Dear Mathlinkers, an outline of my proof... 1. M the midpoint of BC 2. the triangle MXY is M-isoceles; ZM is the Z-inner bissector of the triangle ZXY 3. M is the center of the circumcircle of DXY (Morel's circle) (see http://jl.ayme.pagesperso-orange.fr/ vol. 4 Symetrique de (OI) p. 5) 4. According to the A-Mention circle (Shamrock theorem), D is the incenter of ZXY. Sincerely Jean-Louis

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We have $\angle YAF=A/2$ and $\angle YFA=90^0+(C/2)$.Hence $\angle DYX=\angle AYF=(B/2)$.Also $\angle IBD=(B/2)$ Hence $IBDY$ is cyclic.Now $ID\perp BD \implies IY\perp BY$ since $IBDY$ is cyclic.Also $AZ\perp BZ$,hence $ABZY$ is cyclic. Hence $\angle YZC=\angle BAY=(A/2)$.Similarly we can prove that $\angle XZC=(A/2)$.Hence $DZ$ bisects $\angle XZY$. Now $\angle BDY=90^0+(C/2)$.Hence $\angle ZYD=180^0-(90^0+(C/2))-(A/2)=(B/2)=\angle DYX$.Hence $DY$ bisects $\angle ZYX$.Hence $D$ is the incentre of $\triangle XYZ$.

Let $P\in AC$ such that $AP=AB$. Let $BP\cap AI=X_0$. Let $EX_0$ meet $\odot DEF$ at $D_0\neq E$. Clearly $D_0$ is on the other side of $BP$ than $F$. Now, $AI$ is the perpendicular bisector of $EF$ and $BP$. Again, $\angle FD_0X_0=\angle ED_0F=\angle AFE=\angle ABP=\angle FBX_0$ i.e. $D_0BFX_0$ is cyclic. Thus, $\angle BDF=\angle BX_0F=90^{\circ}-\angle FX_0A=\angle X_0FE=\angle X_0EF=\angle DEF$. So, $BD_0$ is tangent to $\odot DEF$. This means that $D\equiv D_0$. Thus $X\equiv X_0$. So we proved that $BX\perp AI$. Now, $\angle DYE=\angle FYE=180^{\circ}-\angle YFE-\angle YEF$. But $Y$ is on the perpendicular bisector of $EF$. So, $\angle DYE=180^{\circ}-2\angle YFE=180^{\circ}-2\angle DFE=180^{\circ}-2\angle DEC=\angle DCE$. Thus $DECY$ is cyclic, and so $\angle IYE=\dfrac{1}{2}\angle DYE=\dfrac{1}{2}\angle DCE=\angle ICE$ which means $IYCE$ is cyclic too. Thus $\angle IYC=\angle IEC=90^{\circ}$. So $CY\perp AI$ too. Now, $AXZB$ and $AZYC$ both are cyclic. So, $\angle ZXY=90^{\circ}-\angle BXZ=90^{\circ}-\angle BAZ=\angle ABC$ and $\angle DXY=\angle AXE=90^{\circ}-\angle DEF=90^{\circ}-\angle BDF=\angle IBD=\dfrac{1}{2}\angle ABC$. This means $DX$ bisects $\angle ZXY$. Also, $\angle XZD=180^{\circ}-\angle XZB=\angle XAB=\angle IAB=\angle IAC=\angle YAC=\angle YZC$$=\angle YZD$. Thus $DZ$ bisects $\angle XZY$. All these results imply that $D$ is the incenter of $\triangle XYZ$.

Hi Here's my solution. 1)By angle chasing we understand that DXY=B/2 and DYX=C/2 and ZDY=90+B/2 2)If we prove that ZXD=B/2 the problem will be solved. 3)So we need to say that ZXY=B. 4)Because IBD=B/2 so IBDX is cyclic. So XBD=XID but because DI and AZ are parallel we have XID=XAZ so ABZX is cyclic. 5)So ZXY=B so we are done.

EDIIT...

$\angle IXD=180-\tfrac {\angle A}{2} -\angle AED=90-\tfrac {\angle A}{2}-\tfrac {\angle B}{2}=\tfrac {\angle C}{2}$ So $IDXE$ is cyclic. $\angle IXC=180-\angle IFC=90=\angle AZC$ So $AZXC$ is cyclic as well. And now, $\angle DXZ=\angle EDB-\angle CZX=180-\angle AEX-\tfrac{\angle A}{2}=\angle DXY$ Similarly we show $YD$ bisects $\angle XYZ$ and so $I$ is the incenter $XYZ$