Inside a regular triangle $ABC$, points $X$ and $Y$ are chosen such that $\angle{AXC} = 120^{\circ}$, $2\angle{XAC} + \angle{YBC} = 90^{\circ}$and $XY = YB = \frac{AC}{\sqrt{3}}$. Prove that point $Y$ lies on the incircle of triangle $ABC$.
Source: 239 2013 S3
Tags: geometry
Inside a regular triangle $ABC$, points $X$ and $Y$ are chosen such that $\angle{AXC} = 120^{\circ}$, $2\angle{XAC} + \angle{YBC} = 90^{\circ}$and $XY = YB = \frac{AC}{\sqrt{3}}$. Prove that point $Y$ lies on the incircle of triangle $ABC$.