Remark: This problem is also the Problem 1 of Grade 11.

The most beautiful problem in this year's CSMO Grade 10. Here is my solution which use only angle chasing. Obviously, $KA=KD=KE=KF\implies KO_1\perp DE$. Therefore $180^{\circ}-\angle KDO_1=\angle DKO_1+\angle KO_1D=\angle EAD+\angle EMD=\angle AEK+\angle EMB=\angle B$. Similarly $\angle KDO_2=180^{\circ}-\angle C$. Now let $O_2'$ be a point on $KO_2$ such that $O_1O_2'\parallel BC$. Therefore $\Delta ABC, \Delta KO_1O_2'$ are homothetic and therefore $KD$ bisects $O_1O_2'$. Draw a parrallelogram $KO_1LO_2'$. It follows that $K,D,L$ are colinear. Since $\angle LDO_1=\angle KO_1O_2'=\angle LO_2'O_1\implies DO_1LO_2'\text{ is cyclic }$. Thus $\angle KDO_2'=180^{\circ}-\angle LO_1O_2'=180^{\circ}-\angle C=\angle KDO_2$. Hence $O_2'=O_2$ and we are done.

Let $\odot(DEM)$ and $\odot(DFN)$ meet for a second time at $X.$ It is enough to prove that $DX \perp BC.$ After inversion with pole $D$, we obtain the following new problem: Let $D$ be the midpoint of $\overline{BC}.$ Choose points $E, F$ such that $\angle DBE = \angle DCF = 90^{\circ}.$ Let $A$ be the projection of $D$ onto $EF$ and let $K$ be the reflection of $D$ in $EF.$ Let $BC$ meet $\odot(DEK)$ and $\odot(DFK)$ for a second time at $M$ and $N$ respectively. Lines $EM$ and $FN$ meet at $X.$ Prove that $DX \perp BC.$ Proof: Let $W$ be the midpoint of $\overline{EF}$, and note that $DW \perp BC.$ Let $\infty_{EF}$ and $\infty_{DW}$ be points at infinity on $EF$ and $DW.$ Since $EF$ is the perpendicular bisector of $\overline{DK}$, the center of $\odot(DEMK)$ lies on $EF.$ Because the circumcenter and orthocenter are isogonal conjugates, it follows that $ED, EM$ are isogonal WRT $\angle BEF.$ Similarly, $FD, FN$ are isogonal WRT $\angle CFE.$ Since $BE \parallel CF$, there is a line $\ell$ parallel to both a bisector of both $\angle BEF$ and $\angle CFE.$ Reflecting in $\ell$, we find \[ -1 = D(E, F; W, \infty_{EF}) = (EM, FN, D\infty_{EF}, D\infty_{DW}) = X(E, F, \infty_{EF}, \infty_{DW}). \]Of course, we also have $-1 = (E, F; \infty_{EF}, W).$ Therefore, $X \in DW$, so $DX \perp BC.$

My solution : Let $AB$ intersects $(DEM)$ again at $Q$, $AC$ intersects $(DFN)$ again at $P$ and $(DEM)$ intersects $(DFN)$ again at $X$. Let $BB_0$ and $CC_0$ be the altitudes of triangle ABC. It easy to see that $EA-EB = AC_0$ and $FA-FC = AB_0$. Since $\angle DMQ = 90^{\circ}$, then DQ is the diameter of $(DEM)$ so $O_1$ is the midpoint of $DQ$. Similarly, $O_2$ is the midpoint of $DP$. So $O_1O_2$ is parallel to $PQ$. It suffices to prove that $PQ$ is parallel to $BC$. By power of point, $QB \cdot BE = BD \cdot BM$. So $QB = \frac{BD \cdot BM}{BE}$. Similarly, $CP = \frac{CD \cdot CN}{CF}$. By Menelaos on $\triangle ABD$ and transversal $MEK$ we have \[ \frac{DM}{MB} \cdot \frac{BE}{EA} \cdot \frac{AK}{KD} = 1\] So $\frac{BM}{BE} = \frac{DM}{EA}$. Similarly, $\frac{CN}{CF} = \frac{DN}{FA}$. This gives $\frac{DM}{BM} = \frac{EA}{EB}$ and so $\frac{DM}{BD} = \frac{EA}{EB - EA} = \frac{EA}{AC_0}$. Similarly, $\frac{DN}{CD} = \frac{FA}{AB_0}$. Hence \[\frac{BQ}{CP} = \dfrac{\frac{BD \cdot BM}{BE}}{\frac{CD \cdot CN}{CF}} = \frac{BM}{BE} \cdot \frac{CF}{CN} = \frac{DM}{EA} \cdot \frac{FA}{DN} = \frac{AB_0}{AC_0} \] Since $ \frac{AB_0}{AC_0} = \frac{AB}{AC}$. Then, $\frac{BQ}{CP} = \frac{AB}{AC}$. Hence PQ is parallel to BC and we are done.

CQYIMO42 wrote: Let $ABC$ be an acute-angled triangle. In $ABC$, $AB \neq AC$, $K$ is the midpoint of the the median $AD$, $DE \perp AB$ at $E$, $DF \perp AC$ at $F$. The lines $KE$, $KF$ intersect the line $BC$ at $M$, $N$, respectively. The circumcenters of $\triangle DEM$, $\triangle DFN$ are $O_1, O_2$, respectively. Prove that $O_1 O_2 \parallel BC$. Let $\overline{AL}$ be an altitude in $\triangle ABC$. Let $X, Y$ be antipodes of $D$ in $\odot(DEN), \odot(DFM)$ respectively. Observe that $AX:AB=AY:AC$ is equivalent to $LM:LB=LN:LC$. Now $LM:LK=\sin 2B : \sin \angle KMD$ since $\angle LKE=180^{\circ}-2B$. Hence it suffices to show $\tfrac{\sin KMD}{\sin KND}=\left(\tfrac{\sin B}{\sin C}\right)^2$. Now let $H$ be the midpoint of $A$-symmedian chord in $\triangle ABC$ and $O$ be the circumcenter. Then let $J=\overline{HO} \cap \overline{BC}$; it is well known that $\overline{JA}$ is a tangent to $\odot(ABC)$. Observe that $\angle HBA=\angle DAB=\angle KEA$ hence $\overline{BH} \parallel \overline{KE}$. Likewise $\overline{CH} \parallel \overline{KF}$. Now we see $\tfrac{\sin KMD}{\sin KND}=\tfrac{CH}{HB}=\tfrac{JC}{JB}=\tfrac{AC^2}{AB^2}$ hence we're done.

Let $T$ be on $\overline{BC}$ with $\overline{AT} \perp \overline{BC}$, and let $f(X)$ be a linear function of the plane which gives the difference in power of $X$ with respect to $\odot(DEM)$ and $\odot(DFN)$, and where segments are directed. For $X\equiv A$, we have that $$f(A)=2f(K)-f(D)=AD(KN-KM)=AD(NF-EM).$$Since $\triangle CFN$ is similar to the triangle formed by lines $\overline{AC}, \overline{FK}$, and the $A$-midline, we have that $FN/FK=FC/FM_{\overline{AC}}=S_C/S_A$ in Conway's triangle notation; then $$f(A)=\frac{AD^2(S_C-S_B)}{2S_A}=\frac{AD^2(AC^2-AB^2)}{2S_A}=\frac{BC\cdot AD^2\cdot TD}{S_A}.$$ For $X\equiv T$, we have \begin{align*}f(T)=TD(TM+TN)=TD\cdot MN &= TD(BM+BC+CN)\\ &= TD \cdot BC \left(\frac{S_B+4S_A+S_C}{4S_A} \right)\\ &=\frac{TD\cdot BC \cdot AD^2}{S_A}. \end{align*} Hence, $f(A)=f(T)$ and so $\overline{AT} \perp \overline{O_1O_2}$, which yields the desired since $\overline{AT}$ is perpendicular to $\overline{BC}$.

∵∠MKD=2∠AEK=2∠MEB=2∠MED-180°=180°-∠MO₁D, ∴∠EKO₁=∠MDO₁=∠MED-90°=∠MEB=∠AEK, ∴AB∥KO₁. Let D'∈AD, BD'∥O₁D. ∵∠D'BD=∠MDO₁=∠AEK=∠BAD, ∴ΔDBD'∽ΔDAB, ∴CD²=BD²=DA*DD', ∴ΔDCD'∽ΔDAC, ∴O₂D∥CK. Obviously, ΔABD'∽ΔKO₁D, ΔACD'∽ΔKO₂D. Therefore, ΔABC∽ΔKO₁O₂. ∴∠KO₁O₂=∠ABC. ∵AB∥KO₁, ∴O₁O₂∥BC.

Hand-written solution

Let $\overline{AB}$ intersect $(DEM)$ again at $P$ and $\overline{AC}$ intersect $(DFN)$ again at $Q$. Let $(DEM)$ and $(DFN)$ intersect again at $X$. It suffices to show $\overline{DT} \perp \overline{BC}$. We then have $\angle DTX=\angle DMX=\angle DEX=90^\circ$ and $\angle DTY=\angle DNY=\angle DFY=90^\circ$, hence $X,T,Y$ are collinear. It then suffices to show that $DMXT$ and $DNYT$ are rectangles, which is evidently to $MX=NY$ (since this will imply both are equal to $DT$). Let $H$ be the foot of the altitude from $A$ to $\overline{BC}$. By similar triangles, it suffices to show $\frac{BM}{BH}=\frac{CN}{CH}$. I'm pretty sure this is a straightforward coordinate bash by setting $D=(0,0)$.

We will show that the second intersection of $(DEM)$ and $(DFN)$ is the intersection $L$ of tangents at $E$ and $F$ to $(ADEF)$. It suffices to show that $L$ is on the bisector of $BC$ due to angles. We have $-1=(DE,DF;DL,\perp_{DA})=(AB,AC;\perp_{DL},AD)$ by a $90^{\circ}$ rotation, so $DL\perp BC$.

Let the perpendicular at $D$ to $AD$ intersect $AB,AC,EF$ at $X,Y,Z$ respectively. Let the perpendicular at $E$ to $KE$ and at $F$ to $KF$ intersect at $P$. $EF\cap BC=T, PK\cap EF=S, KZ\cap AB=G,KZ\cap AC=H$. We will prove that $PD\perp BC$. $\rule{24cm}{0.1mm}$ Lemma: In triangle $ABC$ let $AD$ be an altitude. $(AD)$ intersects $AB,AC$ at $X,Y$ respectively. $XY$ intersects $BC$ at $S$. $K$ is the midpoint of $AD$. $SK$ intersects $AB,AC$ at $R,Q$ respectively. Then $RK=KQ$. Proof: Let the other altitudes be $BE,CF$ and $EF\cap BC=T$. Let $D'$ be the reflection of $D$ according to $S$. We have $SD^2=SX.SY=SB.SC\implies (D',D;C,B)=-1=(T,D;C,B)\implies SD=ST$ Hence we have $-1=(AT,AD;AC,AB)=(RQ\infty,K;R,Q)\implies RK=KQ$ $\rule{24cm}{0.1mm}$ Take the inversion centered at $K$ with radius $KD$. We have $DZ^*\perp KZ$ since $\angle KDZ=90$. Also $P,D,Z^*$ are collinear since $K,S,D,Z$ are cyclic. Thus we need to prove that $KZ\parallel BC$. Applying the lemma in triangle $AXY$ with altitude $AD$, we get that $GK=KH$. Let the parallel to $BC$ passing through $K$ intersect $AB,AC$ at $G',H'$ respectively. $\frac{GK}{KH}=1=\frac{G'K}{KH'}\implies G=G',H=H'$ Thus $GH\parallel BC$ as desired.$\blacksquare$

The main idea of this solution is to provide an alternative characterization of $(DEM)$. Let $(DEM)$ meet $AB$ again at $P$. Note that $$\angle BDP=\angle BEM=\angle AEK=\angle BAD.$$Thus, if we define $P$ to be the point on $AB$ such that $\angle BDP=\angle BAD$, then the circle can be described as $(PD)$. Define $Q$ similarly. Now, the goal is to show that the radical axis of $(PD)$ and $(QD)$ is perpendicular to $BC$. Since it goes through $D$, this means we want to show that it's the perpendicular bisector of $BC$, which we call $\ell$. In other words, we wish to show that the foot from $P$ to $\ell$ is the same as the foot from $Q$ to $\ell$, which is just saying $PQ\parallel BC$. Now, reflect $B$ over $P$ to $P'$, and reflect $C$ over $Q$ to $Q'$. Let $K=(ABC)\cap AD$. Since $\angle BAD=\angle BDP=\angle BCP'$ and $\angle CAD=\angle CDQ=\angle CBQ'$, $K$ also lies on $BQ'$ and $CP'$. Finally, since $BQ'$ and $CP'$ intersect on the A-median, by Ceva's, $P'Q'\parallel BC$ and we are done.