Let $K=EF\cap PQ$. Obviously $A, I, Q$ are colinear where $I$ is the incenter. This implies that $DR\perp EF$ and $DRKM$ is cyclic. Now let $T=EF\cap BC$. Then $TR\cdot TK=TD\cdot TM$ and it suffices to show that $TB\cdot TC = TD\cdot TM$. To do that, let $\Delta DB_1C_1$ be the orthic triangle of $\Delta BIC$. Recall that $B_1, C_1\in EF$ (Iran) and quadrilaterals $BB_1C_1C$ and $B_1C_1DM$ are cyclic. Hence $TB\cdot TC=TB_1\cdot TC_1=TD\cdot TM$ as desired.

To show $TB \cdot TC = TD \cdot TM$, just notice the harmonic bundle.

Oops, I am a bit hasty . Note that $(B,C;D,T)=-1$, therefore $MD\cdot MT=MB^2$ hence $$TD\cdot TM = TM^2-MD\cdot MT = (TM+MB)\cdot(TM-MB)=TB\cdot TC$$as desired.

It is clear that $\overline{DR} \perp \overline{EF}$. Now we just need $\overline{EF}$ to be the external bisector of $\angle BRC$; however that follows since $(B,C,D, \overline{EF} \cap \overline{BC})=-1$.

Always sad that Korean geo problems are much easier than US ones. Also sad that I cannot take KMO anymore since I moved to the US. I tried this problem and it was really easy to prove by length bash. Not even a bash. Outline: Let $X=PQ\cap EF, Y=EF\cap PQ.$ WTS: $YR\cdot YX=YB\cdot YC$ Since $(R,D,M,X)$ is cyclic where $M$ is the midpoint of the $BC,$ we want to prove that $YM\cdot YD=YB\cdot YC.$ Let $BC=a, CA=b, AB=c.$ Then by Menelaus on $\triangle ABC$ and line $YRX$, we get $YB=\frac{a(c+a-b)}{2(b-c)}.$ $\Leftrightarrow YD\cdot YM=YB YC$ $\Leftrightarrow (YB+BD)(YC-CM)=YB\cdot YC$ $\Leftrightarrow BD\cdot YC=YB\cdot CM+ BD\cdot CM$ $\Leftrightarrow BD(YB+YC)=YB\cdot CM+ BD\cdot CM$ $\Leftrightarrow BD\cdot BC=YB(CM-BD)+ BD\cdot CM$ $\Leftrightarrow BD\cdot BC=YB\cdot DM+ BD\cdot BM$ $\Leftrightarrow BD\cdot BM=YB\cdot DM$ $\Leftrightarrow \frac{(c+a-b)}{2}\cdot \frac{a}{2}=\frac{a(c+a-b)}{2(b-c)}\cdot \frac{b-c}{2}$ And we are done.

Lemma 9.17 EGMO kills it

Sweet problem rkm0959 wrote: Let there be a scalene triangle $ABC$, and its incircle hits $BC, CA, AB$ at $D, E, F$. The perpendicular bisector of $BC$ meets the circumcircle of $ABC$ at $P, Q$, where $P$ is on the same side with $A$ with respect to $BC$. Let the line parallel to $AQ$ and passing through $D$ meet $EF$ at $R$. Prove that the intersection between $EF$ and $PQ$ lies on the circumcircle of $BCR$. Denote $\odot (BCR)$ by $\Omega$. Also, let $T = \overline{RD} \cap \Omega$.Consider the following claim: Lemma: $T$ is the mid-point of arc $\widehat{BC}$ of $\Omega$ not containing $R$. Proof: Let $\overline{EF} \cap \overline{BC} = K$. Then it is obvious that $(KD;BC) = -1$. Also the given condition $\overline{DR} \parallel \overline{AQ} \implies \overline{DR} \perp \overline{EF}$. Now both these combined imply that $\overline{DT}$ is the internal angle bisector of $\angle BRC \iff T$ is the mid-point of arc $\widehat{BC}$ of $\Omega$ not containing $R$. $ $ Hence, $TB = TC \implies T \in \overline{PQ}$. Now $\angle SRT = 90^{\circ} \implies S$ is the antipode of $T$ wrt $\Omega$.

As this is my first solve after 2020 , so I'll post my solution even if it is same as others. 2017 KJMO P3 wrote: Let there be a scalene triangle $ABC$, and its incircle hits $BC, CA, AB$ at $D, E, F$. The perpendicular bisector of $BC$ meets the circumcircle of $ABC$ at $P, Q$, where $P$ is on the same side with $A$ with respect to $BC$. Let the line parallel to $AQ$ and passing through $D$ meet $EF$ at $R$. Prove that the intersection between $EF$ and $PQ$ lies on the circumcircle of $BCR$. Note that $AD,BE,CF$ are concurrent at the Gregonne Point of $\triangle ABC$. Hence, if $EF\cap BC=K$. Then $(K.D;B,C)=-1$. Let $EF\cap PQ=T$ also notice that $PQ$ bisects $BC$ let $PQ\cap BC=M$. Also $\overline{A-I-Q}$ where $I$ is the incenter of $\triangle ABC$. So, $\angle DRT=90^\circ$. So Combining Maclaurin and PoP we get that $$KB.KC=KD.KM=KR.KT\implies T=EF\cap PQ\in\odot(BCR).\blacksquare$$

It's still 2019 in Azerbaijan ! And it's about 15-20 minutes left to $\text{2020}$ , so I hurry up (Sorry for having a similar solution , HnY ) Let $T$ be the midpoint of $BC$ and let $PQ\cap{EF}=X$ $EF\cap{BC}=Z$ $(\xi)$ $ZD.ZT=ZB.ZC$

Also clearly $Q$ is the midpoint of the arc $\overarc{BC}$ , so $\overline{A-I-Q}$ And $AQ\perp{EF}$ , thus $DR\perp{EF}\rightarrow{DRXT}$ is cyclic . $\text{PoP}\rightarrow{ZD.ZT=ZR.ZX}$ $(\omega)$ Combining $(\xi)$ and $(\omega)$ $ZR.ZX=ZB.ZC$ $\blacksquare$

Let $M$ be the midpoint of $\overline{BC},$ and $X,Y,$ and $Z$ the intersection of $\overline{EF}$ with $\overline{PQ},\overline{BC},$ and $\overline{AQ},$ respectively. Notice that $\triangle AFZ\cong\triangle AEZ$ so $$\measuredangle XMD=90=\measuredangle RZQ=\measuredangle XRD$$and $XRDM$ is cyclic. By the Midpoint Lengths Lemma and Power of a Point, $$YB\cdot YC=YD\cdot YM=YR\cdot YX$$and $BRXC$ is cyclic. $\square$