My solution: Let $R=PQ \cap BC$ and $T$ be the midpoint of $BC$ . Let $O$ be the center of $\odot (ABC)$ and $M$ be the midpoint of arc $BAC$ . Let $N \equiv AI \cap \odot (O)$ be the midpoint of arc $BC$ and $G$ be the midpoint of $IM$ . From $(B,C;D,R)=-1 \Longrightarrow RP \cdot RQ=RB \cdot RC=RD \cdot RT \Longrightarrow T \in \odot (DPQ)$ . Since $ID \perp BC, MN \perp BC$ , so $G$ lie on the perpendicular bisector of $DT$ . From $OG \parallel NI \equiv AI \Longrightarrow OG \perp AM \Longrightarrow OG \perp PQ$ , so $G$ lie on the perpendicular bisector of $PQ \Longrightarrow G$ is the center of $\odot (DPQ)$ . Since $O_1O_2$ is the perpendicular bisector of $AI$ , so from $GA=GI$ we get $G$ lie on $O_1O_2$ . i.e. $O_1O_2$ pass through the center of $\odot (DPQ)$ Q.E.D

Nice Solution. TelvCohl wrote: Let $R=PQ \cap BC$. Yeah, when $AB=AC$...

In another thread regarding this problem, izat wrote: Let $M$ - midpoint of $BC$ .Let $PQ$ and $BC$ intersect at $R$(w.l.o.g assume that AC>BC) , then $R;D;B;C$ - harmonic.So, $(1)RB*DC=BC*RC$. Let $RB=x, BD=y,DM = z$ then from (1) we get $(2)xz=y^2+zy$. $RP*RQ= RB*RC$ ,however, $RB*RC=RD*RM$ it follows from (2) . So $D,P,Q,M$ are concyclic.It's well-known that $O_1$ and $O_2$ are intersection of $CI$ and $BI$ with the circumcircle of $ABC$ and that $O_1O_2$ is perpendicular to $AI$. But $PQ$ is also perpendicular to $AI$ , so $PQ$ and $O_1O_2$ are parallel. Let $N$ be midpoint of arc $CAB$ , then it is easy to prove that $NO_2IO_1$ is parallelogarm. Let $K$ be intersection $O_1O_2$ and $NI$ so $K$ is midpoint of $O_1O_2$. $ID$ perpendicular to $BC$ , $NM$ perpendicular to $BC$, so $ID$ and $NM$ are parellel, that's why $KD=KM$. Moreover, $KO$ perpendicular to $PQ$, where $O$ is circumcenter of $ABC$. So perpendicular bisectors of $DM$ and $PQ$ intersect at $K$, so $K$ is circumcenter of $DPQ$ and lies on $O_1O_2$

We claim that the circumcenter of $\triangle DPQ$ must be the midpoint of $O_1O_2$. Case 1. $AB=AC$. From Fact 5, we know that $O_1, O_2$ lie on the circumcircle of $\triangle ABC$. Now take $M=O_1O_2 \cap AD$ and $N=PQ \cap AD$. Note the symmetry wrt $AD$. Let $AD$ hit the circumcircle of $\triangle ABC$ at $K$. We show that $M$ is the circumcenter of $\triangle DPQ$. First, note that $O_1AO_2I$ is a parallelogram by a simple angle chase. This gives that $M$ is the midpoint of $AI$. Since symmetry gives us $MP=MQ$, we just need to show that $MP=MD$. Now let $a=AM$, $b=MN$. We get $NI=MI-MN=AM-MN=a-b$. Also, since $\triangle IFN \sim \triangle IAF$, we have $ID = IF = \sqrt{IN \cdot IA} = \sqrt{2a(a-b)}$ Also, as $\angle AFI = \angle ADB = 90$ and $\angle ANF = \angle ABD = 90$, we have $(F,N,K,B)$ and $(F,I,D,B)$ cyclic. This gives us $AF \cdot AB = AN \cdot AK = AI \cdot AD$. We are ready to compute. We calculate $$MP^2=MN^2+PN^2 = MN^2+PN \cdot NQ = MN^2+AN \cdot NK = MN^2 + AN \cdot AK - AN^2$$ $$=MN^2+AI \cdot AD - AN^2 = b^2+2a \cdot (2a+\sqrt{2a(a-b)}) - (a+b)^2 = 3a^2-2ab+2a\sqrt{2a(a-b)} = (a+\sqrt{2a(a-b)})^2 = MD^2$$ so $MP=MD=MQ$, giving $M$ as the circumcenter of $\triangle DPQ$ as required. Case 2. $AB \not= AC$. Take $R$ as the midpoint of arc $BC$, and $N$ as the midpoint of arc $BAC$. Take $M$ as the midpoint of $BC$. Lemma. Take $D, E, F \in BC, CA, AB$ such that $AD, BE, CF$ concur. Take $M$ as the midpoint of $BC$, let $EF$ hit the circumcircle of $\triangle ABC$ at $P,Q$. Then, $D, M, P, Q$ are cyclic. Proof of Lemma. Taking $BC \cap EF =X$, we have $(X,D,B,C)$ harmonic, so $$XP \cdot XQ = XB \cdot XC = XD \cdot XM$$ giving $D, M, P, Q$ cyclic as required. Now using this lemma, we have $D, M, P, Q$ cyclic. Take $O'$ as the midpoint of $IN$. As $O_1NO_2I$ is a parallelogram by a simple angle chase, we know that $O'$ is also the midpoint of $O_1O_2$. Now since $ID \perp BC$ and $NM \perp BC$, we know that $O'$ lies on the perpendicular bisector of $DM$. Also, as $O'O \parallel IR = AI$ and $AI \perp EF = PQ$, we have $O'O \perp PQ$. Since $O$ lie on the perpendicular bisector of $PQ$, so does $O'$. This gives us that $O'$ is the circumcenter of $\triangle DPQ$, as desired. $\blacksquare$

$EF\cap BC=\{A'\}$ $\implies$ $(B,C;D,A')=-1$ let $M$ be the midpoint of $BC$ $\implies$ $A'B\cdot A'C=A'E\cdot A'F=A'D\cdot A'M$ $\implies$ $M\in \odot PQD$.$PQ||O_1O_2$ and so its perpendicular bisector passes thru the midpoint of $O_1O_2$ say $R$.Let $I_B,I_C$ be the excenters and let $X'$ be the projection of $X$ on $BC$. $$DO_1'=\tfrac{DI_B'}{2}=\tfrac{b}{2}=\tfrac{c}{2}-\tfrac{c-b}{2}=DO_2'-DM=MO_2' \implies MR=RD$$ And hence the perpendicular bisector od $MD$ also passes thru $R$ and so $R$ is the desired circumcenter.

Here's a nice barybash. Let $\omega$ be the circle $(DPQ)$. If $X=EF\cap BC$ and $M$ is the midpoint of $\overline{BC}$, then $$(B,C;D,X)=-1\implies XD\cdot XM=XB\cdot XC=XP\cdot XQ,$$ so $M$ lies on $\omega$. Note that $E=(s-c:0:s-a)$ and $F=(s-b:s-a:0)$. Hence the equation of $EF$ is $$-(s-a)x+(s-b)y+(s-c)z=0.$$ Then $\omega$ has equation $$-a^2yz-b^2zx-c^2xy+k(-(s-a)x+(s-b)y+(s-c)z)(x+y+z)=0,$$ for some $k$. Plugging in $M=(0:1:1)$ gives $k=\frac{a}{2}$. Since $O_1O_2$ is the perpendicular bisector of $\overline{AI}$, it suffices to show that $A,I$ have equal powers with respect to $\omega$. Easily we see that $\operatorname{Pow}(A,\omega)=-\frac{a}{2}(s-a)$ and now $$\begin{align*}\operatorname{Pow}(I,\omega)&=\frac{1}{a+b+c}\left(-abc+\frac{a}{2}(-a(s-a)+b(s-b)+c(s-c))\right)\\&=\frac{1}{a+b+c}\left(\frac{a}{2}\left(2s(s-a)+a^2-b^2-c^2-2bc\right)\right)\\&=-\frac{a}{2}(s-a).\end{align*}$$ The result follows.

Let $M$ and $N$ be the midpoints of $BC$ and $O_1O_2$, respectively, $K'$ be the midpoint of arc $BAC$ and $K$ be the antipode of $K$ in $\odot(ABC)$. Claim: $M, D, P, Q$ are concyclic. Proof: Let $EF$ intersect $BC$ at $R$. Note that $(B, C; D, R) = -1$ and thus, $D$ and $R$ are inverses wrt $\odot(M, MB)$. Therefore, we have $$RD\cdot RM = RB\cdot RC = RP\cdot RQ$$ Note that $O_1, O_2$ are the midpoint of arc $AB$, not containing $A$, and $AC$, not containing $B$, respectively. $I$ is the orthocenter of $\triangle KO_1O_2$. Thus, the midpoint of $IK'$ is the midpoint of $O_1O_2$. The perpendicular biscector of $DM$ passes through the midpoint of $IK'$ which is $N$. The perpendicular bisector of $PQ$ passes through $N$ since $PQ\parallel O_1O_2$. Thus, $N$ is the centre of $\odot(ABC)$. [asy][asy] unitsize(2.5cm); import geometry; pair A = dir(130), B = dir(210), C = -1/B; triangle t = triangle((point) A, (point) B, (point) C); circle c = circumcircle(t); pair I = incenter(t), O_1 = circumcenter(I, A, B), O_2 = circumcenter(I, A, C), K = circumcenter(I, B, C), K_ = -K; pair D = foot(I, B, C), E = foot(I, C, A), F = foot(I, A, B); pair P = intersectionpoints(line(E, F), c)[0], Q = intersectionpoints(line(E, F), c)[1]; pair N = (O_1 + O_2)/2, M = (B + C)/2, R = extension(P, Q, B, C); draw(t); draw(circumcircle(t)); dot(M); dot(A^^B^^C); dot(O_1^^O_2^^K^^K_); dot(I); dot(P^^Q); dot(D^^E^^F); draw(D--P--Q--cycle, linewidth(1.1)); draw(circumcircle(D, P, Q), dashed+linewidth(0.4)); dot(R); draw(P--R--B); dot(N); draw(O_1--O_2^^I--K_, linetype("1 2")+linewidth(0.6)); label("A", A, A); label("B", B, B*dir(10)); label("C", C, C); label("P", P, P*dir(-20)); label("Q", Q, Q); label("K", K, K); label("K$'$", K_, K_); label("O$_1$", O_1, O_1); label("O$_2$", O_2, O_2); label("D", D, (0,-1)); label("M", M, (0,-1)); label("E", E, dir(80)); label("F", F, dir(120)); label("I", I, dir(0)); label("R", R, -dir(0)); label("N", N, dir(120)); [/asy][/asy]

Again it helps to know one's lemmas . Anyway, here's my solution (Nothing new): Let $M$ be the midpoint of $BC$, and let $N$ be the midpoint of $\overarc{BAC}$. From here, we get that if $EF \cap BC=X$, then $XD \cdot XM=XB \cdot XC=XP \cdot XQ$, which means that $M$ lies on $\odot (DPQ)$. Let $J$ be the midpoint of $IN$. Then, as $IO_1NO_2$ is a parallelogram (easy angle chase), we get that $J$ lies on $O_1O_2$ also. And, as $EF \parallel O_1O_2$, so $OJ$ is the perpendicular bisector of $PQ$, where $O$ is the center of $\odot (ABC)$. Also, $ID \parallel NM$, which means that $J$ lies on the perpendicular bisector of $MD$ also. Thus, $J$ is nothing but the center of $\odot (DMPQ)$, which lies on $O_1O_2$. $\blacksquare$

Solution as far in my memory. It suffices to prove that $I, O=$ circumcenter $\odot (DPQ), M=$ midpoint $\overarc{BAC}$ are all collinear. Now, invert in the incircle. Let $H$ be the orthocenter of $\triangle DEF$ and $M'=$ midpoint $DH.$ Note that $\overline{I,M,M'},$ so we have to prove that $\odot DEF, \odot (DH), \odot (AP^*Q^*)$ share a common radical axes where $P^*,Q^* \equiv \odot EIF \cap$ N.P.C. of $\Delta DEF$ which follows easily since $EF,$ line joining alt. feets $,P^*Q^*,$ and the radical axes are all concurrent.

rkm0959 wrote: In a triangle $\triangle ABC$ with incenter $I$, the incircle meets lines $BC, CA, AB$ at $D, E, F$ respectively. Define the circumcenter of $\triangle IAB$ and $\triangle IAC$ $O_1$ and $O_2$ respectively. Let the two intersections of the circumcircle of $\triangle ABC$ and line $EF$ be $P, Q$. Prove that the circumcenter of $\triangle DPQ$ lies on the line $O_1O_2$.

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Let $H$ be the orthocenter of $\triangle DEF$, $M$,$N$ and $G$ be the midpoint of $EF$ and $HD$ respectively. We will show that $M$ is the circumcenter of $\triangle DPQ$. Indeed, since $O_1O_2$ and $PQ$ are both perpendicular to $AI$, $M$ lies on the perpendicular bisector of $PQ$. CLAIM. $G,I$ and the circumcenter of $\triangle DPQ$ are collinear. Proof. Invert w.r.t. the incircle, then $P',Q'$ are the second intersections of the circles $(EIF)$ and the nine-point circle of $\triangle DEF$. Let $P'Q'$ intersect $EF$ at $V$. Let $B$ be the second intersection of $(DEF)$ and $(DBH_DH_E)$, then $V$ is the radical center of the nine-point circle, $(IEF)$ and $(DEF)$. Therefore, $V,B,D$ are collinear. Hence $VBF'Q'$ is cyclic. Therefore from $GD=GB$, $I,G$ and the circumcenter of $(DBP'Q')$ are collinear. $\blacksquare$ [asy][asy] size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -19.740708580371834, xmax = 13.960468482885158, ymin = -22.1023460884845, ymax = 6.321778338532966; /* image dimensions */pen qqwuqq = rgb(0,0.39215686274509803,0); pen zzttff = rgb(0.6,0.2,1); /* draw figures */draw(circle((-1.592707484217268,-12.268506667683056), 8.721686845166335), linewidth(0.8) + blue); draw(circle((-4.128410707846865,-4.385808887405556), 4.309679157122033), linewidth(0.8) + blue); draw(circle((-4.9475581856959066,-1.8393389593838159), 5.413472031345108), linewidth(0.8) + linetype("4 4") + qqwuqq); draw(circle((-1.425692272557547,-3.5484190910458966), 8.619358314244066), linewidth(0.8) + qqwuqq); draw((-6.668010004233001,3.2934697234182364)--(-11.957561564910876,-7.606599262359754), linewidth(0.8) + zzttff); draw((-11.957561564910876,-7.606599262359754)--(-2.5303362432022625,-0.3833730737062355), linewidth(0.8) + zzttff); draw((-11.957561564910876,-7.606599262359754)--(5.98492302129587,-7.950250382205761), linewidth(0.8) + zzttff); draw((5.98492302129587,-7.950250382205761)--(-6.668010004233001,3.2934697234182364), linewidth(0.8)); draw((-11.957561564910876,-7.606599262359754)--(0.12929282178150386,-3.718504540707435), linewidth(0.8) + zzttff); draw((-6.668010004233001,3.2934697234182364)--(-6.8311291431361845,-5.223198683765217), linewidth(0.8)); draw((-6.668010004233001,3.2934697234182364)--(-8.999426705314148,-7.663256207069486), linewidth(0.8)); /* dots and labels */dot((-6.668010004233001,3.2934697234182364),dotstyle); label("$D$", (-6.5480769897730084,3.593302259568209), NE * labelscalefactor); dot((-8.999426705314148,-7.663256207069486),dotstyle); label("$E$", (-8.8867707717428,-7.3505853099058145), NE * labelscalefactor); dot((5.98492302129587,-7.950250382205761),dotstyle); label("$F$", (6.104856035755863,-7.650417846055788), NE * labelscalefactor); dot((-1.425692272557547,-3.548419091045896),linewidth(4pt) + dotstyle); label("$I$", (-1.3010076071484764,-3.3028460718811754), NE * labelscalefactor); dot((-1.507251842009139,-7.806753294637623),linewidth(4pt) + dotstyle); label("$A'$", (-1.3909573679934684,-7.560468085210796), NE * labelscalefactor); dot((-0.34154349146856555,-2.3283903293937622),linewidth(4pt) + dotstyle); label("$B'$", (-0.22161047700857273,-2.1035159272812827), NE * labelscalefactor); dot((-7.833718354773575,-2.1848932418256246),linewidth(4pt) + dotstyle); label("$C'$", (-7.717423880757904,-1.9535996592062959), NE * labelscalefactor); dot((-7.976646107827376,-6.326019892340046),linewidth(4pt) + dotstyle); label("$P'$", (-7.8673401488328905,-6.091288658075927), NE * labelscalefactor); dot((0.12929282178150386,-3.718504540707435),linewidth(4pt) + dotstyle); label("$Q'$", (0.2581215808313845,-3.4827455935711593), NE * labelscalefactor); dot((-11.957561564910876,-7.606599262359754),linewidth(4pt) + dotstyle); label("$V$", (-11.825129626012538,-7.380568563520812), NE * labelscalefactor); dot((-10.044276518023667,-3.663932523734483),linewidth(4pt) + dotstyle); label("$B$", (-11.01558177840761,-3.81256138333613), NE * labelscalefactor); dot((-6.8311291431361845,-5.223198683765217),linewidth(4pt) + dotstyle); label("$H$", (-6.697993257847995,-4.981908274321026), NE * labelscalefactor); dot((-8.408789520505717,-4.887497692815122),linewidth(4pt) + dotstyle); label("$H_D$", (-8.287105699442852,-4.652092484556055), NE * labelscalefactor); dot((-2.5303362432022625,-0.3833730737062355),linewidth(4pt) + dotstyle); label("$H_E$", (-2.4103879909033776,-0.1546044423064564), NE * labelscalefactor); dot((-6.749569573684592,-0.9648644801734902),linewidth(4pt) + dotstyle); label("$G$", (-6.638026750618001,-0.7242862609914056), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Now it suffices to show that $G,I,M$ are collinear. Notice that $\triangle O_1O_2O_3$ and $\triangle EDF$ are homothetic. Since $I$ is the orthocenter of $\triangle O_1O_2O_3$, $NH\|IM$. Meanwhile, $H,N$ and the antipode of $D$ w.r.t. the incircle are collinear, so $GI\|NH$. This completes the proof.