Let $\omega \equiv (I)$ touch $BC$ at $D$ and let $D'$ be the antipode of $D$ on $(I).$ It's well-known that $D' \in AA'.$ WLOG assume that $Z$ is between $D$ and $Y$ and let the Z-excircle $(J)$ of $\triangle XYZ$ touch $BC$ at $A''.$ Since $X$ is the insimilicenter of $(I) \sim (J)$ and $ID' \parallel JA'',$ then $X,D',A''$ are collinear $\Longrightarrow$ $A'' \equiv A'.$ Therefore $XY+XZ=DA'=AC-AB=\text{const}.$

Define $\Omega$ the A-excircle of $\triangle ABC$, $\Gamma$ the Y-excircle of $\triangle XYZ$ and let $D\equiv BC\cap \omega$ and $D$, $Y$, $Z$ lie on $BC$ in this order. $\Longrightarrow$ The insimilicenter of $\omega$ and $\Gamma$ is $X$, the insimilicenter of $\Omega$ and $\Gamma$ lie in $BC$, the exsimilicenter of $\Omega$ and $\omega$ is $A$ $\Longrightarrow$ by Monge-D'alemberg theorem we get the insimilicenter of $\Omega$ and $\Gamma$ is $AX\cap BC=A'$ hence the $\Gamma$ is tangent to $BC$ in $A'$ $\Longrightarrow$ $ZA'=YD$ $\Longrightarrow$ since $ZA'=YD$ we get $XZ+XY=A'D$ which it is fixed.

I think the problem is based on the following lemma. Lemma. Triangle $ABC$ with $\omega$ is incircle. The A-excircle is tangent to $BC$ at $A'$. $X$ is in line $AA'$. Tangents from $X$ to $\omega$ intersect $BC$ at $Y,Z$ then $BY=CZ$. I have a proof for this but it requires to consider the position of point $X$ on line $AA'$, which gives a lot of case work. I hope someone can have a better proof for this.

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Not as smart as the above solution, but a very harmonic one. Let $D'$ be the anti-pole of $D$. Then $A,D',A'$ are collinear. Let $G = D'F \cap ED$, $H = DF \cap D'E$, $R = EF \cap DD'$. Then by Brokard's theorem, $HG$ is the polar line of $R$, in particular $HG \parallel BC$ . Since the polar line $EF$ of $X$ passes through R, by La Hire's theorem, $X$ lies on $HG$. Let $S = HG \cap DD'$, $P = HG \cap EF$, $M=XR \cap BC$ then \[ -1 = (E,F; P,R) \stackrel{X}{=} (Y,Z; \infty, M)\]Therefore $M$ is the midpoint of $YZ$. Moreover, \[-1 = (D',D; S,R) \stackrel{X}{=} (D,A'; \infty, M) \]Hence $M$ is the midpoint of $DA'$ as well. As a result, $XY+XZ = YD+ZD = DA'$ which is fixed.

shinichiman wrote: I think the problem is based on the following lemma. Lemma. Triangle $ABC$ with $\omega$ is incircle. The A-excircle is tangent to $BC$ at $A'$. $X$ is in line $AA'$. Tangents from $X$ to $\omega$ intersect $BC$ at $Y,Z$ then $BY=CZ$. I have a proof for this but it requires to consider the position of point $X$ on line $AA'$, which gives a lot of case work. I hope someone can have a better proof for this. I like it. Let $D,E,F$ be points of tangency of the incircle with sides $BC,CA,AB$ respectively. Place the unit circle on Argand plane in such way that it's a unit circle centered at $0$. Take $P,Q$ as two different points of incircle such that lines $PX,QX$ are tangent to $\omega$. Denote wlog $Y,Z$ as intersections of line $BC$ with lines $PX,QX$ respectively. We have$$x=\frac{2pq}{p+q},\ b=\frac{2df}{d+f},\ c=\frac{2de}{d+e},\ a=\frac{2ef}{e+f},\ y=\frac{2dp}{d+p},\ z= \frac{2dq}{d+q}.$$Let $g=-d$. It is a known lemma following by homothethy centered at $A$ that $G\in AA'$. Because $A\neq G, A'$ we have $$X\in AA'\iff X\in AG\iff \frac{x-g}{a-g}=\overline{\left(\frac{x-g}{a-g}\right)}\iff d(2ef+de+df)\cdot\overline{x}=(2d+e+f)x+2(d^2-ef)\iff$$$$d(2ef+de+df)\cdot \frac{2}{p+q}=(2d+e+f)\cdot \frac{2pq}{p+q}+2(d^2-ef)\iff q=\frac{d(2ef+de+df)+(ef-d^2)p}{(2d+e+f)p+d^2-ef}.$$Move on to the length equality. $$y-b=\frac{2d^2(p-f)}{(p+d)(d+f)},\ z-c=\frac{2d^2(q-e)}{(q+d)(d+e)}.$$We're left with proving $$\left|\frac{p-f}{(p+d)(d+f)}\right|=\left|\frac{q-e}{(q+d)(d+e)}\right|$$Since $$q-e=\frac{(d+e)^2(f-p)}{(2d+e+f)p+d^2-ef},\ q+d=\frac{(p+d)(d+e)(d+f)}{(2d+e+f)p+d^2-ef}$$we're done. QED #1679

Denote by $D,E,F,U,V$ touch-points of $\omega$ with $BC,AC,AB,XY,XZ$ respectively. Let $AA'\cap \omega =\left\{S,R\right\}$ (where $S$ lies on segment $AR$), and let $M$ be an intersection of $BC$ with tangent to $\omega$ through $R$. Note that $URVS,RESF$ are harmonic, and as well-known tangent to $\omega$ through $S$ is parallel to $BC$, so by duality wrt $\omega$ we obtain $$(YZM\infty)=(DU,DV,DR,DS)=-1=(DF,DE,DR,DS)=(BCM\infty),$$i.e. $M$ is midpoint of $YZ,BC$ and thus also is midpoint of $DA'$. Hence $$|XY|+|XZ|=|XU|\pm |UY|+|ZV|\mp |XV|=|YD|+|ZD|=|DA'|$$which not depends to $X$.

After so many misreads I finally solved this one... Let $D'$ be the antipode of $D$ in $\omega$ let $XY \cap \omega = E, XZ \cap \omega = F$, let $DD' \cap EF = K$, the line through $X$ parallel to $BC$ meet $DD',EF,KZ$ at $R,P,G$. Let $H = D'E \cap DF$, we first claim $DD',EF,HY,GZ$ concur, this follows from repeated application of brocard's and la hire's, now $(EF;PK) = (YZ,M \infty )$ so $M$ is mid pt. of $YZ$ but it is also $DA'$ so we have $XY + XZ = DA' = AB-AC$ so done. Funny Remark: I was trying to solve the case where the tangents are to excircle for 30+ min before realizing I misread, I somehow got some progress there also maybe some version of this is true.