liekkas wrote: $E,F$ lie on $AB,AD$, respectively Did you mean to say "$E,F$ lie on $AB,CD$, respectively"? Also, can you please provide a diagram? I was having a hard time trying to draw it.

NikoIsLife wrote: Did you mean to say "$E,F$ lie on $AB,CD$, respectively"? Sorry, it should be $E,F$ lie on $AD,BC$ ,respectively.

Can't draw diagram either , can someone please post a diagram

Hmm I don't think the diagram is drawable in its current formulation. If you draw the circle $ESF$ with center $T$ first and then $O$, $EF$ a chord on a circle, then $S$ a variable point on the circle, then $M$ $N$ should be on the circles $\omega_E, \omega_F$ with diameters $OE,OF$ respectively, so the intersection of $MF, NE$ cannot lie on the circle with center $T$ (to see this consider drawing tangents from $F$ to $\omega_E$, $E$ to $\omega_F$). Did I miss something?

Bumping this thread

I think this should be the actual question: Cyclic quadrilateral $ABCD$ has circumcircle $(O)$, midpoints of $BC,CD$ are $M,N$ respectively. $E,F$ are on $AB,AD$ respectively, such that $O$ is the midpoint of $EF$. $EN$ meets $FM$ at $P$. Denote $S$ as the circumcenter of $\triangle EPF$. $PO$ intersects $AD,BA$ at $Q,R$ respectively. Suppose $OSPC$ is a parallelogram, prove that $AQ=AR$.

No solutions for this?

By some trigonometry bash, we conclude CB=CD hence by directed angle, it's easy to show that AQ=AS

Let $H$ be the orthocenter of $\triangle PEF$. Since $CP\parallel OS\perp EF$ $\implies C$ lies on $HP$. Note circumradii $R$ of $(PEF),(HEF)$ are equal, combine $EO=OF$ and $OC=SP=R$ $\implies OC$ is a diameter of the $9-$point circle of $\triangle HEF$. Since $\angle ONC=90^{\circ}\implies N$ lies on $(OC)$ as well as $EP$ extended$\implies N$ is the foot of perpendicular from $E$ to $HF$. Similarly $M\in HE$, and $C$ is the circumcenter of $MPNH$. Let $T$ be the circumcenter of $\triangle HEF$. Note $OTCP$ is a parallelogram$\implies TC\parallel OP$. Now $180^{\circ}-\angle EAF=\angle MCN=2\angle EHF=\angle ETF$ $\implies AETF$ cyclic, combine $TE=TF\implies AT$ bisects $\angle EAF$. Since $CM=CN\implies CB=CD\implies AC$ bisects $\angle BAD$, thus $A,T,C$ collinear $\implies AC\parallel OP$, from which the conclusion follows.

#below: Thanks! My solution is so ugly that I very surprised to see this beautiful solution!

nguyenhaan2209 wrote: #above: I think you have miss the most important part to prove H lies on EM, FN. Why so many people still don't recognize it? Well, this directly follows from observing $\angle OMC=\angle ONC=\frac{\pi}{2}$ with reference triangle $\triangle PMN$.

Note that $CP \parallel OS \perp EF$ and $CP = OS = \tfrac12PH$ where $H$ is the orthocenter of $\triangle PEF$ hence $C$ is actually the midpoint of $PH$. Therefore, $CO$ is a diameter of the nine-point circle $\omega$ of $\triangle PEF$ and since\[\angle CMO = \angle CNO = 90^{\circ}\]we have that $M, N \in \omega$. Furthermore, $M \in PF$ and $N \in PE$ so $M$ and $N$ are actually the feet from $E, F$ to $PF, PE$, respectively. Thus $EM \cap FN = H$. Furthermore, $C$ is the center of $(HMNP)$ with diameter $PH$, so\[CM = CP = CN = CH.\]Most importantly, $CM = CN$ so $CB = CD$. Thus $AC$ bisects the angle $\angle BAD$ so it just suffices to show that $OP \parallel AC$ since that would imply\[\angle ASR = \angle BAC = \angle DAC = \angle ARS\]which would finish. Let $S'$ be the reflection of $S$ over $EF$; this point is the circumcenter of $\triangle HEF$ since $(PEF), (HEF)$ are reflections of each other over $EF$. Check that\[\angle ES'F = 2\angle EHF = 2\angle MHN = \angle MCN = \angle BCD\]hence $AES'F$ is actually cyclic and since $S'E = S'F$ by circumradius it actually holds that the line $AS'$ bisects $\angle EAF$ implying $S' \in AC$. In $\triangle HEF$, we see that $OS'CP$ is clearly a parallelogram so $OP \parallel S'C \implies OP \parallel AC$ and we are done. $\blacksquare$

Very Nice liekkas wrote: Cyclic quadrilateral $ABCD$ has circumcircle $(O)$. Points $M$ and $N$ are the midpoints of $BC$ and $CD$, and $E$ and $F$ lie on $AB$ and $AD$ respectively such that $EF$ passes through $O$ and $EO=OF$. Let $EN$ meet $FM$ at $P$. Denote $S$ as the circumcenter of $\triangle PEF$. Line $PO$ intersects $AD$ and $BA$ at $Q$ and $R$ respectively. Suppose $OSPC$ is a parallelogram. Prove that $AQ=AR$. Solution. Let the conditions: $OSPC$ is a parallelogram, $OE=OF$, and $S$ is the center of $\odot(PEF)$, be respectively denoted by $(\spadesuit), (\clubsuit)$, and $(\star).$ And, let $H$ be the orthocenter of $PEF.$ So, $P, E, F, H$ forms an orthocentric system.$-(\diamondsuit)$ Now, by $(\clubsuit)$ $SO \perp EF$, and from $(\spadesuit)$ $CP \parallel OS$, so $CP \perp EF$, and thus $C \in PH.$ Next, as we know that $\odot(PEF)$ and $\odot(HEF)$ have the same radius, say $R$. So, $$OC \overset{(\spadesuit)}{=} SP \overset{(\star)}{=} R.$$So, its clear that in $\triangle HEF,$ we have $OC$ as the diameter of it's NPC. Next, we have that $\angle OMC = 90^{\circ} = \angle ONC,$ hence, $\{M, N\} \in \odot(OC)$; also $M \in PF$ and $N \in PE$, so along with $(\diamondsuit)$, we get $M, N$ as the foot of altitudes from $F$ and $E$ onto $HF$ and $HE$ respectively, ergo $EN \cap FM = P$ in $\triangle HEF.$ We also observe that $$PC \overset{(\spadesuit)}{=} OS \overset{(\diamondsuit)}{=} \frac{1}{2} PH,$$so, $C$ is the center of $\odot(PMNH). - (\heartsuit)$ So, $CM=CN$, but as $M,N$ are midpoints of $BC,CD$ respectively, thus $BC=CD$. Hence, by Fact 5, we get $AC$ as the bisector of $\angle BAD. - (\#)$ Now, we let $K$ to be the center of $(HEF).-(\bigstar)$ As $KO, PC$ both are perpendicular to $EF$, hence $KO \parallel CP$, and it's well known that $KO = \tfrac 12 PH$, which is $PC$, thus $KOPC$ is a parallelogram, using $(\spadesuit)$ throughout. So, $KC \parallel OP.$ Now, as $\angle BCD = 180^{\circ} - \angle BAD$, so $$\angle 180^{\circ} - \angle EAF = \angle MCN \overset{(\heartsuit)}{=} 2\angle EHF \overset{(\bigstar)}{=} \angle EKF,$$so $A,E,K,F$ are concyclic. And as $KE=KF$ due to $(\bigstar)$, hence again by Fact 5, we get $AK$ as the bisector of $\angle EAF,$ which is $\angle BAD$. So, at last using $(\#)$, we get that $A, K, C$ are collinear, whence $AK \parallel OP.$ Now, as $PO \cap \{AD, AB\} = \{Q, R\}.$ So, $$\angle AQR = \angle EQP = \angle EAK \overset{(\#)}{=} \angle KAF = \angle PRF \overset{\text{ver.opp.}}{=} \angle ARQ \implies AQ=AR.\ \blacksquare$$

Oops, the construction of the diagram took me 1 hour and a half, but great problem Let $H$ be the orthocenter of $EPF$. We know that $CP\perp EF$, which means $C$ lies on the $P$ altitude. Furthermore, since $SO = CP$, and it is well known that $SO = \frac{1}{2} HP$ (draw in median and use euler line), we have $C$ is the midpoint of $PH$. Since $\angle OMC = \angle ONC = 90$, and $O$ is the midpoint of $EF$, we have $(OMNC)$ is the euler circle. Furthermore, since $M$ is on the euler circle and it lies on $PF$, it must be the foot of the altitude from $E$, so $\angle EMF = 90$. Similarly, $\angle ENF = 90$. Now, consider $\triangle EHF,$ and let $K$ be its circumcenter. Since $\angle HMP = \angle HNP = 90$, and $C$ is the midpoint of $PH$, we have $C$ is the center of $(HMPN)$, so $CM = CN$. Since $CM = MB, CN = ND$, we have $CD = CB$, and since $(ABCD)$ lies on a circle, this means $AC$ bisects $\angle BAD$. Furthermore, since $KE = KF$, and $\angle EKF = 2\angle EHF = \angle MCN = 180 - \angle BAD$, this means $(KEAF)$ lie on a circle, and $KA$ bisects $\angle BAD$. Thus, $A, K, C$ are collinear. I claim $KC || OP$. Since $O$ is the midpoint of $EF$, and $P$ is the orthocenter of $\triangle EFH$, reflecting $P$ over $O$ to get $P'$ results in $P'$ being the antipode of $H$. Since $K$ is the midpoint of $HP'$, and $C$ is the midpoint of $HP$, this means $CK || P'P = PO$. However, $CA$ bisects $\angle BAD$ and $A, K, C$ are collinear, so \[\angle AQR = \angle CAQ = \angle CAB = \angle ARQ\]This implies $AQ = AR$, which is the result we wanted to prove.

interesting how we all chose the point names H and K

Let $Q$ be the reflection of $P$ over $C$ and let $T$ be the reflection of $S$ over $O.$ $\textbf{Claim: }$ $P$ is the orthocenter of $\triangle QEF.$ $\emph{Proof: }$ Let $Q'$ be the point on $\overline{PC}$ such that $P$ is the orthocenter of $\triangle Q'EF.$ It is well known that $(Q'EF)$ is the reflection of $(PEF)$ across $\overline{EF}.$ Therefore, since $O$ is the midpoint of $\overline{EF},$ circles $(Q'EF)$ and $(PEF)$ are symmetrical with respect to $(ABCD).$ However, since $SP=OC$ and $\overline{SP}\parallel\overline{OC},$ we know $P$ and $C$ are corresponding points on $(PEF)$ and $(ABCD)$ respectively. Thus, $Q'=Q,$ as needed. $\blacksquare$ $\textbf{Claim: }$ $T$ is the circumcenter of $\triangle QEF.$ $\emph{Proof: }$ Shift the diagram by $P-C=S-O.$ Then, $T$ goes to $O$ and $(QEF)$ goes to $(ABCD),$ so we are done. $\blacksquare$ $\textbf{Claim: }$ $M,N$ are the feet of the altitudes from $F,E$ respectively in $\triangle QEF.$ $\emph{Proof: }$ Just note that $\overline{EN}\perp\overline{FQ}$ and $\angle ONC = 90^\circ$. $\blacksquare$ \ To finish, write \begin{align*} \angle FAE &= \angle DAB\\ &= 180^\circ-\angle BCD\\ &= 180^\circ-\angle MCN\\ &= 180^\circ-2\angle MQN\\ &= 180^\circ-\angle ETF, \end{align*}so quadrilateral $AEFT$ is cyclic. In particular, $AT$ bisects $\angle DAB$ and thus passes through $C.$ Since $OTCP$ is a parallelogram, $\overline{OP}$ is parallel to the bisector of $\angle DAB,$ implying the desired statement.

Very cool problem. I somehow guessed the cyclicity of $MNEF$ and more importantly that $C$ is the midpoint of arc $BD$ immediately after drawing the sketch and well L567 carried me for the rest of the solution. Notice that as $OS$ is the perpendicular bisector of $EF$, $CP$ is also perpendicular to $EF$ and we also know that the circumcircle of $OMCN$ is the nine-point circle of $\triangle HEF$ where $H$ is the orthocenter of $\triangle PEF$. Then we have that $C$ is the midpoint of $PH$ as it lies on the perpendicular from $H$ to $EF$ and the nine-point circle. Then, we have that $M$ and $N$ are the feet of perpendiculars from $E$ to $HF$ and $F$ to $HE$ meaning that $MEFN$ is cyclic and is also tangent to $BC$ and $CD$ as $OM$ and $ON$ are perpendicular to $BC$ and $CD$, respectively, and our also radii meaning that $CM = CD$ and consequently, $BC = CD$ and therefore $C$ lies on the angle bisector of $\angle BAD$. Now, let $O_1$ be the center of $\triangle HEF$. Then $O_1OPC$ is a parallelogram. Moreover, $$\angle EO_1F = 2\angle EHF = \angle BCD = 180^{\circ} - \angle BAD = \angle EAF $$meaning that $AEO_1F$ is cyclic and $EO_1 = EF$ meaning that $O_1$ also lies on the angle bisector of $\angle BAD$. Then $O_1C$ which coincides with the angle bisector of $\angle BAD$ is parallel to $PO$ and the conclusion follows readily. $\blacksquare$

Ahhh .... The problem involved a lot of construction as expected from a China TST problem. Claim 1- $H \in PC$ Let $H$ be the orthocentre of $\triangle PEF$ then $CP || OS \bot EF$ which implies $H \in PC$ as desired. Claim 2:- $FN \cap EM=H$ As $CP=OS=\frac{1}{2}PH$ which implies $C$ is the midpoint of $PH$ and $C$ is the circumcentre of $\odot (PMHN)$ as $\angle HMP=\angle HNP=90^\circ$ Constructing the nine-point circle of $\triangle HEF$,we see that $OMCN$ is the nine-point circle which imples $FN\cap EM=H$ as desired. Claim 3:- $A,O',C$ is collinear. Construct the circumcentre($O'$) of the $\triangle HEF$ $180^{\circ}-\angle EAF=\angle MCN=2\angle EHF=\angle EO'F$ $\implies AEO'F$ cyclic, and as $O'E=O'F \Rightarrow$ $AO'$ bisect $\angle EAF=\angle BAD$ which implies $AEO'F$ is cyclic. As $CM=CN$ which implies $CB=CD$ which implies $AC$ bisects $\angle BAD$ Hence $A,O',C$ are collinear. Also $OO'PC$ is a parallelogram. Hence $O'C||OP \Rightarrow AC||OP \Rightarrow \angle ARQ=\angle AQR \Rightarrow AQ=AR$ as desired.

I might be missing something, but how do people conclude $M$ and $N$ are the feet of the altitudes from them being on the Euler circle and on the altitudes, can't they be the midpoints of $FP$ or $EP$?

[asy][asy] unitsize(0.8cm); pair A, B, C, D, E, F, H, M, N, O, P, S, O1, Q, R; O=(0,0); A=(-4sqrt(31)/7,27/7); E=(-4,0); F=(4,0); B=intersectionpoints(circle(O,5),E--2*E-A)[0]; D=intersectionpoints(circle(O,5),F--2*F-A)[0]; C=extension(A,incenter(A,B,D),O,B+D); M=(B+C)/2; N=(C+D)/2; P=extension(E,N,F,M); S=circumcenter(P,E,F); H=orthocenter(P,E,F); Q=extension(P,O,A,D); R=extension(P,O,A,B); O1=E+F-S; draw(circle(O,5)); draw(A--B--C--D--A--E--F--M--E--N--F--H--E--H--P--S--O1--O--C--A); draw(P--R--A); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SW); label("$D$", D, dir(0)); label("$E$", E, W); label("$F$", F, NE); label("$H$", H, dir(-90)); label("$M$", M, SW); label("$N$", N, SE); label("$O$", O, dir(150)); label("$P$", P, NNE); label("$S$", S, dir(90)); label("$O'$", O1, W); label("$Q$", Q, SW); label("$R$", R, dir(90)); [/asy][/asy] Let $H$ be the orthocenter of $PEF$. Then, since $OSPC$ is a parallelogram, this means that $C$ is the midpoint of $PH$, so the circle with diameter $CO$ is the nine-point circle of $PEF$. Since $M$ and $N$ lie on the circle with diameter $CO$, this means that $M$ is either the midpoint of $PF$ or $M$ is the foot of the altitude from $E$ to $PF$. Suppose $M$ is the midpoint of $PF$. Then, if $E$ and $F$ are inside the circumcircle of $ABCD$, then segments $EN$ and $FM$ intersect, so $M$ cannot be the midpoint of $PF$. Therefore, $E$ and $F$ are outside the circumcircle of $ABC$. However, since $BPCF$ and $OSPC$ are parallelograms, we get $OS\parallel PC\parallel BF$, so $\angle BFE=90^{\circ}$, contradiction. Therefore, $M$ is the foot from $E$ to $PF$. Similarly, $N$ is the foot from $F$ to $PE$, so $CM=CN$ implies $CB=CD$. Now, $AQ=AR$ follows from $OP\parallel AC$. Let $O'$ be the circumcenter of $HEF$. Then, $OO'CP$ is a parallelogram, so it suffices to show $AC$ passes through $O'$. Since $\angle EO'F=2\angle EHF=\angle MCN=180^{\circ}-\angle BAD$, we get $AEO'F$ is cyclic. Therefore, $O'E=O'F$ implies $\angle EAO'=\angle O'AF$, so the angle bisector of $\angle BAD$ passes through $A$, $O'$, and $C$.

Solution from Twitch Solves ISL: We let $H$ denote the orthocenter of $\triangle PEF$. [asy][asy]size(12cm); pair E = dir(142.15); pair F = dir(37.85); pair H = dir(275); pair P = E+H+F; pair M = extension(F, P, E, H); pair N = extension(E, P, F, H); pair C = midpoint(H--P); pair D = 2*N-C; pair B = 2*M-C; pair A = extension(B, E, D, F); filldraw(E--F--H--cycle, invisible, deepgreen); pair O = midpoint(E--F); filldraw(circumcircle(C, M, N), invisible, red); draw(CP(O, C), blue); draw(circumcircle(P, E, F), blue); filldraw(A--B--C--D--cycle, invisible, deepcyan); pair S = circumcenter(P, E, F); draw(M--F, lightcyan); draw(N--E, lightcyan); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$H$", H, dir(H)); dot("$P$", P, dir(P)); dot("$M$", M, dir(M)); dot("$N$", N, dir(N)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$B$", B, dir(B)); dot("$A$", A, dir(A)); dot("$O$", O, dir(O)); dot("$S$", S, dir(210)); /* TSQ Source: E = dir 142.15 F = dir 37.85 H = dir 275 P = E+H+F M = extension F P E H N = extension E P F H C = midpoint H--P D = 2*N-C B = 2*M-C A = extension B E D F E--F--H--cycle 0.1 lightgreen / deepgreen O = midpoint E--F circumcircle C M N 0.1 lightred / red CP O C blue circumcircle P E F blue A--B--C--D--cycle 0.1 lightcyan / deepcyan S = circumcenter P E F R210 M--F lightcyan N--E lightcyan */ [/asy][/asy] Let $\omega$ denote the circle with diameter $\overline{OC}$, passing through $M$ and $N$. Claim: The circle $\omega$ is the nine-point circle of $\triangle PEF$ (or $\triangle HEF$ if you prefer). Proof. We observe a few facts: Clearly $\omega$ has radius half that of $(O)$. Since $SP = OC$, the circles $(S)$ and $(S)$ are congruent, hence the radius of $\omega$ is half that of $(PEF)$ too. Point $O$ is the midpoint of $\overline{EF}$, The antipode of $O$ --- namely $C$ --- is known to lie on the $P$-altitude (because $\overline{SO} \perp \overline{CP}$ and $\overline{SO} \parallel \overline{EF}$). $\blacksquare$ Claim: $\overline{AC}$ bisects $\angle BAD$. Proof. We have $OM = ON$, so $BC = CD$. $\blacksquare$ Claim: We have $\overline{CA} \parallel \overline{OP}$. Proof. From $ABCD$ is cyclic, we can compute \[ \measuredangle EAF = \measuredangle BAD = \measuredangle BCD = \measuredangle MCN = \measuredangle MON = 2 \measuredangle FHE \]so $A$ lies on the circle through $E$, $F$, and the circumcenter of $\triangle HEF$. Denote this circumcenter by $W$. As $WE = WF$, so this implies $\overline{AW}$ bisects $\angle EAF$, and hence $AWC$ are collinear. Since $OWCP$ is a parallelogram, this completes the proof. $\blacksquare$ This completes the solution.

This problem makes me happy Claim: $M$ and $N$ are the feet of the altitudes from $F$ and $E$ to $\overline{EP}$ and $\overline{FP}$, respectively. Proof: Let $C'$ be the reflection of $P$ over $C$. Because $\overline{CP} \perp \overline{EF}$ and $CP=SO$, $(CO)$ is the nine-point circle of triangle $C'EF$. But $\angle OMC = \angle ONC = 90^\circ$, so $M$ and $N$ must be the feet of the altitudes. $\blacksquare$ It follows that $(EMNF)$ has center $O$, which implies $CB=CD$. Now, let $O'$ be the circumcenter of triangle $C'EF$. Claim: $EO'FA$ is cyclic. Proof: Because $\angle FAE = 2\angle DAC = 2\angle CMN = 180^\circ - 2\angle EC'F = 180^\circ - \angle EO'F$. $\blacksquare$ But $EO' = O'F$, so $\overline{AO'}$ bisects $\angle EAF$. It follows that $O'$ lies on $\overline{AC}$, so $\overline{AC} \parallel \overline{OP}$ by homothety at $C'$. Thus $\overline{OP}$ is parallel to the bisector of $\measuredangle(\overline{AB}, \overline{AD})$ and desired result follows.