Here is my solution. Problem. Let ABC be a triangle and H its orthocenter. Let P be a point on the circumcircle of triangle ABC (distinct from the vertices A, B, C), and let E be the foot of the B-altitude of triangle ABC. Let the parallel to the line BP through the point A meet the parallel to the line AP through the point B at a point Q. Let the parallel to the line CP through the point A meet the parallel to the line AP through the point C at a point R. The lines HR and AQ intersect at some point X. Prove that EX || AP. Solution. We will use directed angles modulo 180°. Since H is the orthocenter of triangle ABC, the lines AH, BH, CH are the altitudes of this triangle; hence, the point E, being the foot of the B-altitude, must lie on the line BH, and we have $AH\perp BC$, $BH\perp CA$ and $CH\perp AB$, so that < (AH; BC) = 90°, < (BH; CA) = 90° and < (CH; AB) = 90°. Thus, < CHA = < (CH; AH) = < (CH; AB) + < (AB; BC) - < (AH; BC) = 90° + < (AB; BC) - 90° = < (AB; BC) = < ABC. But since the point P lies on the circumcircle of triangle ABC, we have < ABC = < APC, and since CR || AP and AR || CP, we have < (AP; CP) = < (CR; AR), or, equivalently, < APC = < CRA. Thus, < CHA = < ABC = < APC = < CRA, and it follows that the points C, A, H and R lie on one circle. Thus, < HRC = < HAC. In other words, < (HR; CR) = < (AH; CA). Since CR || AP, we have < (HR; CR) = < (HR; AP), and this becomes < (HR; AP) = < (AH; CA). But < (AH; CA) = < (AH; BC) + < (BC; CA) = 90° + < (BC; CA) = 90° + < BCA, and since the point P lies on the circumcircle of triangle ABC, we have < BCA = < BPA. Thus, < (AH; CA) = 90° + < BPA, so that < (AH; CA) - < BPA = 90°. Hence, < (HR; BP) = < (HR; AP) - < (BP; AP) = < (AH; CA) - < BPA = 90°. Thus, $HR\perp BP$. Since AQ || BP, this becomes $HR\perp AQ$. Thus, < HXA = 90°. On the other hand, since E is the foot of the B-altitude of triangle ABC, we have < HEA = 90°. Thus, the points X and E lie on the circle with diameter HA. This yields < EXA = < EHA, or, equivalently, < (EX; AQ) = < (BH; AH). But < (BH; AH) = < (BH; CA) + < (CA; BC) - < (AH; BC) = 90° + < (CA; BC) - 90° = < (CA; BC) = < ACB, and since the point P lies on the circumcircle of triangle ABC, we get < ACB = < APB. Since AQ || BP, we have < (AP; BP) = < (AP; AQ). Thus, < (EX; AQ) = < (BH; AH) = < ACB = < APB = < (AP; BP) = < (AP; AQ), so that EX || AP, and the problem is solved. Darij

Darij, your enunciation is complicated. The points $R$ and $Q$ are the reflections of the point $P$ w.r.t. the midpoints of the sides $AC$, $AB$ repectively ! I think I make no mistakes. I wish you a happy New Year - 2000 !

thank you for solved problem Happy new year

First, we use vectors to show that $BP\perp RH$. Put the origin at the circumcenter of ABC, so that $\vec{H}=\vec{A}+\vec{B}+\vec{C}$ and $|\vec{A}|=|\vec{B}|=|\vec{C}|=|\vec{P}|$. Because of parallelogram APCR, $\vec{R}=\vec{A}+\vec{C}-\vec{P}$, and now we calculate $\vec{BP}\cdot\vec{RH}=(\vec{P}-\vec{B})\cdot(\vec{P}+\vec{B})=0$, i.e. $BP\perp HR$. This means that <AXH = 90 (since AX || BP), and so <AXH = <AEH, i.e. AEXH is cyclic. From here, a simple angle chase (directed, modulo 180 degrees) gives <AXE = <AHE = <BCA = <BPA = <XAP, and so EX || AP, as desired.

I remember I had solved this a long way back.Sorry for posting late.Its pure euclidean geometry. Let $D,E,F$ be the foot of altitudes from $A,B,C$ respectively and wlog let $P$ belongs to minor arc $BC$ (the proof is exactly same in all the cases).Now $APBQ$ is a parallelogram so $\angle{BQP}=C$.So points $A,H,B,Q$ are concyclic $\Rightarrow \angle{HQA}=\angle{HBA}=90-A$.Similarly points $H,A,R,C$ are concyclic.So $\angle{HRC}=\angle{HAC}=90-C$.Also $\angle{QHX}=\angle{QHA}+\angle{AHX}=\angle{QBA}+\angle{ACR}=\angle{BAP}+\angle{CAP}=A$ so from $\triangle{QHX}$ we get that $\angle{HXA}=90^{\circ}$.Hence points $A,X,E,H$ are concyclic $\Rightarrow \angle{HXE}=\angle{HAE}=90-C \Rightarrow XE \parallel RC \parallel AP$.

Let the reflection of $\bigtriangleup ABC$ across $AH$ be $\bigtriangleup AB'C'$ and the reflection of $E$ across $AH$ be $D$. $1)\ EX \| AP \iff AEHD$ is cyclic $\iff \angle HXA = 90^{\circ}$(simple angle chasing) so problem now reduces in proving $RX \perp XA$ $2)\ \angle {B}'DA = 90^{\circ}$, so, $RX \perp XA \iff \angle XAD = \angle RHB'$ (simple transformation geometry tricks) $3)\ AHCR{B}'$ and $AH{C}'QB$ are cyclic(simple angle chasing) $4)\ \angle XAD = \angle QA{C}' = \angle RA{B}' = \angle RH{B}' ($from $(3)$ and some simple angle chasing) Now, $(2)$ and $(4)$ solves the problem.

Dear Mathlinkers, by applying a spacial case of the Reim's theorem http://jl.ayme.pagesperso-orange.fr/Docs/6%27.pdf we avoid an angle chasing. Sincerely Jean-Louis

$RH\cap PB={K}, RC\cap BP={T}, AH\cap BC={L}$ $\angle{AHC}=\angle{ARC}=180-\angle{C}$, therefore $(ARHC)$ cyclic. $\rightarrow \angle{KRT}=\angle{HAC}=90- \angle{C}$, also $\angle{RTK}=\angle{APB}= \angle{C}$ So $HR \perp QA$ and $HR \perp BP$. This gives a new pair of cyclic quadrilaterals, namely $(XAEH),(BHKL).$ $\rightarrow \angle{EAP}=\angle{CAP}=\angle{CBP}=\angle{LBK}=\angle{LHK}=\angle{XHA}=\angle{XEA} \leftrightarrow AP \parallel XE \blacksquare$

WLOG assume that $P$ lies on arc $AC$ not containing $B$. Claim 1- $AHRC$ is cyclic. Proof- $\angle AHC=180-\angle CAH-\angle ACH=180-(90-\angle C)-(90-\angle A)=\angle A+\angle C=\angle APC=\angle ARC$. Claim 2-$APHX$ is cyclic. Proof- As $AH$ is diameter, we need $\angle AXH=90$. This can be achieved- $$\angle AXH=180-\angle XAH- \angle XHA=180-(\angle XAB+\angle BAH)-(\angle ACR)$$$$=180-(\angle ABP+(90-\angle B))-(\angle PAC)=180-(90-\angle CBP)-(\angle CBP)=90^{\circ}$$ Moving to the problem, We have $$\angle AXE=\angle AHE=90-\angle HAC=\angle C=\angle APB=\angle AQB$$. Hence, $EX \parallel QB \parallel AQ$, as desired. $\blacksquare$

Assume $P$ lies on arc $BC$ not containing $A$. Then, $\angle AQB=\angle C=180^{\circ}-\angle AHB$ $\implies$ $AQBH$ & similarly $AHCR$ are cyclic. $BQ$ $=$ $AP$ $=$ $CR$ $\implies$ $AH$ $\perp$ $QR$ and then note that, $$\begin{cases} \angle QHR=\angle QBA+\angle RCA=\angle BAP+\angle CAP=\angle BAC \\ \angle QAR =360^{\circ} - (\angle BAC+180^{\circ})=180^{\circ} -\angle BAC \end{cases} \implies A \text{ is the orthocenter of } \Delta QHR$$Hence, $AHEX$ is cyclic and By Reim's Theorem $\implies$ $XE||QB||AP$

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%201.pdf p. 25... Sincerely Jean-Louis

Here's another solution: Animate $P$ on $\odot (ABC)$. Then taking homothety from the midpoint of $AB$ with ratio $-1$, we get that $P \mapsto Q$ is a projective map. Similarly $P \mapsto R$ is also projective. Also, taking perspectivity from $A$ to the line at infinity, we see that $P \mapsto \infty_{AP}$ is a projective map. Let $$X_1=AQ \cap \odot (AH), X_2=RH \cap \odot (AH), X_3=E \infty_{AP} \cap \odot (AH) $$As $A, E, H$ are fixed, so we get that the points $X_1, X_2, X_3$ are all projective related to $P$. So it suffices to show $X_1=X_2=X_3$ for three positions of $P$. One can easily see this to be true for $P=A, C$ and when $P$ is the antipode of $B$ in $\odot (ABC) $ (It requires some work, but I am really not in the mood of writing all that ). Hence, done. $\blacksquare$

[asy][asy] size(10cm); defaultpen(fontsize(10pt)); pair A = dir(125), B = dir(210), C = dir(330), P = dir(280), Q = A+B-P, R = A+C-P, H = orthocenter(A,B,C), E = foot(B,C,A), X = extension(A,Q,R,H); dot("$A$", A, dir(90)); dot("$B$", B, dir(260)); dot("$C$", C, dir(330)); dot("$P$", P, dir(280)); dot("$Q$", Q, dir(90)); dot("$R$", R, dir(90)); dot("$X$", X, dir(0)); dot("$H$", H, dir(270)); dot("$E$", E, dir(45)); draw(unitcircle); draw(A--B--C--A, linewidth(1.2)); draw(R--H--B--P--C--H--A--X--E--H^^C--R--A--Q--B^^A--P); draw(circumcircle(A,B,H)^^circumcircle(A,C,H)^^circumcircle(A,E,H), dotted); [/asy][/asy] Let $\measuredangle$ denote a directed angle modulo $180^\circ$. Claim 1: $AHBQ$ and $AHCR$ are cyclic. Proof. We have $$\measuredangle AQB = \measuredangle BPA = \measuredangle BCA = \measuredangle AHB,$$so $AHBQ$ is cyclic. The other case is analogous. $\square$ Claim 2: $\overline{QA} \perp \overline{RH}$. Proof. Since $\overline{QB} \parallel \overline{AP} \parallel \overline{RC}$, we have $$\angle QXR = \angle XQB + \angle XRC = \angle APB + \angle CAH = \angle ACB + \angle CAH = 90^\circ$$as desired. $\square$ It follows from claim 2 that $AXEH$ is cyclic. Therefore $$\angle HRC = \angle HAC = \angle HXE,$$and we are done. $\blacksquare$

Solved with Smileyklaws and pretty big hints from tree_3 and mathgirl199 Note that $AHRC$ is cyclic since $$\measuredangle AHC = \measuredangle CBA = \measuredangle CPA = \measuredangle ARC$$Now, since $QA \parallel BP$ and $AR \parallel PC$, we have $$\measuredangle XAR =\measuredangle QAR = \measuredangle BPC = \measuredangle BAC$$But remark that $$\measuredangle ARX = \measuredangle ARH = \measuredangle ACH = 90^{\circ}-\measuredangle BAC$$so it follows that $RX \bot AQ$. Then $AXHE$ is cyclic, so $$\measuredangle AXE = \measuredangle AHE = -\measuredangle ACB = \measuredangle BPA = \measuredangle AQB$$which means $XE \parallel QB \parallel AP$, as needed. $\blacksquare$

Notice that $PBC$ and $AQR$ are congruent and corresponding sides are parallel. Let the orthocenter of $PBC$ be $H'$; it's well known that $PH'=AH$ so that a translation bringing $PBC$ to $AQR$ brings $H'$ to $H$ so that $H$ is the orthocenter of $AQR$. Thus $AQ\perp HR$ and $A,X,H,E$ are concyclic. So $\measuredangle EAP=\measuredangle PBC=\measuredangle XHA=\measuredangle XEA$ as desired. pls point out any issues with my directed angles, thanks!

I'll just outline my solution.