37th Iranian Mathematical Olympiad

Let us denote by $K$ the intersection point of lines $BT$, $XY$. Note that $△XYB\sim △XZC$. Using the fact that $BT\parallel YZ$, we get $$\frac{XK}{YK}=\frac{XT}{ZT}.$$ Therefore, $K$ and $T$ are corresponding points in triangles $△XYB$ and $△XZC$. Which gives us $$\angle TCX=\angle KBX=\angle TBX,$$ and $TB=TC$. So, $T$ lies on the perpendicular bisector of $BC$ and since $AB=AC$, $T$ also lies on the angle bisector of $\angle A$.