If $O$ is the center of $(ACX)$, then $MAOY$ is a kite so $AY \perp MO$, and $TC \perp MO$ since $TC$ is the radical axis of $(BIC), (ACX)$, so $TC \parallel AY$. $\square$

Note that $\angle BTC=90^{\circ}-\dfrac{1}{2}\angle BAC=180^{\circ}-\angle XAC$ Which means that $T\in \odot(BIC)$ So $MT=MC$ Which means that $YT=AC\Rightarrow AY//CT$

Solved with rama1728. Quite a nice and simple problem. We immediately observe the following key claim. Claim : Point $T$ lies on $(ACX)$ Proof : Simply note that, \[\measuredangle XTC = \measuredangle BTC = \measuredangle CIB = \measuredangle CAX\]since $X$ lies on the external angle bisector of $\angle A$. Now, armed with this claim. We attack the problem. Let $O$ be the center of $(AXCTY)$. Note that this implies that $CT$ is the radical axis of circles $(BIC)$ and $(AXC)$ which implies that $CT \perp OM$. Further, $M$ lies on the perpendicular bisector of $AY$ (since $MY=MA$). So does $O$ (by the definition of the circumcenter). Thus, $MO \perp AY$ as well. But this implies that, $AY \parallel CT$ which was the desired result.

An easy solution First that $\angle BTC=90^{\circ}-\dfrac{1}{2}\angle BAC=180^{\circ}-\angle XAC$ so $XACYT$ is cyclic pentagon. It is a well known that $M$ is the circumcenter of the triangle $BIC$. So $M$ lies on the perpendicular bisector of $TC$. From $MA=MY$, $M$ is also lies on the perpendicular bisector of $AY$. Since $M$ isn’t circumcenter of the triangle (because circumcirles of the triangles $ACT$ and $BTC$ are intersect) then perpendicular bisector of $AY$ and perpendicular bisector of $TC$ are the same line, so $AY||TC$

alinazarboland wrote: In triangle $\triangle ABC$ , $I$ is the incenter and $M$ is the midpoint of arc $(BC)$ in the circumcircle of $(ABC)$not containing $A$. Let $X$ be an arbitrary point on the external angle bisector of $A$. Let $BX \cap (BIC) = T$. $Y$ lies on $(AXC)$ , different from $A$ , st $MA=MY$ . Prove that $TC || AY$ (Assume that $X$ is not on $(ABC)$ or $BC$) $M$ is the center of $(BIC)$ so it is enought to prove that $T,B,X$ are collinear since then $OM$ wiil be perpedicular to both $AY,CT$.(I consider as $T$ the intersection of the two circles)

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Claim: $T$ lies on $(AXC)$ $$\measuredangle(TX,TC)=\measuredangle(TB,TC)=\measuredangle(IB,IC)=\measuredangle(AX,AC)$$ Now to finish let $O$ be the center of $(AXC)$. Then since we have that $MT=MC$ and $MA=MY$ we have that $TC\perp MO$ and $AY\perp MO$, as desired.

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