A,C); pair F=foot(C,A,B); pair DD=2*D-A; pair AA=-A; pair PP=extension(AA,DD,B,C); pair EE=extension(A,B,AA,C); pair FF=extension(A,C,B,AA); draw(circumcircle(A,B,C)); draw(A--B--C--cycle); draw(B--EE); draw(C--FF); draw(C--EE); draw(B--FF); draw(A--DD); draw(EE--PP); draw(FF--PP); draw(DD--PP); dot("$A$",A,dir(A)); dot("$B$",B,dir(200)); dot("$C$",C,dir(-10)); dot("$D$",D,dir(135)); dot("$E'$",EE,dir(EE)); dot("$F'$",FF,dir(FF)); dot("$P'$",PP,dir(90)); dot("$D'$",DD,dir(DD)); dot("$A'$",AA,1.5*dir(-70)); [/asy][/asy] We will solve the problem with a $\sqrt{bc}$ inversion. Let $\psi$ be $\sqrt{bc}$ inversion (the unique involutive Mobius transformation viewing the plane as $\mathbb{CP}^1$ swapping $(A,\infty)$ and $(B,C)$). Note that \[\psi(P)=\psi((ABC))\cap\psi((ADO))=BC\cap A'D'\]since $\psi(O)=D'$, the reflection of $A$ in $D$, and $\psi(D)=A'$, the antipode of $A$. Furthermore, $(ABE)\perp AB$, so $C\psi(E)\perp AC$, so $\psi(E)=E'$ is the unique point on $AB$ such that $E'C\perp AC$, and similarly we can find $F'=\psi(F)$. Finally, \[X'=\psi(X)=\psi(PE)\cap\psi((ABC))=(AP'E')\cap BC\]and $Y'=\psi(Y)=(BP'F')\cap BC$. Showing that $XY\parallel BC$ is equivalent to showing $AX$ and $AY$ are isogonal, or equivalently, $AX'$ and $AY'$ are isogonal. Note that \[\angle X'AB=\angle X'AE'=\angle X'P'E'=\angle BP'E'\]and $\angle Y'AC=\angle CP'F'$. Thus, it suffices to show that $\angle BP'E'=\angle CP'F'$. By DDIT with $BCE'F'$, there exists a projective involution on the pencil of lines through $P$ swapping $(P'B,P'C)$, $(P'E',P'F')$, and $(P'A',P'A)$. Note that reflection in line $BC$ certainly swaps $(P'B,P'C)$ and $(P'D',P'A)$, so we must have that $P'E'$ and $P'F'$ are reflections in $BC$. Thus, $\angle BP'E'=\angle CP'F'$, as desired. $\blacksquare$

Really Nice Configuration, also it took me a lot of time to solve. ELMOSL 2018 G5 wrote: Let scalene triangle $ABC$ have altitudes $AD, BE, CF$ and circumcenter $O$. The circumcircles of $\triangle ABC$ and $\triangle ADO$ meet at $P \ne A$. The circumcircle of $\triangle ABC$ meets lines $PE$ at $X \ne P$ and $PF$ at $Y \ne P$. Prove that $XY \parallel BC$. Proposed by Daniel Hu [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -13.13, xmax = 9.61, ymin = -3.65, ymax = 9.65; /* image dimensions */ pen qqqqcc = rgb(0,0,0.8); /* draw figures */ draw((-2.39,5.63)--(-3.75,-0.85), linewidth(1.2) + qqqqcc); draw((2.43,-0.97)--(-3.75,-0.85), linewidth(1.2) + qqqqcc); draw((-2.39,5.63)--(2.43,-0.97), linewidth(1.2) + qqqqcc); draw((-2.39,5.63)--(-2.45,-0.87), linewidth(0.4)); draw((2.43,-0.97)--(-3.4791912408759122,0.4403240875912413), linewidth(0.4)); draw((-3.75,-0.85)--(0.14679197034393177,2.1563844389481437), linewidth(0.4)); draw(circle((-0.6059639489077233,1.8728566312522381), 4.159195946925585), linewidth(0.4)); draw(circle((-4.348688565703188,2.397803279068029), 3.7793592764144286), linewidth(0.4)); draw((-3.752828183131605,4.59244417650331)--(2.595608977427613,4.527822115449499), linewidth(1.2)); draw((-3.3547667310674094,-1.2485203388866422)--(-3.752828183131605,4.59244417650331), linewidth(0.4)); draw((2.595608977427613,4.527822115449499)--(-3.3547667310674094,-1.2485203388866422), linewidth(0.4)); /* dots and labels */ dot((-2.39,5.63),dotstyle); label("$A$", (-2.53,5.91), NE * labelscalefactor); dot((-3.75,-0.85),dotstyle); label("$B$", (-4.15,-1.05), NE * labelscalefactor); dot((2.43,-0.97),dotstyle); label("$C$", (2.63,-1.13), NE * labelscalefactor); dot((-2.43,0.19),dotstyle); label("$H$", (-2.35,0.39), NE * labelscalefactor); dot((-2.45,-0.87),dotstyle); label("$D$", (-2.47,-1.31), NE * labelscalefactor); dot((-3.4791912408759122,0.4403240875912413),dotstyle); label("$F$", (-3.85,0.53), NE * labelscalefactor); dot((0.14679197034393177,2.1563844389481437),dotstyle); label("$E$", (0.39,2.01), NE * labelscalefactor); dot((-0.6059639489077233,1.8728566312522381),dotstyle); label("$O$", (-0.95,1.95), NE * labelscalefactor); dot((-3.752828183131605,4.59244417650331),dotstyle); label("$Y$", (-4.09,4.63), NE * labelscalefactor); dot((2.595608977427613,4.527822115449499),dotstyle); label("$X$", (2.67,4.73), NE * labelscalefactor); dot((-3.3547667310674094,-1.2485203388866422),dotstyle); label("$P$", (-3.49,-1.71), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let $H$ be the orthocenter of $\triangle ABC$. Now perform a $\sqrt{AH\cdot AD}$ Inversion around $A$. The problem becomes equivalent as follows. Inverted Problem wrote: $AF'E'$ is a triangle with $AH',EB',FC'$ are the altitudes of this triangle and $D'$ be the orthocenter of the triangle $AF'E'$. Let $O'$ be the reflection of $A$ over $B'C'$ and $O'D'\cap B'C'=P'$. Let $B'C'\cap \odot(AP'E')=X'\neq P'$ and let $B'C'\cap\odot(AP'F')=Y'\neq P'$. Then $\odot(AB'C')$ and $\odot(AX'Y')$ are tangent to each other. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(25cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -13.13, xmax = 9.61, ymin = -3.65, ymax = 9.65; /* image dimensions */ pen qqqqcc = rgb(0,0,0.8); pen ttffqq = rgb(0.2,1,0); /* draw figures */ draw((-2.39,5.63)--(-3.75,-0.85), linewidth(1.2) + qqqqcc); draw((2.43,-0.97)--(-3.75,-0.85), linewidth(1.2) + qqqqcc); draw((-2.39,5.63)--(2.43,-0.97), linewidth(1.2) + qqqqcc); draw((-2.39,5.63)--(-2.45,-0.87), linewidth(0.4)); draw((2.43,-0.97)--(-3.4791912408759122,0.4403240875912413), linewidth(0.4)); draw((-3.75,-0.85)--(0.14679197034393177,2.1563844389481437), linewidth(0.4)); draw((-3.4791912408759122,0.4403240875912413)--(0.14679197034393177,2.1563844389481437), linewidth(0.4)); draw((1.2246701122129298,-2.0076877600013514)--(-3.111079153822597,0.614539609464802), linewidth(0.4)); draw((0.14679197034393177,2.1563844389481437)--(-6.43,-0.93), linewidth(0.4)); draw((-6.43,-0.93)--(-3.75,-0.85), linewidth(0.4)); draw(circle((0.470700567309705,2.6591479900655726), 4.124265922650659), linewidth(0.4) + red); draw(circle((-14.842118976010424,4.860691636940459), 12.475860785941475), linewidth(0.4) + ttffqq); draw((0.14679197034393177,2.1563844389481437)--(4.3506882680007735,4.05745584001262), linewidth(0.4)); draw((-6.43,-0.93)--(-11.45,-3.23), linewidth(0.4)); draw((-3.111079153822597,0.614539609464802)--(-3.75,-0.85), linewidth(0.4)); draw((-3.111079153822597,0.614539609464802)--(2.43,-0.97), linewidth(0.4)); draw((-3.111079153822597,0.614539609464802)--(-2.45,-0.87), linewidth(0.4)); draw((1.2246701122129298,-2.0076877600013514)--(-2.39,5.63), linewidth(0.4)); /* dots and labels */ dot((-2.39,5.63),dotstyle); label("$A$", (-2.75,5.85), NE * labelscalefactor); dot((-3.75,-0.85),dotstyle); label("$F'$", (-4.23,-1.21), NE * labelscalefactor); dot((2.43,-0.97),dotstyle); label("$E'$", (2.57,-1.33), NE * labelscalefactor); dot((-2.43,0.19),dotstyle); label("$D'$", (-2.35,0.39), NE * labelscalefactor); dot((-2.45,-0.87),dotstyle); label("$H'$", (-2.47,-1.31), NE * labelscalefactor); dot((-3.4791912408759122,0.4403240875912413),dotstyle); label("$B'$", (-3.85,0.53), NE * labelscalefactor); dot((0.14679197034393177,2.1563844389481437),dotstyle); label("$C'$", (0.23,2.35), NE * labelscalefactor); dot((1.2246701122129298,-2.0076877600013514),dotstyle); label("$O'$", (1.39,-2.27), NE * labelscalefactor); dot((-3.111079153822597,0.614539609464802),dotstyle); label("$P'$", (-2.93,0.85), NE * labelscalefactor); dot((-6.43,-0.93),dotstyle); label("$X_A$", (-7.11,-0.79), NE * labelscalefactor); label("$d$", (-21.07,15.05), NE * labelscalefactor,ttffqq); dot((4.3506882680007735,4.05745584001262),dotstyle); label("$X'$", (4.43,4.25), NE * labelscalefactor); dot((-11.45,-3.23),dotstyle); label("$Y'$", (-11.37,-3.03), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Note that to prove the tangency we just need to prove that acute $\angle Y'AF'=\angle E'AX'\implies\angle X_AP'F'=\angle C'P'E'$. Where $B'C'\cap E'F'=X_A$. So we can easily eliminate the circles $\odot(AP'E')$ and $\odot(AP'F')$ from the Inverted Diagram. Claim:- $H'P'\perp B'C'$. Now restate this inverted Problem in terms of the Orthic triangle of $\triangle AF'E'$. Restated Inverted Problem wrote: Let $H'B'C'$ be a triangle and $A$ be it's $H'-\text{Excenter}$ of $\triangle H'B'C'$ and let $O'$ be the reflection of $A$ over $B'C'$ and let $O'D'\cap B'C'=P'$ where $D'$ is the incenter of $\triangle H'B'C'$. Then $H'P'\perp B'C'$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.63, xmax = 12.62333333333334, ymin = -1.4966666666666704, ymax = 9.33; /* image dimensions */ /* draw figures */ draw(circle((-2.75,1.89), 3.041870769407829), linewidth(0.4)); draw((-4.93,4.99)--(-1.09,4.91), linewidth(0.4)); draw((-5.637506737408078,2.8467040395026757)--(-3.59,8.83), linewidth(0.4)); draw((-3.59,8.83)--(-0.05347709113927701,3.2977435774199364), linewidth(0.4)); draw((-3.59,8.83)--(-2.75,1.89), linewidth(0.4)); draw((-3.59,8.83)--(-3.5510151843817788,4.96127114967462), linewidth(0.4)); draw((-3.5510151843817788,4.96127114967462)--(-2.536666666666658,7.97), linewidth(0.4)); /* dots and labels */ dot((-2.75,1.89),dotstyle); label("$A$", (-2.6966666666666628,2.0233333333333348), NE * labelscalefactor); dot((-2.643333333333329,4.93),dotstyle); label("$D$", (-2.59,5.06333333333334), NE * labelscalefactor); dot((-4.93,4.99),dotstyle); label("$B'$", (-4.883333333333329,5.1166666666666725), NE * labelscalefactor); dot((-1.09,4.91),dotstyle); label("$C'$", (-1.043333333333329,5.036666666666672), NE * labelscalefactor); dot((-5.637506737408078,2.8467040395026757),dotstyle); dot((-3.59,8.83),dotstyle); label("$H'$", (-3.5366666666666626,8.956666666666678), NE * labelscalefactor); dot((-0.05347709113927701,3.2977435774199364),dotstyle); dot((-2.536666666666658,7.97),dotstyle); label("$O'$", (-2.4833333333333294,8.103333333333344), NE * labelscalefactor); dot((-3.5510151843817788,4.96127114967462),dotstyle); label("$P'$", (-3.4966666666666626,5.09), NE * labelscalefactor); dot((-3.2351426510922887,5.8982023792624805),dotstyle); label("$D'$", (-3.1233333333333295,5.836666666666674), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let $H'A\cap B'C'=T$. Then $$-1=(H'T;D'A)\overset{P'}{=}(H'P'\cap O'A,D;O',A)\implies O'A\|H'P'\implies H'P'\perp B'C'$$ Now coming to the main Problem we get that $H'P'\perp B'C'$ and as $(X_AH';F'E')=-1$ we get that $\angle F'P'H'=\angle H'P'E'\implies\angle X_AP'F'=\angle E'B'C'$ which we were supposed to prove. Hence $\odot(AB'C')$ and $\odot(AX'Y')$ are tangent to each other at $A'$. Now Inverting back we get that $XY\|BC$. $\blacksquare$

Huh, I guess I never got around to solving this problem Let $H$ be the orthocenter; we begin by inverting about $A$ swapping $H$ and $D$. We claim that $O$ maps to $O'$, the reflection of $A$ in $\overline{EF}$. This is true because if we let $A'$ be the antipode of $A$, we have 1) $(ABA'C)$ swaps with $EF$, and 2) $O$ is the midpoint of $AA'$. Then $P'$ is the intersection of $O’H$ and $EF$; we now claim that $P'$ is the foot from $D$ onto $EF$. This is true because $\angle{AP'E} = \angle{O’P'E}$, so it follows by the right-angle angle-bisector condition for harmonic divisions. Next, we note that $X$ maps to the intersection of $(AP'C)$ with $EF$ -- call it $X'$ -- and that $Y$ maps to $Y'$, the intersection of $(AP'B)$ and $EF$. It suffices to show that $\angle{Y'AB} = \angle{X'AC}$. But we have $\angle{Y'AB} = \angle{FP'B}$ and $\angle{X'AC} = \angle{EP'C}$, so we just need to show that $P'D$ bisects $\angle{BPC}$. But this is also true by the right-angle angle-bisector condition, and we are done.

Define $Z \equiv AH \cap \odot (ABC) $ . Let $A'$ denote the antipode of $A$ in $\odot (ABC)$ . Let $T$ denote the point on $\odot (ABC)$ such that $AT \parallel BC$ . Also let $H$ denote the orthocenter. We claim that $P,H,T$ are collinear . The proof is by complex numbers with $\odot (ABC)$ as the unit circle . First note that $h=a+b+c$ , $t= \frac {bc}{a}$ , $z= -\frac {bc}{a}$ and $d= \left(\frac {h+z}{2} \right)$ . Note that $P$ is the center of spiral similarity mapping $\overline{DO} \mapsto \overline{ZA'}$ . So we can compute $p$ as follows $$ p = \frac {da'-oz}{d+a'-o-z} = \frac {-ad}{d-a+ \frac {bc}{a}}$$$$ \implies p = \frac{\frac{1}{2}\left(a+b+c - \frac{bc}{a}\right)(-a)}{\frac{1}{2}\left(a+b+c - \frac{bc}{a}\right) -a + \frac{bc}{a}} = \frac{-a(a+b+c)+bc}{b+c-a + \frac{bc}{a}} = \frac{-a^2(a+b+c) + abc}{ab+bc+ca - a^2} = \frac {abc-a^2h}{ab+bc+ca-a^2}$$ Now we compute $$h-p = h- \frac {abc-a^2h}{ab+bc+ca-a^2} = \frac {(a+b+c)(ab+bc+ca)-abc}{ab+bc+ca-a^2} = \frac {(a+b)(b+c)(c+a)}{ab+bc+ca-a^2}$$ Also $$h-t = a+b+c- \frac {bc}{a} = \frac{a^2+ab+ac-bc}{a}$$. We now compute $$ \left( \frac {h-p}{h-t} \right) = \frac {a(a+b)(b+c)(c+a)}{(ab+bc+ca-a^2)(a^2+ab+ac-bc)}$$. We have $$ \overline { \left( \frac {h-p}{h-t} \right)} = \frac {\frac 1a (\frac 1a +\frac 1b)(\frac 1b + \frac 1c)(\frac 1c +\frac 1a )}{( \frac {1}{ab} +\frac {1}{bc} +\frac {1}{ca} - \frac {1}{a^2})( \frac {1}{a^2} + \frac {1}{ab} +\frac {1}{ac} - \frac {1}{bc})} $$ $$\implies \overline { \left( \frac {h-p}{h-t} \right)} = \frac {(a+b)(b+c)(c+a)}{a^3b^2c^2 \left(\frac {ac+a^2+ab-bc}{a^2bc}\right) \left(\frac {bc+ac+ab-a^2}{a^2bc}\right) } = \frac {h-p}{h-t}$$ So $P,H,T$ are collinear as desired. So , we get that $\overline {PA}$ and $\overline {PH}$ are isogonal in $\angle {BPC}$ . Next note that by DDIT on $BCEF$ ,with $P$ , we have that $\{\overline {PB},\overline {PC}\}$ , $\{\overline {PA},\overline {PH}\}$ , $\{\overline {PE},\overline {PF}\}$ are swapped under some involution . Considering the first two pairs , we know that this involution is just isogonal conjugation in $\angle {BPC}$ . Hence $\overline {PE},\overline {PF}$ are isogonal in $\angle {BPC}$ To finish , note that $$ \measuredangle {YCB} =\measuredangle {YPB} = \measuredangle {XPC}=\measuredangle {XBC} \implies BY = CX \implies \boxed {XY \parallel BC} $$. We are done . $\blacksquare$

Aryan-23 wrote: Define $Z \equiv AH \cap \odot (ABC) $ . Let $A'$ denote the antipode of $A$ in $\odot (ABC)$ . Let $T$ denote the point on $\odot (ABC)$ such that $AT \parallel BC$ . Also let $H$ denote the orthocenter. We claim that $P,H,T$ are collinear . The proof is by complex numbers with $\odot (ABC)$ as the unit circle . First note that $h=a+b+c$ , $t= \frac {bc}{a}$ , $z= -\frac {bc}{a}$ and $d= \left(\frac {h+z}{2} \right)$ . Note that $P$ is the center of spiral similarity mapping $\overline{DO} \mapsto \overline{ZA'}$ . So we can compute $p$ as follows $$ p = \frac {da'-oz}{d+a'-o-z} = \frac {-ad}{d-a+ \frac {bc}{a}}$$$$ \implies p = \frac{\frac{1}{2}\left(a+b+c - \frac{bc}{a}\right)(-a)}{\frac{1}{2}\left(a+b+c - \frac{bc}{a}\right) -a + \frac{bc}{a}} = \frac{-a(a+b+c)+bc}{b+c-a + \frac{bc}{a}} = \frac{-a^2(a+b+c) + abc}{ab+bc+ca - a^2} = \frac {abc-a^2h}{ab+bc+ca-a^2}$$ Now we compute $$h-p = h- \frac {abc-a^2h}{ab+bc+ca-a^2} = \frac {(a+b+c)(ab+bc+ca)-abc}{ab+bc+ca-a^2} = \frac {(a+b)(b+c)(c+a)}{ab+bc+ca-a^2}$$ Also $$h-t = a+b+c- \frac {bc}{a} = \frac{a^2+ab+ac-bc}{a}$$. We now compute $$ \left( \frac {h-p}{h-t} \right) = \frac {a(a+b)(b+c)(c+a)}{(ab+bc+ca-a^2)(a^2+ab+ac-bc)}$$. We have $$ \overline { \left( \frac {h-p}{h-t} \right)} = \frac {\frac 1a (\frac 1a +\frac 1b)(\frac 1b + \frac 1c)(\frac 1c +\frac 1a )}{( \frac {1}{ab} +\frac {1}{bc} +\frac {1}{ca} - \frac {1}{a^2})( \frac {1}{a^2} + \frac {1}{ab} +\frac {1}{ac} - \frac {1}{bc})} $$ $$\implies \overline { \left( \frac {h-p}{h-t} \right)} = \frac {(a+b)(b+c)(c+a)}{a^3b^2c^2 \left(\frac {ac+a^2+ab-bc}{a^2bc}\right) \left(\frac {bc+ac+ab-a^2}{a^2bc}\right) } = \frac {h-p}{h-t}$$ So $P,H,T$ are collinear as desired. So , we get that $\overline {PA}$ and $\overline {PH}$ are isogonal in $\angle {BPC}$ . Next note that by DDIT on $BCEF$ ,with $P$ , we have that $\{\overline {PB},\overline {PC}\}$ , $\{\overline {PA},\overline {PH}\}$ , $\{\overline {PE},\overline {PF}\}$ are swapped under some involution . Considering the first two pairs , we know that this involution is just isogonal conjugation in $\angle {BPC}$ . Hence $\overline {PE},\overline {PF}$ are isogonal in $\angle {BPC}$ To finish , note that $$ \measuredangle {YCB} =\measuredangle {YPB} = \measuredangle {XPC}=\measuredangle {XBC} \implies BY = CX \implies \boxed {XY \parallel BC} $$. We are done . $\blacksquare$ Same solution,but i have found the calculations of $P-H-T$ collinear in a different way and i think is a bit easier like this. Take a point $T'$ on $(ABC)$ such that $AT'$ parralel to $BC$, and take line $T'H$ that intersect $(ABC)$ at $P'$,Now prove that $(AODP')$-Cyclic. Let $(ABC)$ be unit circle ,and let $a=1$, $o=0$, from here we get that $t'=bc$, $h=b+c+1$, and since $bc$ is on unit circle $d=\frac {b+c+1-bc}{2}$, also we know that $P-H-T'$ collinear and $P$ lies on unit circle(we use that advantage to deal with conjugates), now let $= \frac{p-t'}{p-h} = \overline\frac{p-t'}{p-h}$ from this we get that $p=\frac{b+c+1-bc}{1-b-c-bc}$, Now we use the cyclic lemma,on $(apdo)$ (its important to choose the right direction,because getin $p-d$ gets a lot longer,so we try to avoid it.)and we get easy and short calculations which gives us that $p=p'$. The rest follows from $DDIT$

See my solution to this problem on my Youtube channel here: https://www.youtube.com/watch?v=uy7vD75Tcsw

Complex numbers with unit circle $\Omega:=(ABC)$. Since $d=\tfrac12(a+b+c-bc/a)$, we know \begin{align*} \angle APO=\angle ADO &\Rightarrow \frac{p-a}{p}\div \frac{d-a}{d} \in \mathbb{R} \\ &\Rightarrow \frac{(p-a)(a^2+ab+ac-bc)}{p(-a^2+ab+ac-bc)} \in \mathbb{R} \\ &\Rightarrow \frac{(p-a)(a^2+ab+ac-bc)}{p(-a^2+ab+ac-bc)} = \frac{-\frac{p-a}{ap}\cdot \frac{bc+ac+ab-a^2}{a^2bc}}{\frac{1}{p}\cdot \frac{-bc+ac+ab-a^2}{a^2bc}} \\ &\Rightarrow p = \frac{a(a^2+ab+ac-bc)}{a^2-ab-bc-ca}. \end{align*}Since $E\in \overline{PX}$, we know $p+x=e+px\overline{e}$, so \[ x = \frac{p-e}{p\overline{e}-1}.\]The numerator is \begin{align*} p-e &= \frac{a(a^2+ab+ac-bc)}{a^2-ab-bc-ca} - \frac{ab+b^2+bc-ac}{2b} \\ &= \frac{2ab(a^2+ab+ac-bc)-(ab+b^2+bc-ac)(a^2-ab-bc-ca)}{2b(a^2-ab-bc-ca)} \\ &= \frac{a^3 b + a^3 c + 2 a^2 b^2 + a^2 b c - a^2 c^2 + a b^3 + a b^2 c + b^3 c + b^2 c^2}{2b(a^2-ab-bc-ca)} \\ &= \frac{(a + b) (b + c) (a^2 + a b - a c + b c)}{2b(a^2-ab-bc-ca)}, \end{align*}and the denominator is \begin{align*} p\overline{e}-1 &= \frac{a(a^2+ab+ac-bc)}{a^2-ab-bc-ca} \cdot \frac{bc+ac+ab-b^2}{2abc} - 1 \\ &= \frac{a(a^2+ab+ac-bc)(ab+bc+ca-b^2)-2abc(a^2-ab-bc-ca)}{2abc(a^2-ab-bc-ca)} \\ &= \frac{a^3 b + a^3 c + a^2 b c + a^2 c^2 - a b^3 + a b^2 c + 2 a b c^2 + b^3 c + b^2 c^2}{2bc(a^2-ab-bc-ca)} \\ &= \frac{(a + b) (b + c) (a^2 - a b + a c + b c)}{2bc(a^2-ab-bc-ca)}. \end{align*}Therefore, their quotient is \[ x = \frac{c(a^2+ab-ac+bc)}{a^2-ab+ac+bc},\quad \text{hence also} \quad y = \frac{b(a^2-ab+ac+bc)}{a^2+ab-ac+bc} \]where $y$ followed by symmetry. Note that $xy=bc$. Since these 4 are all on the unit circle, WLOG by rotation $c=e^{i\theta}$ and $b=e^{-i\theta}$, so $xy=bc=1$, which means $x=e^{i\alpha}$ and $y=e^{-i\alpha}$ for some $\alpha$. Hence $XY\parallel BC$.

Easy problem

By DDIT/Isogonal Lines Lemma it suffices to prove $PH,PA$ isogonal in $\measuredangle BPC$ or equivalentaly $\measuredangle HPC = \measuredangle C$ Let $H_A$ be the reflection of orthocenter in $BC$ Claim 1: $\measuredangle HPH_A = 90^\circ$

Now, $\measuredangle HPC = \measuredangle HPH_A - \measuredangle H_APC = 90^\circ - (90^\circ-\measuredangle C)= \measuredangle C$, and we're done.

Nice to find multiple working approaches. Here is one using spiral sim. First note that as $\measuredangle FBP = \measuredangle ECP,$ it suffies to proof $PE:PF= BE:CF$. Now introduce $Q$ the Miquel Point of $FEBC.$ Let $\Gamma$ be the $Q-$ appolonius circle of $Q$ with respect to $EF$. By above it suffies to show $P \in \Gamma.$ Now note that $H\in \Gamma$ as $PEHF$ is harmonic (well known) and we can easily compute the arc $QH$ by intercepting $QM,HN$ where $M,N$ are the midpoints of arc $EF$ in $(AEF)$. Now to finish: Let $H',O'$ be the reflection of $H$ with respect to $BC$ and the $A$ antipode respectively. Then by definition $P$ is the Miquel point of $AH,AO,HO,H'A'$, and so is the center of the spiral sim sending $DH'$ to $OA'$. Therefore $\triangle PHH' \sim \triangle POA'$ as $H=2D-H'$ and $A=2O-A'$. Therefore $\measuredangle APH=\measuredangle A'P'H'\Rightarrow \measuredangle QPH = \measuredangle QBA + \measuredangle A'BH'$ and by an easy computation the right hand side is the semi-arc measure of $QH$ with respect to $\Gamma$. So we are done. Sorry for not posting the complete angle chase Remark: One can see the above approach only guarantee the angles have the same sines, and so are either equal or complementary. To deal with this edge case, just see the problem can clearly bee rewritten (in complex numbers) as a certain polynomial in $a,b,c$ been zero, so it suffies to show, that for fixed $b,c$ there are infinitely many $a$ satisfying the problem. For this sake, just see that if $P$ is on the arc $BC$ not containing $A,$ and $ABC$ is acute we are done, since both angles are acute. But now just see that if $A$ is some $BC$ arc midpoint, $P$ is the other arc midpoint, then we are done as, by continuity , for any $A$ sufficiently near the arc midpoint, $P$ is on the desired arc.

a1267ab wrote: Let scalene triangle $ABC$ have altitudes $AD, BE, CF$ and circumcenter $O$. The circumcircles of $\triangle ABC$ and $\triangle ADO$ meet at $P \ne A$. The circumcircle of $\triangle ABC$ meets lines $PE$ at $X \ne P$ and $PF$ at $Y \ne P$. Prove that $XY \parallel BC$. Proposed by Daniel Hu Let $H'$ be the reflection $H$ with respect to $D$ then $<OPD=<OAD=<OH'D$ so $DP=DH=DH'$ wich givew that $P,H,P'$ lie on the same line. Now be this lemma https://artofproblemsolving.com/community/q3h3050791p27485591 we are done

Let $A'$ be the unique point on the circumcircle of $ABC$ such that $AA'$ is parallel to $BC$. And let $A_0$ be the intersection of $AD$ and $(ABC)$ besides $A$. $\textit{Claim:}$ $P-H-A'$ are collinear. $\textit{Proof:}$ It's obvious that the desired collinearity is equivalent to $PH \perp PA_0$. Since $\measuredangle{POA_0}=2\measuredangle{PAD}=2\measuredangle{POA_0}$ we obtain $\measuredangle{POD}=\measuredangle{DOA_0}$ which means $PD=DA_0$. Since $HD=DA_0$ we conclude the desired perpendicularity. Therefore, the desired collinearity has proven. From Dersargues Invoution Theorem, there exists unique involution with pairs $(PA, PH), (PF,PE), (PB,PC)$. From projecting these lines to the circumcircle of $(ABC)$, we obtain that there exists unique involution with pairs $(A,A'),(B,C),(X,Y)$ which means $AA', BC$ and $XY$ are concurrent. Since $AA'\parallel BC$ we obtain that $XY\parallel BC$. Therefore the proof has been completed.

Let $A_1$ be point on $(ABC)$ such that $\overline{AA_1} \parallel \overline{BC}$. Then easy angle chase to get that $P$, $H$, $A_1$ is collinear (consider $\overline{AD} \parallel (ABC)$). Then apply DDIT from $P$ to $BECF$, and project from $P$ to $(ABC)$ to finish.