Let the midpoint of arc $BAC$ of $(ABC)$ be $M$. We claim that this is the required point. This lies on $(ABC)$ and $(ADE)$ due to $MB = MC$ and spiral similarity. Now we need to show that $M$ is on $(AO_1O_2)$. For this, invert at $A$ with power $\sqrt{AB \cdot AC}$ and then reflect about the perpendicular bisector of $\angle BAC$. Then $M$ inverts to the point where $AM$ meets $BC$, say $X$. So we need to show that if an arbitrary line through $X$ meets $AB, AC$ at $Y, Z$, then the reflections of $A$ over $BZ$ and $CY$ are collinear with $X$. Note that $BZ, CY$ meet on the internal bisector of $A$. So rephrasing the problem one final time, we need to prove the following : Modified problem wrote: Let $P$ be a point on the internal bisector of $\angle BAC$ of a triangle $ABC$. Reflections of $A$ over $BP, CP$ are $M, N$ respectively. Prove that $MN$ passes through $X$. Now note that the circumcenter of $AMN$ is $P$, and the line $AX$ is tangent to its circumcircle. Now since the lines $NZ, MY$ determine an isosceles triangle with $MN$, we simply need to prove the following Lemma : Let $ABC$ be a triangle with a point $Q$ on the perpendicular bisector of $BC$. Lines $BQ, CQ$ meet the perpendicular bisectors of $AB, AC$ respectively at $Y, Z$. Then the tangent at $A$ to $(ABC)$, $YZ$ and $BC$ concur. To see this, let the midpoints of $AB, AC$ be $M_c, M_b$. Also let the feet of $Q$ on $AB, AC$ be $Q_c, Q_b$. Then we have $$\frac{QY \cdot CZ}{YB \cdot ZQ} = \frac{Q_cM_c \cdot M_bC}{M_cB \cdot Q_bM_b} = \frac{AC \cdot Q_cM_c}{AB \cdot Q_bM_b}$$But we also have, due to projections of $QO$ on $AB, AC$, that (where $O$ is the circumcenter of $ABC$) $$\frac{Q_cM_c}{Q_bM_b} = \frac{AC}{AB}$$This means that $\frac{QY \cdot CZ}{YB \cdot ZQ} = \frac{AC^2}{AB^2}$, so by converse of Menelaus' theorem, we are done.

Let $M$ be the midpoint of arc $BAC$ in $(ABC)$,$O$ the center of $(ABC)$ and $T$ the second intersection of $(AEB)$ and $(ADC)$.Since $\triangle{MDB}=\triangle{MEC} \implies EMD=A$ so $M \in (ADE)$ also we have(by trivial angle chasing): $\triangle{TDB}=\triangle{TEC} \implies TD=TC \implies$ $T$ lies on angle bisector of $A$,so $$O_1 O_2 \perp AT ,AM \perp AT \implies AM||O_1 O_2$$so all we need to prove is that $OO_1O_2$ is isosceles,which is immediate by angle chasing,so $MAO_1 O_2$ is isosceles trapezoid(because $AM$ and $O_1 O_2$ have the same perpendicular bisector and they are parallel) and cyclic,which completes the proof.

Amir Hossein wrote: Let $ABC$ be an acute scalene triangle. Let $D$ and $E$ be points on the sides $AB$ and $AC$, respectively, such that $BD=CE$. Denote by $O_1$ and $O_2$ the circumcentres of the triangles $ABE$ and $ACD$, respectively. Prove that the circumcircles of the triangles $ABC, ADE$, and $AO_1O_2$ have a common point different from $A$. Proposed by Patrik Bak, Slovakia Let $M$ be the common point of $(ABC)$ and $(ADE)$. Since $M$ is the center of spiral similarity that carries $BD$ to $CE$ and $|BD|=|CE|$, we get that $M$ is midpoint of arc $BAC$. Combining this fact with a little angle chasing gives that $$\angle O_1OA=\angle O_2OM.$$Claim: $OO_1=OO_2$ Proof: We observe that $\triangle BOO_1\sim \triangle BEC$ and $COO_2\sim \triangle BED$ $\implies$ \[\frac{OO_1}{OO_2}=\frac{|DC||BO_1|}{|BE||CO_2|}.\]Which is equals to $1$ because of simple result of Sine's Law in triangles $ABE$ and $ADC$. $\blacksquare$ Finally, considering the similar triangles {$\triangle AOO_1$ and $\triangle MOO_2$} and {$\triangle MOO_1$ and $\triangle AOO_2$} we conclude that $AO_1O_2M$ is isosceles trapezoid which is indeed cyclic. $\blacksquare$ Yaghi wrote: ...so all we need to prove is that $OO_1O_2$ is isosceles,which is immediate by angle chasing. Was that really solvable with angle chasing

$OO_1$ is perpendicular to $AB$,$O_1 O_2$ is perpendicular to $A$ angle bisector,so $\angle OO_1O_2=\frac{A}{2}$(and similar for $OO_2O_1$)

Yaghi wrote: $OO_1$ is perpendicular to $AB$,$O_1 O_2$ is perpendicular to $A$ angle bisector,so $\angle OO_1O_2=\frac{A}{2}$(and similar for $OO_2O_1$) Nice! After concluding $O_1O_2$ is perpendicular to $A$ angle bisector I did’t see how to use it) But anyways please include such details in your solution, it is not that trivial.

You are right.I was in a hurry when writing that solution,that's why some details are missing.Sorry about it.

we have dealt with some similar configuration previously but i ve no link consider the similarity $t_1$ with center $S_1=(ABC)\cap (ADE$ that send $DB\mapsto EC $ then $SDB\sim SEC$ but $BD=CE\implies S $ is on the mid arc of $BC$ , let $K,L$ the midpoints of $DB,CE$ we know that $t_1(K)=L$ so $AKLS$ are cyclic Moreover let $t_2 $ be the similarity with center $S_2=(ABE)\cap (ADC)$ that maps $DB \to CE $ it sends also $K\mapsto L $ hence $AKLS_2,AKLS_1S_2$ are cyclic so the circle $(AKL)$ with center $O'$ is midcircle of $(ABC)$ (with centers $O$) , $(ADE)$ (with centers $O_3$) and also of $(ADC),(ABE)$ thus $O'$ is the common midpoint of $OO_3$ and $O_1O_2$ beside $O'K=O'L,S_1K=S_1L,S_2K=S_2L$ implies $O',S_1,S_2$ are collinear thus $S_1AS_2=90^\circ$ moreover $O_1O_2\perp AS_2,OO'\perp S_1A \implies OO'\perp O_1O_2$ thus $OO'$ is perpendicular bisector of $O_1O_2$ there fore $S_1$ is on $(AO_1O_2)$ RH HAS

Solution,but there are not a lot details. Let $O$ be the circumcenter of $(ABC).$ Let $M=(ADE)\cap (ABC),$ then we can find with Spiral similarity $M$ is midpoint of arc $BAC.$ Then if we show $OO_1=OO_2,$ the rest is easy.(Use $OA=OM$ and a little angle chasing.) We can find with many angles(f.e. in my solution I marked approximately 5 angle ) $\triangle AO_1O\sim \triangle BEC,$ and $\triangle AO_2O\sim \triangle CDB.$ Then we have $\frac{AO}{OO_1}=\frac{BC}{EC}=\frac{BC}{BD}=\frac{AO}{OO_2},$ Then $OO_1=OO_2.$ $ Q.E.D.$

Bravo to everyone! Excellent job.

Amir Hossein wrote: Let $ABC$ be an acute scalene triangle. Let $D$ and $E$ be points on the sides $AB$ and $AC$, respectively, such that $BD=CE$. Denote by $O_1$ and $O_2$ the circumcentres of the triangles $ABE$ and $ACD$, respectively. Prove that the circumcircles of the triangles $ABC, ADE$, and $AO_1O_2$ have a common point different from $A$. Proposed by Patrik Bak, Slovakia Let $O, O’$ be the circumcentres of $\triangle ABC, \triangle ADE$ respectively. Let $P, Q$ and $R,S$ denote the projections of $O, O’$ and $O_1, O_2$ on $AB$ and $AC$ respectively. Note that $$O_1O’ = PQ = \frac{BD}{2} = \frac{CE}{2} = RS = O_2O$$So, $O, O’, O_1, O_2$ form a rhombus. Thus, $OO’$ is the perpendicular bisector of $O_1O_2$. Hence circumcentres of $\triangle AO_1O_2, \triangle ABC, \triangle ADE$ are collinear and hence their circumcircle are concurrent. $\blacksquare$

By spiral similarity we know $(ADE)$ intersects $(ABC)$ at the topmost point which is suppose $M$. We want to show $AMO_2O_1$ is an isosceles trapezoid. Let $P,Q$ be the centers of $(ABC), (ADE)$ respectively. Claim: $O_1PO_2Q$ is a rhombus. Proof. Notice that $O_1P\parallel O_2Q$ because both are perpendicular to $AB$. Similarly $O_1Q\parallel O_2P$. Which means $O_1PO_2Q$ is a parallelogram. Now it is sufficient to prove $PQ\perp O_1O_2$. To prove this we will prove a sub-claim. Sub-claim: $F$ lies on the radical axis of $(ABE), (ACE)$ where $F$ is the foot of the $A-$angle bisector on $BC$. Proof of sub-claim. We will use linearity. It is sufficient to prove that $F$ lies on the radical axis of those two circles $(ABE)=\omega_1$ and $(ACD)=\omega_2$. Let $f:\mathbb{R}^2\to \mathbb{R}$ such that $f(\bullet)=P(\bullet,\omega_1)-P(\bullet,\omega_2)$. It is known that $f$ is linear. Let $BF=m, CF=n$. We want to show $f(F)=0$ which is equivalent to show \[ 0=f(F)=\frac{n}{a}f(B)+\frac{m}{a}f(C)\iff \frac{f(B)}{f(C)}=-\frac{m}{n}=-\frac{c}{b} \]Now \begin{align*} \frac{f(B)}{f(C)}&=\frac{0-BD\cdot BA}{CE\cdot CA-0}=-\frac{c}{b} \end{align*}and we are done for the sub-claim. $\square$ Clearly $PQ\perp AM$. From the sub-claim we know $AM\perp AF$. Which means $PQ\perp O_1O_2$. $\blacksquare$ Since $AM\parallel O_1O_2$ and $\triangle AQ_1\cong \triangle MQO_2$ therefore $AMO_2O_1$ is an isosceles trapezoid therefore $AMO_2O_1$ is cyclic.

Amir Hossein wrote: Let $ABC$ be an acute scalene triangle. Let $D$ and $E$ be points on the sides $AB$ and $AC$, respectively, such that $BD=CE$. Denote by $O_1$ and $O_2$ the circumcentres of the triangles $ABE$ and $ACD$, respectively. Prove that the circumcircles of the triangles $ABC, ADE$, and $AO_1O_2$ have a common point different from $A$. Proposed by Patrik Bak, Slovakia RMO Mock Solution: Let $\odot(ABC)\cap \odot(ADE)=F$, so we just need to show $\odot(AO_1O_2F)$. Claim I: The desired point is arc-midpoint of $\widehat{BAC}$. Proof: Consider the spiral sim. taking $DB \to EC$, we get $F$ is the arc-midpoint of $\widehat{BAC}$ beacuse we also have $BD=CE$. Claim II: $OO_1=OO_2$ Proof: We have $\triangle OO_1A\sim \triangle CEB$ and $\triangle OO_2A\sim \triangle BDC$. Note that $\measuredangle FOO_2=\measuredangle ECB=\measuredangle AOO_1 $, the first angle chasing step comes from the fact that $FO\perp BC$, and we get $\odot(AO_1O_2F)$, so we are done!