seems nice... Does anyone have a solution?

My solution: Let $E,F,D$ feet of $B,C,A$ on $AC,AB,BC$: First assume that $X,Y,Z$ are collinear then let $XC\cap YB=T$ in triangle $CZB$: $CZ^2=CD.CB$(1) and in triangle $CXA$: $CE.CA=CX^2$(2) because $CE.CA=CD.CB$ from (1),(2) we get that $CZ=CX$ similarly we get that $BZ=BY$ note that $AX^2=AE.AC=AF.AB=AY^2\longrightarrow AX=AY$ because $X,Y,Z$ are collinear $\angle YZB+\angle CZX+90=180\longrightarrow \angle YZB+\angle CZX=90$* so because $\angle BZY=\angle BYZ,\angle CZX=\angle CXZ$ from * :$\angle TYX+\angle TXY=90\longrightarrow \angle XTY=90$ so $AXTY$ is square($AX=AY$) and $ZBTC$ is rectangle$\longrightarrow TC=BZ=BY\longrightarrow BY+CX=AX$** note that the lengths of tangents from $A,B,C$ WRT nine point circle are equal $x=\sqrt{AE.\frac{AC}{2}},y=\sqrt{BF.\frac{AB}{2}},z=\sqrt{CE.\frac{AC}{2}}$ but observe that $AE.AC=AX^2,BF.AB=BY^2,CE.AC=CX^2$ so $x=\frac{AX}{\sqrt{2}},y=\frac{BY}{\sqrt{2}},z=\frac{CX}{\sqrt{2}}$ but according to ** $y+z=x$. You can prove the inverse in a similar way. DONE

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If $X,Y,Z$ are collinear is trivial solution, the part hard of the problem is when $YB+XC=AY$ (is equivalent to "the length of the tangent drawn from $A$ to the nine point circle of $\triangle ABC$ is equal with the sum of the lengths of the tangents drawn from $B$ and $C$ to the nine point circle of $\triangle ABC$") $\gamma$ is the semiplane that determine $BC$ and where lies on $A$ If we fixed $\tau_1$ and $\tau_2$ the circles of centers $B,C$ and radius $YB,CX$, then only exist a point $P$ on $\gamma$ and on the radical axis of $\tau_1$ and $\tau_2$ such that the common power of $P$ is $(YB+XC)^2$, and this is $A$. The externally bisector angle of $\angle BZC$ cut again to $\tau_1$ and $\tau_2$ at $Y'$ and $X'$ the tangents to $\tau_1$ and $\tau_2$ from $Y'$ and $X'$ cut at $A'$ and since $\angle Y'A'X' =\angle Y'BZ=\angle X'CZ=90^{\circ} \wedge \angle BYZ=\angle CXZ=45^{\circ}$ is easy that $A'Y'=BZ+ZC \Rightarrow A' \equiv A, Y' \equiv Y, X' \equiv X$ is sufficient.