I have an ugly computational proof.I hope someone has a nicer solution.Let $K=XY \cap BT, L=AC \cap BT$.$BX$ is the external angle bisector of $\triangle XTK$ so by angle bisector theorem $\frac{XT}{XK}=\frac{BT}{BK}$ combining this with sin law in triangles $\triangle ZTL,\triangle YKB,\triangle XTK $ we obtain $\frac{BT}{TL}=\frac{BY}{ZL}$ which by Thales implies $\frac{BT}{TL}=\frac{AB}{AL}$.Hence by angle bisector Theorem we are done.

My solution: We need to prove that $\angle TBC = \angle TCB$ Let $P$ be intersection of $BC$,$YZ$. $YZ||BT \longrightarrow \angle CBT = \angle ZPC$ As $\angle YXP = TXC$ therefor it's enough to prove that $\triangle YXP$ ~ $\triangle TXC$ or $\frac {YX}{XP} = \frac {XT}{XC}$. $\triangle BXY$ ~ $\triangle XCZ \longrightarrow BX.XZ=YX.XC$ (*) $\triangle BXT$ ~ $\triangle XPZ \longrightarrow BX.XZ=XP.XT$ (**) (*),(**) $\longrightarrow \frac {YX}{XP} = \frac {XT}{XC} \blacksquare$

Let $XY \cap BT = D$. $DT \parallel YZ \implies XY/XZ = XD/XT$. $\triangle BXY \sim \triangle CXZ \implies XY/XZ = XB/XC$. It means $\triangle BXD \sim \triangle CXT$ So $\angle CBT = \angle BCT$ and result follows.

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Let the angle bisector of $\angle{A}$ intersects $XZ$ at $T$ We wanna show that, $YZ||BT$ Let $AT\cap YX=N$ And $AT\cap BC=M$ Let, $\angle{ABC}=\angle{ACB}=x$ $\angle{ZXC}=\angle{YXB}=\angle{BXT}=y$ So, $M$ is the midpoint of $BC$ As well as $NT$ So, $\angle{TBM}=\angle{MBN}=z$ Define, $BN \cap YZ=K$ And $J=BN\cap AC$ $\angle{AYK}=\angle{ABT}=x+z$ $\angle{AJK}=\angle{JCK}+\angle{JBC}=x+z$ So, $\angle{YKJ}=\angle{BKP}=180^o-2z$, where $P=YZ\cap BC$ So, $\angle{KPB}=z=\angle{PBT}$ We are done.

Iran Round 2 2016 P2 wrote: $ABC$ is an isosceles triangle ($AB=AC$). Point $X$ is an arbitary point on $BC$. $Z \in AC$ and $Y \in AB$ such that $\angle BXY = \angle ZXC$. A line parallel to $YZ$ passes through $B$ and cuts $XZ$ at $T$. Prove that $AT$ bisects $\angle A$. Solution: WLOG, assume, $XB>XC$ and $T$ outside $\Delta ABC$ to avoid config. issue. Let $Y',T'$ be reflections of $Y,T$ over $BC$. Hence, $T'$ $\in$ $YX$ & $Y'$ $\in$ $TX$. Let $YZ$ $\cap$ $BC$ $=$ $W$ & WLOG, $WB > WC$. So, $\Delta BXT$ $\sim$ $\Delta WXZ$ & $\Delta BXY'$ $\sim$ $\Delta CXZ$ gives, $CT||WY'$. Then, $\angle TBC$ $=$ $\angle YWB$ $=$ $\angle TCB$ $\implies$ $TB=TC$. So, $AT$ bisects $\angle A$ $\qquad \blacksquare$

$Y\mapsto Z$ is projective and since $f(A)\neq A$, it follows that $YZ$ is tangent to conic. When $Y={\infty}_{AB}$ we see that $YZ$ is line at infinity so this conic i parabola and $$YZ\mapsto {\infty}_{YZ} \mapsto B{\infty}_{YZ} \mapsto T$$is projective and we need to check $3$ cases. But this is trivial when: $Y={\infty}_{AB}$; when $Y$ is intersection of $AB$ and line ortogonal to $BC$ at $X$ and when $Y=B$ so we are done .

It's same as the above solutions, but with an easier end. Let $YZ \cap BC :{P}$. Also, let $\measuredangle TBP = \measuredangle BPY = \alpha$ and $\measuredangle XYP= \beta$. So we get $$\measuredangle BXY = \measuredangle CXZ = \alpha + \beta$$$\Delta BXT \sim \Delta PXZ$. $\Rightarrow \frac{BX}{XP} = \frac{TX}{XZ} \Rightarrow BX\cdot XZ = PT\cdot XY$ Also from $\Delta BXY \sim \Delta CXZ$, we get $BX\cdot XZ = CX\cdot XY$ With these two equalities we get $$PX\cdot XT = CX\cdot XY$$So, $$\frac{sin \hat{(CTX)}}{sin \hat{(XCT)}} = \frac{CX}{XT} = \frac{PX}{XY} = \frac{sin \beta }{sin \alpha}$$We see that $\measuredangle CTX + \measuredangle XCT = \alpha + \beta$. So by Masmis Theorem, we get $\measuredangle \hat{CTX} = \beta$ and $\measuredangle \hat{XCT} = \alpha$. This means that $BT=TC$ so $AT \perp BC$ QED

we have triangles BXY and CXZ are similar. Let YX meet BT at S. By Thales theorem XY/XZ = XS/XT and from similarity of BXY and CXZ we have XY/XZ = XB/XC. now we can easily prove XBS and XCT are similar so ∠XBT = ∠XCT so T lies on perpendicular bisector of BC which is angle bisector of A. we're Done.

Another solution Let $YZ$ meet $BC$ at $N$. we have $\frac{XT}{TZ} = \frac{\sin{TCX}}{\sin{TCZ}} . \frac{CX}{CZ} \implies \frac{XB}{BN} = \frac{\sin{TCX}}{\sin{TCZ}} . \frac{BX}{BY} \implies \frac{\sin{TCX}}{\sin{TCZ}} = \frac{\sin{BNY}}{\sin{BYN}}$ we also have $\angle BNY + \angle BYN = \angle TCX + \angle TCZ$ and $\angle B , \angle C < 90$ so $\angle TCX = \angle BNY = \angle CBT \implies \angle TBC = \angle TCB \implies T$ is on angle bisector of $A$.