Question 91
Triangle $\triangle ABC$ is given. Let $A_{1}$ be the image of the intersection point of the bisector of $\measuredangle BAC$ and side $BC$ under the central symmetry whose center is the midpoint of side $BC$. Points $B_{1}$ (on side $CA$) and $C_{1}$ (on side $AB$) are defined analogously. The intersection of the circle circumscribed about $\triangle A_{1}B_{1}C_{1}$ with line $AB$ is the set $\{Z, C_{1}\}$, with line $BC$ the set $\{X, A_{1}\}$, and with line $CA$ the set $\{Y, B_{1}\}$. If the perpendiculars from the points $X, Y$ and $Z$ to $BC, CA$ and $AB$, respectively, meet at one point, prove that $\triangle ABC$ is isosceles.