The "hidden" diagram in this problem is right triangles $PEB, PFC$ are inversely similar, which leads to $PQ\perp EF$.

See http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=496141 as well. Best regards, sunken rock

my solution: Since Cheva to the triangle $BPC$ we get that \[ \frac{\sin \angle BPQ}{\sin \angle CPQ}=\frac{AC}{BD} .\] So if $l$ be line with passes through midpoints of the $BC,AD$ then $l \parallel PQ$. Hence $PQ\bot EF$.

Let $R=AB \cap CD$, and $l$ the line perpendicular to $EF$ through $P$, which intersects $AB,CD$ in $X,Y$. Some quick angle chasing yields that $\triangle BPX \sim \triangle CPR$, and $\triangle BPR \sim \triangle CPY$. Let $m$ be the angle bisector of $\angle BPC$ and $f$ the transformation consisting of a reflection about $m$ and expansion by a factor of $CD/AB$. Then we easily see that $f$ sends $R,X,A,E,B$ into $Y,R,D,F,C$. So we have that $(R,X;E,B)=(Y,R;F,C)=(R,Y;C,F)$, which means that $XY,CE,BF$ are concurrent.

Let $S_1$ and $S_2$ be midpoints of $DP$ and $AP$, $M$ and $N$ be midpoints of $AD$ and $BC$. Triangles $MS_1F$ and $ES_2M$ are congruent so $ME=MF, NE=NF$ and $MN$ is perpendicular to $EF$. Thus $MN$ is the Gauss's line of quadrilaterals $XDPA$ and $XFQE$ ($X$ is intersection of $AB$ and $CD$) so it passes thru midpoints of $XP$ and $XQ$ so $MN$ and $PQ$ are parallel, so we are done.

Here's a solution with inversion: We compose an inversion $\mathcal{I}: X \to X'$ with center $P$ and arbitrary radius $r.$ Then by basic inversive properties, we find that $\mathcal{I}$ takes \[Q \equiv BF \cap CE \mapsto Q' \equiv \odot (PB'F') \cap \odot (PC'E'), \quad EF \mapsto \odot (PE'F'), \quad PQ \mapsto PQ'.\] Therefore, since inversion preserves angles between intersecting curves, $EF \perp PQ$ is equivalent to $\odot (PE'F') \equiv \omega$ and $PQ'$ intersecting at an angle of $0$, i.e. $PQ'$ passing through the center of $\omega.$ To see this, let $P'$ be the antipode of $P$ w.r.t. $\omega$ and let $X', Y'$ be the second intersections of $B'E', C'F'$ with $\odot (PB'Q') \equiv \Gamma_1, \odot (PC'Q') \equiv \Gamma_2$, respectively. Since the center of $\omega$ clearly lies on the line $PP'$, it suffices to prove that $P, P', Q$ are collinear. Observe that $PE \perp BE \implies PB' \perp E'B' = X'B'$, so $\overline{PX'}$ is a diameter of $\Gamma_1.$ Then since $\overline{PP'}$ is a diameter of $\omega$, it follows that $PF' \perp P'F'$ and $PF' \perp X'F'$, so $F', P', X'$ are collinear. Similarly, we find that $E', P', Y'$ are collinear. Now, since $PQ'$ is just the radical axis of $\Gamma_1, \Gamma_2$, it suffices to show that $P'$ has equal power with respect to each circle. Because \[\measuredangle X'E'Y' = \measuredangle B'E'Y' = \measuredangle B'E'P + \measuredangle PE'Y' = \measuredangle P'F'C' + \measuredangle X'F'P = \measuredangle X'F'C' = \measuredangle X'F'Y',\] where the angles are directed and we have used the fact that $\measuredangle PBE = \measuredangle FCP \implies \measuredangle B'E'P = \measuredangle P'F'C'$, it follows that the points $E', F', X', Y'$ are concyclic. Therefore, by Power of a Point, $P'E' \cdot P'Y' = P'F' \cdot P'X'$, so $P'$ has equal power w.r.t. $\Gamma_1$ and $\Gamma_2$, as desired. $\square$

There is a well known lemma that states that if A, B, C, D, E are points and $A(C, D ; B, E)= B(C, D ; A, E)$ then $CDE$ are collinear. If $R = AB \cap DC$ and $\ell$ is the reflection of $PR$ through the angle bisector of $BPC$, then by symmetry and since $PAEB \sim PDFC$ we have $B( \ell \cap DC, \ell \cap BA; Q, C) = P(\ell, PR; PF, PC) = P(PR, \ell ; PF, PC) = C(\ell \cap DC, \ell \cap BA; Q, B)$ and thus $\ell \cap DC, \ell \cap BA, Q$ are collinear, that is, $\boxed{Q \in \ell}$. Now we finish easily, since $PQ, PR$ are isogonal at the angle at $P$. Notice that the above result didn't use $PE \perp BA$, only that $PAEB \sim PDFC$.

Note that $\measuredangle FCP = \measuredangle DCA = \measuredangle DBA = \measuredangle PBE.$ Therefore, $\triangle FCP$ and $\triangle EBP$ are inversely similar right triangles $\implies \measuredangle CPF = \measuredangle EPB \implies PB, PC$ are isogonal WRT $\angle FPE.$ Moreover, if $X$ is a point at infinity on the $P$-altitude of $\triangle PEF$, then $EB, EX$ are isogonal WRT $\angle PEF$ and $FX, FC$ are isogonal WRT $\angle EFP.$ Thus, by Jacobi's Theorem, $PX, EC, FB$ are concurrent, as desired. $\square$

Does anyone have a complex number bash solution?

Yes! Amusingly, points $A$ and $D$ are useless and we only use that the angles $PBE$ and $PCF$ are equal. Let $P=0$, $E=e$, $F=f$. Now, note that we have $B=e(1+ki)$ for some real $k$ due to $PEB$ being right. We get that $C=f(1-ki)$, because $PFC$ and $PEB$ are inversely similar. In other words if we reflect $C$ over $F$ to $C'$ we get $PEB$ and $PFC'$ are directly similar using the cyclic property, S$C'=f(1+ki)$ and $C=f(1-ki)$. Use prime for conjugate due to laziness*, and say $Q=x$. From $Q$ on $BF$ we get $x(e'-f'-f'ki)+ef'(1+ki)=x'(e-f+fki)+e'f(1-ki)$ and also $x(f'-e'+e'ki)+e'f(1-ki)=x'(f-e-eki)+ef'(1+ki)$. Add to get $x(e'ki-f'ki)+(ef'+e'f)+(ki)(ef'-e'f)=x'(fki-eki)+(e'f+ef')+(ki)(ef'-e'f)$. Cancel and divide through by $ki$ to get $x(e'-f')=x'(f-e)$. This in fact is exactly the condition of the line through $x=Q$ and $0=P$ is perpendicular to the line through $e=E$ and $f=F$. Done. * - I am in college. I am not going to spend time making my complex bashes looking nice with LaTeX.

First let $X$ and $Y$ be circumcenters of $ABP$ and $DCP$ resp. and $T$ the intersection of $(AD)$ and $(BC)$ . $U$ the intersection of $(AB)$ and $(DC)$ It's easy to get $(OT) \parallel (XY)$ and it's well-knowen that $(OT) \bot (UP)$ . So $(XY) \bot (UP)$ $(1)$ we have $(PE)$ isogonal to $(PF)$ . $(PB)$ isogonal to $(PC)$ wrt $PBC$ and since $U = (EB) \cap (FC)$ and $P = (CE) \cap (BF)$ we get $(PQ)$ isogonal to $(PU)$ wrt $PBC$ therefore $ \angle FPU = \angle EPQ $ $(2)$ Now we have \[ \frac{PE}{PF}=\frac{PB}{PC}=\frac{PX}{PY} \implies PE*PY=PF*PX \]and easy to see that $P,X,F$ and $P,Y,E$ are collinear . thus $XEFY$ cyclic This combined with $(1)$ and $(2)$ gives the desired result

a similar solution with different ending: isogonal lines property: If $(AR,AS);(AP,AQ)$ are couples of isogonals wrt $\angle BAC$, and $U=RP\cap SQ; V=RQ\cap PS$, then $(AU,AV)$ are isogonal wrt $\angle BAC$. now back to the problem: on one hand: $\angle EPA=90-\angle BAC=90-\angle BDC=\angle DPF \rightarrow (PE,PF)$ are isogonal wrt $\angle DPC$. on the other hand $(PB,PC)$ are isogonals wrt $\angle DPC$. applying the above property and letting $J=AB\cap BC$ yellds that $(PJ,PQ)$ are isogonals wrt $\angle DPC$. which implies that $(PQ,PJ)$ are isogonals wrt $\angle EPF$ as well. and since the circumcenter of $\Delta EPF$ lays on $PJ$, necessarly $PQ$ is an alltitude of $\Delta EPF$ as desired.

One liner:- Let $AB,CD$ meet at point $R$ then, we know that $PR$ is the diameter of the circumcircle of triangle $PEF$. Now, we just need to show that $PQ,PR$ are isogonal rays in angle $BPC$. Indeed, this is true, since, we know that $\angle BPE=\angle CPF=90-\angle ABD$ and so rays $PE,PF$ are isogonal in angle $BPC$ and by a forgotten isogonality Lemma, we see that $BF \cap CE$ and $BE \cap CF$ are isogonal in angle $BPC$ that is $PQ$ is perpendicular to $EF$.

Sorry, but what is this forgotten isogonality lemma? Thanks!

MathPanda1 wrote: Sorry, but what is this forgotten isogonality lemma? Thanks! This is the Isogonality lemma.

Is this lemma citable on Olympiads without proof (national, not the IMO)? Thanks!

MathPanda1 wrote: Is this lemma citable on Olympiads without proof (national, not the IMO)? Thanks! No as far as I know, since it is a forgotten Lemma

Let $G,H$ be the projections of $P$ onto $BC$ and $DA$. It's easy to see that $GFHE$ is circumscribed to a circle $\omega$ centered at $P$. Let $X,Y,Z,T$ be the tangency points of $\omega$ with $FH,HE,EG,GF$. Let $G^\prime$ be the inverse of the point $G$ wrt $\omega$ and analogous for $F,H,E$. Let $\{M\}=TZ\cap XY,\ \{N\}=XT\cap YZ,\ \{O\}=XZ\cap YT$ and $\{K\}=G^\prime F^\prime\cap YZ,\ \{L\}=G^\prime E^\prime \cap XT$. Poles and polars are taken only wrt $\omega$. As $N$ is the pole of $EF$, it is enough to prove that $P-Q-N$ are collinear. $G^\prime F^\prime$ is the polar of $C$ , $YZ$ is the polar of $E$ , hence $K$ is the pole of $CE$. In the same way, $L$ is the pole of $BF$; as, $\{Q\}=CE\cap BF$, we get that $KL$ is the polar $Q$. Thereby, it is enough to prove that $PN\perp KL$. As $OM$ is the polar of $N$, it is enough to prove that $KL\parallel OM$. As $\widehat{LF^\prime K}=\widehat{TXZ}=\widehat{TYZ}=\widehat{LE^\prime K}$, we infer $LKE^\prime F^\prime$ is cyclic $(*)$. Let $\{I\}=OM\cap XT,\ \{J\}=OM\cap YZ$. As $(N,I, T, X)=(N, J, Z, Y)=-1$ and $F^\prime,\ E^\prime$ are the midpoints of $TX, ZY$, we get $NF^\prime\cdot NI=NT\cdot NX= NZ\cdot NY=NE^\prime \cdot NJ$, so $F^\prime IJE^\prime$ is cyclic, which combined with $(*)$ implies $KL\parallel OM$.

[asy][asy] unitsize(0.3inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); 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 = -23.291994431684675, xmax = 17.877126599901022, ymin = -14.29873495745406, ymax = 10.166521393283418; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-4.14,1.61), 7.868721214680655), linewidth(2) + wrwrwr); draw((-9.119018338843699,7.703123167622951)--(-7.251751509846343,-5.6172938293205155), linewidth(2) + wrwrwr); draw((-7.251751509846343,-5.6172938293205155)--(0.46218712765321435,-4.772526709417554), linewidth(2) + wrwrwr); draw((0.46218712765321435,-4.772526709417554)--(3.663887224956042,0.601973346318015), linewidth(2) + wrwrwr); draw((3.663887224956042,0.601973346318015)--(-9.119018338843699,7.703123167622951), linewidth(2) + wrwrwr); draw((-7.251751509846343,-5.6172938293205155)--(3.663887224956042,0.601973346318015), linewidth(2) + wrwrwr); draw((0.46218712765321435,-4.772526709417554)--(-9.119018338843699,7.703123167622951), linewidth(2) + wrwrwr); draw((-1.4344929468808592,-2.3028670594510214)--(1.0009793983320736,2.0812698445048765), linewidth(2) + wrwrwr); draw((-1.4344929468808592,-2.3028670594510214)--(-1.144764239496572,-4.948506807528793), linewidth(2) + wrwrwr); draw((-9.119018338843699,7.703123167622951)--(-1.144764239496572,-4.948506807528793), linewidth(2) + wrwrwr); draw((1.0009793983320736,2.0812698445048765)--(-7.251751509846343,-5.6172938293205155), linewidth(2) + wrwrwr); draw((1.0009793983320736,2.0812698445048765)--(-1.144764239496572,-4.948506807528793), linewidth(2) + wrwrwr); draw((0.46218712765321435,-4.772526709417554)--(11.218236070887414,-3.594612687242165), linewidth(2) + wrwrwr); draw((11.218236070887414,-3.594612687242165)--(3.663887224956042,0.601973346318015), linewidth(2) + wrwrwr); draw(circle((4.891871562003277,-2.948739873346593), 6.359248351102369), linewidth(2) + wrwrwr); draw((1.0009793983320736,2.0812698445048765)--(0.46218712765321435,-4.772526709417554), linewidth(2) + wrwrwr); /* dots and labels */ dot((-9.119018338843699,7.703123167622951),dotstyle); label("$B$", (-9.5577056940983,8.24304606639785), NE * labelscalefactor); dot((-7.251751509846343,-5.6172938293205155),dotstyle); label("$C$", (-7.937936997773616,-6.2673818381774815), NE * labelscalefactor); dot((0.46218712765321435,-4.772526709417554),dotstyle); label("$D$", (0.5995938391044017,-5.626223395882293), NE * labelscalefactor); dot((3.663887224956042,0.601973346318015),dotstyle); label("$A$", (3.805386050580337,0.9540869329367535), NE * labelscalefactor); dot((-1.4344929468808592,-2.3028670594510214),linewidth(4pt) + dotstyle); label("$P$", (-1.9987851112497776,-3.3990414384358463), NE * labelscalefactor); dot((-1.144764239496572,-4.948506807528793),linewidth(4pt) + dotstyle); label("$F$", (-1.526352574821745,-5.524987852361999), NE * labelscalefactor); dot((-3.140516627011106,-1.7821265524869094),linewidth(4pt) + dotstyle); label("$Q$", (-4.090986344002493,-2.1167245538454678), NE * labelscalefactor); dot((11.218236070887414,-3.594612687242165),linewidth(4pt) + dotstyle); label("$G$", (11.364306633428859,-3.331551076088984), NE * labelscalefactor); dot((1.0009793983320734,2.081269844504877),linewidth(4pt) + dotstyle); label("$E$", (1.139516737879296,2.3376393610474246), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $f$ be the involution on lines through $P$ that is reflection in the angle bisector of $\angle BPC$. Let $G=AB\cap CD$. It is easy to see that $G\in(PEF)$, and in fact, $PG$ is a diameter. We see that $f(PG)$ is perpendicular to $EF$ (the isogonal conjugate of circumcenter is orthocenter), so it suffices to show that $f(PG)=PQ$. Note that \[\angle APE = \pi-\angle PEA-\angle EAP = \pi-\angle PFG-\angle BDC=\angle DPF,\]so $f(PE)=PF$. Thus, by DDIT, $f$ is the unique involution sending $PB\to PC$, $PE\to PF$, $P(BE\cap CF)=PQ\to P(BF\cap CE)=PG$, so $f(PG)=PQ$, as desired.

yugrey wrote: Yes! Amusingly, points $A$ and $D$ are useless and we only use that the angles $PBE$ and $PCF$ are equal. Let $P=0$, $E=e$, $F=f$. Now, note that we have $B=e(1+ki)$ for some real $k$ due to $PEB$ being right. We get that $C=f(1-ki)$, because $PFC$ and $PEB$ are inversely similar. In other words if we reflect $C$ over $F$ to $C'$ we get $PEB$ and $PFC'$ are directly similar using the cyclic property, S$C'=f(1+ki)$ and $C=f(1-ki)$. Use prime for conjugate due to laziness*, and say $Q=x$. From $Q$ on $BF$ we get $x(e'-f'-f'ki)+ef'(1+ki)=x'(e-f+fki)+e'f(1-ki)$ and also $x(f'-e'+e'ki)+e'f(1-ki)=x'(f-e-eki)+ef'(1+ki)$. Add to get $x(e'ki-f'ki)+(ef'+e'f)+(ki)(ef'-e'f)=x'(fki-eki)+(e'f+ef')+(ki)(ef'-e'f)$. Cancel and divide through by $ki$ to get $x(e'-f')=x'(f-e)$. This in fact is exactly the condition of the line through $x=Q$ and $0=P$ is perpendicular to the line through $e=E$ and $f=F$. Done. * - I am in college. I am not going to spend time making my complex bashes looking nice with LaTeX. What is motivation for using $B=e(1+ki)$ instead of normal ortogonality criteria ?

Construct $H$, the orthocenter of $EQF$, $H'=AB\cap CD$, $J=EH\cap BP$, $N$ the foot of the altitude of $P$ on $EF$, and $K=FH\cap CP$. Also, denote $\angle PBE=\angle PCF=\alpha$. Then, $$\frac{EJ}{JH}=\frac{EJ}{H'F-EJ}=\frac{EB}{EH'}=\frac{EP\cot\alpha}{FH}=\frac{EN\cot\alpha}{FH\sin F}$$So, $$\frac{EJ}{JH}\cdot\frac{HK}{KF}\cdot\frac{FP}{PE}=\frac{EN\cot\alpha}{FH\sin F}\cdot\frac{EH\sin E}{FN\cot\alpha}\cdot\frac{FN}{EN}=1$$and by Ceva's, $Q$ must lie on $PN$, as desired.

Suppose $AB$ and $CD$ meet at point $X$. Note that $(PE,PF)$ and $(PB,PC)$ are pairs of isogonal lines with respect to $\angle EPF$. Applying Desargues Dual Involution Theorem on self-intersecting quadrilateral $ECFB$, we note that $(PE,PF)$, $(PB,PC)$ and $(PQ,PX)$ are pairs of the same involution, which turns out to be the isogonality condition because of $(PE,PF)$ and $(PB,PC)$. So $PQ$ and $PX$ are isogonal with respect to $\angle EPF$. This was the main part. Now see that $\angle EPQ=\angle FPX=\angle FEX=90^{\circ}-\angle PEF$, so\[\angle (PQ,EF)=180^{\circ}-\angle EPQ-\angle PEF=90^{\circ}.\]

Almost same as above solutions v_Enhance wrote: In cyclic quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at $P$. Let $E$ and $F$ be the respective feet of the perpendiculars from $P$ to lines $AB$ and $CD$. Segments $BF$ and $CE$ meet at $Q$. Prove that lines $PQ$ and $EF$ are perpendicular to each other. Observe that $PA,PB$ are isogonal in $\angle{EPF}$.Then by ddit on $EBFC$ we have $(PE,PF),(PB,PC),(P(AB\cap CD),PQ)$ are pair of some involution.The first two is just isogonal conjugation wrt $\angle{EPF}$ thus so is the third one.Hence $P(AB\cap CD),PQ$ are isogonal.But since $P$ is anti-pode of $AB\cap CD$ in $\odot{PEF}$ we have $P(AB\cap CD)$ passes through the center of $\odot{EPF}$.Thus $PQ$ is the Altitude in $EPF$ and so $PQ\perp EF$.

Maybe same as others. v_Enhance wrote: In cyclic quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at $P$. Let $E$ and $F$ be the respective feet of the perpendiculars from $P$ to lines $AB$ and $CD$. Segments $BF$ and $CE$ meet at $Q$. Prove that lines $PQ$ and $EF$ are perpendicular to each other. Notice that $\{PE,PF\},\{PB,PC\}$ are isogonal WRT $\odot(PBC)$. So, by Isogonal Line Lemma we get that $P(BF\cap CE)$ and $P(BE\cap CF)$ are isogonal. So, if $BE\cap CF=K$. Then $\{PQ,PK\}$ are isogonal. So, $$\angle QPF=\angle EPK=90^\circ-\angle EKP=90^\circ-\angle PFE\implies PQ\perp EF. \blacksquare$$

Hopefully something new Work in the extended Euclidean plane. Let $PQ \cap BC \equiv X, PQ \cap AB \equiv Y, AB \cap CD \equiv Z$. Taking perspectivity at $B$, projecting onto $CD$, $(Y, P; Q, X) = (Z, D; F, C)$. Taking persectivity at $C$, projecting onto $AB$, $(Y, P; Q, X) = (Y, A; E, B)$.Note that $DF/FC = AE/EB$. $Y$ is a point at infinity iff $Z$ is, and in that case $ABCD$ is an isosceles trapezium with $AB \parallel CD$, in which case the problem is trivially solved. Otherwise, we get $\triangle PYE \sim \triangle PZF$. So $\angle PYE = \angle PZF = \angle PEF$. This gives $PQ$ perpendicular to $EF$, and we're done.

Difficult solution, but this might be something new Step 1. Let the orthocenters of $\triangle APB$, $\triangle CPD$ be $H_1, H_2$ resp. Since $\triangle APB \sim \triangle DPC$, we get $PH_1 / PE = PH_2 / PF$. So $EF \parallel H_1H_2$. Step 2. Let the perpendicular feet from $A,C$ to $BD$ be $A_1, C_1$ resp. and from $B,D$ to $AC$ be $B_1, D_1$ resp. Let midpoint of $AD$, $BC$ be $M,N$ resp. Then $M,N$ are centers of $\odot(ADA_1D_1),\odot(BCB_1C_1)$ resp. Since $(A,B,A_1,B_1)$, $(C,D,C_1,D_1)$ are cyclic and , $H_1$, $H_2$ lie on the radical axis of $\odot(ADA_1D_1),\odot(BCB_1C_1)$. Therefore, $H_1H_2 \bot MN$. Step 3. Let $AB \cap CD=X$, $L=$ midpoint of $PX$. $G,H\in\odot(ABCD)$ are points satisfying $AB\parallel DH$, $CD\parallel AG$. $AG\cap DH=Y$, $CF\cap BG=T$. Then, $\angle BPE=\angle CPF$, $\angle PCF=\angle BCT (\widehat{AD}=\widehat{BH})$, $\angle PBE=\angle CBT (\widehat{AD}=\widehat{CG})$. By Jacobi's thm on $\triangle PBC$, we get $P,Q,T$ are colinear. By Pascal's thm on $(A,B,H),(D,C,G)$, we get $P,Y,T$ are colinear. By Gauss Newton line, $M,N,L$ are colinear. Since $M,L$ are midpoints of $XP,XY$ resp, $ML\parallel PY$. Therefore, $MN\parallel PQ$. Summing up, $EF\parallel H_1 H_2 \bot MN \parallel PQ$

See my solution on my Youtube channel here: https://www.youtube.com/watch?v=AhuKm17HQPM

Let $G = AB \cap CD$. Clearly, we have $$\angle PEG = 90^{\circ} = \angle PFG$$so $EPFG$ is cyclic. We claim that $\angle EPF$ and $\angle APD$ share an internal angle bisector. Indeed, observe $$\angle APE = 90^{\circ} - \angle PAE = 90^{\circ} - \angle CAB = 90^{\circ} - \angle CDB = 90^{\circ} - \angle FDP = \angle DPF$$which clearly implies the aforementioned result. Now, consider the inversion about $P$ with radius $\sqrt{PE \cdot PF}$ followed by a reflection in the internal angle bisector of $\angle EPF$. Trivially, we know $E$ and $F$ swap. In addition, the previous claim implies $B^* \in PC, C^* \in PB$. But we have $$\angle PB^*F = \angle PF^*B = \angle PEB = 90^{\circ}$$and $$\angle PC^*E = \angle PE^*C = \angle PFC = 90^{\circ}$$so $B^*$ is the projection of $F$ onto $PC$, and $C^*$ is the projection of $E$ onto $PB$. Now, recall that $Q = BF \cap CE$, so $Q^*$ is the second intersection between $(PB^*E)$ and $(PC^*F)$. Now, let $B^*F \cap AB = X$ and $C^*E \cap CD = Y$. By Thales' Theorem, we know $PEXB^*$ and $PFYC^*$ are cyclic. Now, observe $$\angle XEY = 90^{\circ} - \angle PEC_1 = 90^{\circ} - \angle PBE = 90^{\circ} - \angle DBA = 90^{\circ} - \angle DCA$$$$= \angle CPF = \angle FPB_1 = 90^{\circ} - \angle PFB_1 = \angle XFY$$so $EFYX$ is cyclic. As a result, the Radical Axis Theorem implies $G, P, Q^*$ are collinear. It's easy to see that $PQ, PQ^* \equiv PG$ are isogonal wrt $\angle EPF$, and Thales' implies the circumcenter of $PEF$ lies on $PG$. But it's well-known that the circumcenter and orthocenter of a triangle are isogonal conjugates, so $PQ$ must contain the orthocenter of $PEF$, which clearly finishes. $\blacksquare$ Remarks: Thinking about isogonality should be fairly natural, as the right angles motivate us to construct the $P$-antipode of $(PEF)$. After drawing a diagram, I pondered $\sqrt{PA \cdot PC}$ inversion and anti-parallel lines. Soon after, I realized $\sqrt{ef}$-inversion looked promising, as $G$ lies on $(PEF)$ and the foot of the $P$-altitude lies on $EF$. (In the end, this reason was insignificant.) Once I decided to shift my focus towards inversion, I realized the main issue with inverting was pinpointing $B^*$ and $C^*$. Fortunately, a good eye and some angle chasing skills led me to find the first claim, and the rest of my solution followed naturally.

Let $B'$ denote the reflection of $B$ over $E$ and $C'$ the reflection of $C$ over $F$. Let $B'C \cap C'B = R$. Note that $\triangle PCC'\sim \triangle PB'B$. As $P$ is the spiral center mapping $C'C \to BB'$, this means $PRCC'$ and $PRBB'$ are cyclic quadrilaterals with circumcenters $O_2,O_1$ respectively. Note that by the similarity $\triangle PCC'\sim \triangle PB'B$, we have $PR\perp O_1O_2\parallel EF$. Now, note that the map $X\mapsto CX\cap PR\mapsto B(CX\cap PR)\cap CC'$ is projective, so because it takes $B$ to $C$, $B'$ to $C'$, and $A$ to $D$ (the last follows from similar triangles), it ought to take $E$ to $F$, hence $Q$ is on line $PR$ and we are done.

Too easy by Isogonal line lemma - Note that as $\triangle PEB \sim \triangle PFC$ , we have $\angle EPB = \angle FPC$ , also let $P_{1} , P_{2}$ be reflection of $P$ about $AB$ and $CD$ respectively , then we have $AB \cap CD = O$ is the circumcenter of $\triangle PP_{1}P_{2}$ , Now in $\angle P_{1}P P_{2}$ we have $PB , PC$ are isogonal so by Isogonal line lemma , we have $PQ , PO$ are isogonal , but this implies that the orthocenter of $\triangle P_{1}PP_{2}$ lies on $PQ$ implies $PQ \perp P_{1}P_{2}$ , but as $P_{1}P_{2} \parallel EF$ , hence $PQ \perp EF$ , and we are done.

[asy][asy] import olympiad; unitsize(10); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); 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 = -12.975305494143722, xmax = 25.20458294849801, ymin = -9.909211849792026, ymax = 9.898967896685921; /* image dimensions */ draw((-0.5017216219212224,7.98425171284678)--(7.683549477221493,2.2277942973015534)--(7.583846113745577,-2.5466209225218646)--(-7.58946638440411,-2.529822128134704)--cycle, linewidth(0.6)); /* draw figures */ draw(circle((0,0), 8), linewidth(0.6)); draw((-7.58946638440411,-2.529822128134704)--(-0.5017216219212224,7.98425171284678), linewidth(0.6)); draw((-0.5017216219212224,7.98425171284678)--(7.583846113745577,-2.5466209225218646), linewidth(0.6)); draw((7.683549477221493,2.2277942973015534)--(-7.58946638440411,-2.529822128134704), linewidth(0.6)); draw((6.095813729616395,3.34440156754638)--(4.644841229465244,1.2812223561332161), linewidth(0.6)); draw((4.644841229465244,1.2812223561332161)--(4.640606925807296,-2.5433623809241452), linewidth(0.6)); draw((7.683549477221493,2.2277942973015534)--(4.640606925807296,-2.5433623809241452), linewidth(0.6)); draw((6.095813729616395,3.34440156754638)--(7.583846113745577,-2.5466209225218646), linewidth(0.6)); draw((6.095813729616395,3.34440156754638)--(4.640606925807296,-2.5433623809241452), linewidth(0.6)); draw((6.748277144262428,0.7613417560652703)--(4.644841229465244,1.2812223561332161), linewidth(0.6)); draw((-0.5017216219212224,7.98425171284678)--(14.48328824029306,-2.5542594859327292), linewidth(0.6)); draw((4.644841229465244,1.2812223561332161)--(14.48328824029306,-2.5542594859327292), linewidth(0.6)); draw(circle((6.754672815729366,-0.1410567719605384), 2.5444581428290216), linewidth(0.6)); draw((7.583846113745577,-2.5466209225218646)--(14.48328824029306,-2.5542594859327292), linewidth(0.6)); draw(circle((9.564064734879151,-0.6365185648997568), 5.279819138608925), linewidth(0.6)); /* dots and labels */ dot((-0.5017216219212224,7.98425171284678),dotstyle); label("$A$", (-0.6896978269570258,8.462496846368817), NE * labelscalefactor); dot((7.683549477221493,2.2277942973015534),dotstyle); label("$B$", (7.853524735455877,2.6032070358648447), NE * labelscalefactor); dot((7.583846113745577,-2.5466209225218646),dotstyle); label("$C$", (7.777920995965498,-3.293884644384315), NE * labelscalefactor); dot((-7.58946638440411,-2.529822128134704),dotstyle); label("$D$", (-8.514684864211322,-3.0670734259131933), NE * labelscalefactor); dot((4.644841229465244,1.2812223561332161),linewidth(4pt) + dotstyle); label("$P$", (4.413554588643602,1.8093677712159193), NE * labelscalefactor); dot((6.095813729616395,3.34440156754638),linewidth(4pt) + dotstyle); label("$E$", (5.850025638960815,3.6994612584752655), NE * labelscalefactor); dot((4.640606925807296,-2.5433623809241452),linewidth(4pt) + dotstyle); label("$F$", (3.959932151701324,-3.293884644384315), NE * labelscalefactor); dot((6.748277144262428,0.7613417560652703),linewidth(4pt) + dotstyle); label("$Q$", (7.021883601061701,0.6375098091151248), NE * labelscalefactor); dot((14.48328824029306,-2.5542594859327292),linewidth(4pt) + dotstyle); label("$G$", (14.846870638315997,-3.0670734259131933), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Define $G = AB \cap CD.$ Note \[\angle PCF = \angle ACD = \angle ABD = \angle PBE,\]so it readily follows $PE$ and $PF$ are isogonal with respect to $\angle BPC.$ It follows by Isogonal Lines Lemma that $PQ$ and $PG$ are isogonal. From here, angle chase \begin{align*} \angle QPF &= \angle GPE \\ &= 90^\circ - \angle PGE \\ &= 90^\circ - \angle PFE, \end{align*}from where the conclusion follows.

v_Enhance wrote: In cyclic quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at $P$. Let $E$ and $F$ be the respective feet of the perpendiculars from $P$ to lines $AB$ and $CD$. Segments $BF$ and $CE$ meet at $Q$. Prove that lines $PQ$ and $EF$ are perpendicular to each other. XmL wrote: The "hidden" diagram in this problem is right triangles $PEB, PFC$ are inversely similar, which leads to $PQ\perp EF$. Here is an easy verification of XmL's idea. (solution from my colleague) Let $X,Y,H$ be perpendicular feet of $P$ onto $BF, CE,EF$ resp. $\angle BPF = \angle CPE \implies \angle PFB + \angle PBX = \angle PEC + \angle PCY$ $\implies \angle XEY = \angle PEY - \angle PEX = \angle PEC - \angle PBX = \angle PFB - \angle PCY = \angle PFB - \angle PFY = \angle XFY$ $\implies E,F,X,Y$ are cyclic $\implies Q \in \overleftrightarrow{PH}=$ radical axis of $\odot(EHPY)$, $\odot(FHPX)$.

Let $Z$ be the foot from $P$ to $EF$. Let $G=(EZP)\cap DE$. Let $H=(FZP)\cap AF$. We want to show $Q$ is on the radical axis of these two circles. It suffices to check $EFHG$ is cyclic. Observe that because of right angles we also have $AEPH$ and $PFDG$ are also cyclic. Now let $f(\ell)$ be the unit complex number with the same direction as $\ell$. We now repeatedly use the fact that $f(\ell_1)f(\ell_2)=f(\ell_3)f(\ell_4)$ if these lines form a cyclic quadrilateral with $\ell_1$ and $\ell_2$ on opposite sides. Now observe that $$f(EP)f(FP)=f(AP)f(DP)$$$$\rightarrow f(AF)f(EP)f(DE)f(PF)=f(DE)f(AF)f(AP)f(DP)$$$$\rightarrow f(HE)f(AP)f(GF)f(DP)=f(DE)f(AF)f(AP)f(DP)$$$$\rightarrow f(HE)f(GF)=f(DE)f(HF),$$so $EFGH$ is cyclic.

Let $R=\overline{AB} \cap \overline{CD}$. By DDIT on $P$ with respect to $EBFC$, $(PB, PC)$, $(PE, PF)$, and $(PQ, PR)$ are swapped by some involution. Since $(PB, PC)$ and $(PE, PF)$ are isogonal pairs of lines, $(PQ, PR)$ is also an isogonal pair. Since the isogonal conjugate of $PR$ is perpendicular to $EF$ (circumcenter-orthocenter conjugacy), we have that $PQ$ is perpendicular to $EF$, as desired.

Here is a slick solution via Jacobi’s Theorem. Denote $M,N$ by the midpoints of $AD,BC$, respectively. By USA TST 2000, $EF\perp MN$. So, it is sufficient to show that $MN\parallel PQ$. Define $R=AF\cap DE$. By Pappus’s Theorem, $R,P,Q$ are collinear. Let $\angle DBA= \alpha$ and $\angle CAB=\beta$ The key is to introduce the point $X$ as a point on $ON$ such that $N$ lies strictly between $O,X$ and $\angle CBX =\alpha$. Consider triangle PBC and the exterior points $E,F$ and $X$. By Jacobi’s Theorem, $P,Q,X$ are collinear. Similarly, introduce the point $Y$ as a point on $OM$ such that $M$ lies strictly between $O,Y$ and $\angle DAY=\beta$. By similar argument, it follows that $P,R,Y$ are collinear. So, we have $Y,R,P,Q,X$ are collinear. Now we’re left to prove that $XY\parallel MN$ which is followed by $$\frac{OM}{MY}=\frac{OM}{MA} / \frac{MY}{MA}=\frac{cot \alpha}{tan \beta}=\frac{cot \beta}{tan \alpha}=\frac{ON}{NB} / \frac{NX}{NB}=\frac{ON}{NX} $$$\square$

Let $BE \cap CF = T$. Then apply DDIT on quadrilateral $BEFC$ from point $P$. Then $(PE, PF)$, $(PB, PC)$, and $(PQ, PT)$ are involutions. However $PB$ and $PC$ are isogonal wrt $\angle BPC$ and $\angle EBP = PCF \implies \angle EPB = \angle FPC$ so $PE$ and $PF$ are isogonal as well. So it follows that $PT$ and $PQ$ are isogonal. However this implies that $PQ$ passes through the orthocenter of $\triangle PEF$ since $PT$ is a diameter of $(PEF)$, so $PQ \perp EF$. identical to 2@above : (

Suppose \(AB\) and \(CD\) meet at point \(X\). Note that \((PE, PF)\) and \((PB, PC)\) are pairs of isogonal lines with respect to \(\angle EPF\). By applying Desargues' Dual Involution Theorem to the self-intersecting quadrilateral \(ECFB\), we observe that \((PE, PF)\), \((PB, PC)\), and \((PQ, PX)\) are pairs in the same involution. This results in the isogonality condition due to \((PE, PF)\) and \((PB, PC)\). Therefore, \(PQ\) and \(PX\) are isogonal with respect to \(\angle EPF\). The key point here is that \(\angle EPQ = \angle FPX = \angle FEX = 90^\circ - \angle PEF\). Consequently, we have: \[ \angle (PQ, EF) = 180^\circ - \angle EPQ - \angle PEF = 90^\circ. \]

Great problems require great solutions, so I came up with one! Let $AB$ and $CD$ meet at $M$. Notice that proving that the Newton-Gauss line $l$ of $MBQC$ is perpendicular to $EF$ is sufficient, as then the midpoint of $MP$ also lies on $l$ and the result follows from the midline of $MPQ$. We need to prove that the projection foots of $B$ and $C$ to $EF$ are symmetric with respect to the midpoint of $EF$. But that is true because $BEcos \angle BEF=BEsin \angle PEF=PEcot \angle ABD sin \angle PEF=d(P,EF)cot \angle ABD=CFcos \angle CFE$