Restate the Problem as follow: Problem. Let $I$ be the incenter of the triangle $\triangle{ABC}$, lines $AI$ meet $BC$ at $A_1$. the perpendicular bisector of $AA_1$ meet $BI$ and $CI$ at $Q$ and $R$ respectively. Prove that $A$ , $Q$ , $I$ and $R$ lie on a circle. Solution. Call $(M_a)$ the perpendicular bisector of $AA_1$. Let the perpendicular to $IB$ and $IC$ trough $D$ meet $AB$ and $AC$ at $B_1$ and $C_1$ respectively. Denote $B_2\equiv (IB)\cap(A_1B_1)$ and $C_2\equiv(IC)\cap(A_1C_1)$. It's obvious that $BB_1=BA_1$ and $CC_1=CA_1$, therefore: $\frac{BB_1}{BA}=\frac{BA_1}{BA}=\frac{CA_1}{CA}=\frac{CC_1}{CA}$ implying that $(B_1C_1)\parallel(BC)$. Denote $M\equiv(M_a)\cap(AA_1)$, then: $(MB_2)\parallel(AB)$ and $(MC_2)\parallel(AC)$ and since triangles $\triangle{AC_1B_1}$ and $\triangle{MC_2B_2}$ are perspective, we conclude that $(B_2C_2)\parallel(B_1C_1)$; Consequently, $(B_2C_2)\parallel(BC)$. So $\triangle{MB_2C_2}$ is directly homothetic to $\triangle{ABC}$. Let $M_c=(M_a)\cap(AC)$ It's obvious that $M_cA=M_cA_1$ and then $\angle{CM_cA_1}=\angle{CAB}$ thus $(A_1M_c)\parallel(AB)$ or again $(A_1M_c)\parallel(AB)$ it follows from Desargue's theorem that $M_cM$ , $A_1B_2$ and $CC_2$ are concurrent which implies that $R$ lies on $A_1B_1$ similary $Q$ lies on $A_1C_1$. From here it's not hard to see that $BCQR$ is a cyclic quadrilateral, and using the fact that $A_1$ is the symmetric of $A$ with respect to $QR$ we can easly get that $\angle{QAR}+\angle{QIR}=180^{\circ}$ implying that points $A$ , $Q$ , $I$ and $R$ are concyclics. This problem was discovred by Jean-louis Ayme. He found out that on the same configuration, if we let $A_2\equiv(B_1C_1)\cap(QR)$. And we denote $O$, $O_1$ and $O_2$ circumcenters of $\triangle{AB_1C_1}$ , $\triangle{A_2C_1Q}$ and $\triangle{A_2B_1R}$, then points $I$ , $O$ , $O_1$ and $O_2$ are concyclics, this problem is much more interesting. If anyone feels interested, I can add here my solution.

How about this solution? Let $M^{\prime}$ be the intersection of circumstance of $\triangle ABD$　and $BI$. Similarly $N^{\prime}$ be the intersection of circumstance of $\triangle ACD$　and $CI$. It is obvious that $AM^{\prime}=M^{\prime}D,AN^{\prime}=N^{\prime}D$,so $M^{\prime}N^{\prime} $is perpendicular bisector of$AD$． Therefore, we can conclude that$ M^{\prime}=M$ and $N=N^{\prime}$. Thus,$\angle MAD=\angle MBD,\angle NAD=\angle NCD$,so $\angle MAN=\angle MAD+\angle NAD=\angle MBD+\angle NCD=180-\angle MIN$ $\Rightarrow A,I,M,N$　lie on a circle. (I'm sorry for my poor English.) I feel interested in Jean-louis Ayme's problem.

It may appear that my solution for the previous problem was complicated. In fact, I have just copied from my notebook the appropriate part of my solution for jayme's problem. Problem. Let $I$ be the incenter of the triangle $\triangle{ABC}$, lines $AI$ meet $BC$ at $A_1$. the perpendicular bisector of $AA_1$ meet $BI$ and $CI$ at $Q$ and $R$ respectively. Let $A_2\equiv(B_1C_1)\cap(QR)$. Denote $O$, $O_1$ and $O_2$ circumcenters of $\triangle{AB_1C_1}$ , $\triangle{A_2C_1Q}$ and $\triangle{A_2B_1R}$, Prove that points $I$ , $O$ , $O_1$ and $O_2$ are concyclics. Solution. Call $(M_a)$ the perpendicular bisector of $AA_1$. Let the perpendicular to $IB$ and $IC$ trough $D$ meet $AB$ and $AC$ at $B_1$ and $C_1$ respectively. Denote $B_2\equiv (IB)\cap(A_1B_1)$ and $C_2\equiv(IC)\cap(A_1C_1)$. It's obvious that $BB_1=BA_1$ and $CC_1=CA_1$, therefore: $\frac{BB_1}{BA}=\frac{BA_1}{BA}=\frac{CA_1}{CA}=\frac{CC_1}{CA}$ implying that $(B_1C_1)\parallel(BC)$. Denote $M\equiv(M_a)\cap(AA_1)$, then: $(MB_2)\parallel(AB)$ and $(MC_2)\parallel(AC)$ and since triangles $\triangle{AC_1B_1}$ and $\triangle{MC_2B_2}$ are perspective, we conclude that $(B_2C_2)\parallel(B_1C_1)$; Consequently, $(B_2C_2)\parallel(BC)$. So $\triangle{MB_2C_2}$ is directly homothetic to $\triangle{ABC}$. Let $M_c=(M_a)\cap(AC)$ It's obvious that $M_cA=M_cA_1$ and then $\angle{CM_cA_1}=\angle{CAB}$ thus $(A_1M_c)\parallel(AB)$ or again $(A_1M_c)\parallel(AB)$ it follows from Desargue's theorem that $M_cM$ , $A_1B_2$ and $CC_2$ are concurrent which implies that $R$ lies on $A_1B_1$ similary $Q$ lies on $A_1C_1$. From here it's not hard to see that $BCQR$ is a cyclic quadrilateral, and using the fact that $A_1$ is the symmetric of $A$ with respect to $QR$ we can easly get that $\angle{QAR}+\angle{QIR}=180$ implying that points $A$ , $Q$ , $I$ and $R$ are concyclics. From another hand, we have: $\angle{AQR}=90-\angle{QAI}=90-\angle{MB_2I}$ and since $(MB_2)\parallel (AB)$ we get that: $\angle{AQR}=\angle{RB_1B}$ thus, $A$ , $Q$ , $R$ and $B_1$ are concyclics. Similarly we can prove that the hexagone $AC_1QIRB_1$ is cyclic. Now let $U_1$ and $V_1$ be the intersections of $\odot(A_2C_1Q)$ with $BI$ and $AC$ respectively, let $Z_1$ be the second intersection of $ IC_1$ with $\odot(A_2C_1Q)$ and $W\equiv(U_1V_1)\cap(IC_1)$. Remark first that: $\angle{V_1U_1I}=\angle{V_1BQ}=\angle{AIU_1}$, then $(U_1V_1)\parallel(AI)$. Furthermore, if $X\equiv(AI)\cap(B_1C_1)$ we get that: $\angle{WU_1Z_1}=\angle{AXB}=\angle{C_1XI}$ and thus $\angle{U_1WZ_1}=\angle{C_1IX}$ hence $\triangle{U_1WZ}\sim \triangle{XIC_1}$; Consequently $(U_1V_1)\parallel(B_1C_1)$. Let $Y\equiv(CI)\cap(B_1C_1)$. It's not hard to see that $AXYR$ is cyclic, it follow that $\angle{A_2ZI}=\angle{IAR}=\angle{A_2YI}$; So $(A_2Z_1)\parallel (IR)$. $\left(\Psi\right)$ Hence $\angle{O_1OI}=\angle{QAB_1}$ thus $\angle{O_1OI}=\angle{QIB_2}=\angle{IZ_1O_1}$ and so, points $I$ , $O$ , $O_1$ and $Z_1$ are concyclic. Lemma. On the cyclic quadrilateral $ABCD$, denote $P\equiv(AB)\cap(CD)$, and $(O)$ , $(O_1)$ , $(O_2)$ circumcircles of triangles $\triangle{ABC}$ , $\triangle{PBC}$ and $\triangle{PAD}$ respectively. Assume that $P$ is nearer to $BC$ than $AD$, If $X$ and $Y$ be the intersections of $\odot(OO_1O_2)$ with $\odot(PBC)$. Then $X$, $Y$ lie both on $A_2O_2$ Proof Left to the reader since it's very easy (only angle-chasing) Return to our problem, let $Z\equiv\odot(OO_1O_2)\cap(A_2O_2)$. From the lemma above, $Z \in (O_1)$. Moreover, $\angle{ZA_2B_1}=\angle{RQB}=\angle{A_2YI}$ hence $(ZA_2)\parallel(YA)$. It follows, using $\left(\Psi\right)$ that $Z\equiv Z_1$ and consequently $O$ , $O_1$ , $O_2$ and $I$ are concyclic.

Because of $CM$ is the angle bisector of $\angle{ACD}$ and $MA = MD$ so we have $A, C, D, M$ are concylic. Similarly, we have $A, B, D, N$ are concylic, too. From that we obtain: $(MA,MI) \equiv (MA,MC) \equiv (DA, DC) \equiv (DA, DB) \equiv (NA, NB) \equiv (NA,NI) (mod \pi)$ That means $A, M, I, N$ are concylic.

$CM$ bisects $\angle ACD$ and also $AM=DM$. So $M$ is on the circumcircle of $ACD$. Similarly $N$ is on the circumcircle of $ABD$. So $\angle ANI = \angle ANB = \angle ADB = 180^\circ- \angle ADC =180^\circ - \angle AMC = 180^\circ - \angle AMI$. Or $ANIM$ cyclic.

Hi ; It's Very easy when we know that $ AMBD $ is cyclic . Best Regard